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ProblemSolvinginDataStructures&AlgorithmsUsingCFirstEdition
ByHemantJain
BookTitle:ProblemsSolvinginDataStructures&AlgorithmsUsingCBookAuthor:HemantJainPublishedbyHemantJainHIG 4, Samarth Tower, Sarvadharm,D Sector, Bhopal, India, Pin Code:462042PublishedinBhopal,IndiaThiseditionpublishedMarch2017
Copyright©HemantJain2017.AllRightReserved.Hemant Jain asserts themoral right tobe identified as the author of thiswork.Allrightsreserved.Nopartofthispublicationmaybereproduced,storedinorintroducedintoaretrievalsystem,ortransmitted,inanyform,orbyanymeans (electrical,mechanical, photocopying, recording or otherwise)without the priorwritten permission of the author, except in the case ofvery brief quotations embodied in critical reviews and certain other non-commercial uses permitted by copyright law. Any person who does anyunauthorized act in relation to this publicationmaybe liable to criminalprosecutionandcivilclaimsfordamages.
ACKNOWLEDGEMENTTheauthorisverygratefultoGODALMIGHTYforhisgraceandblessing.DeepestgratitudeforthehelpandsupportofmybrotherDr.SumantJain.This book would not have been possible without the support andencouragementheprovided.Iwouldliketoexpressprofoundgratitudetomyguide/myfriendNaveenKaushik for his invaluable encouragement, supervision and usefulsuggestionthroughoutthisbookwritingwork.Hissupportandcontinuousguidanceenablemetocompletemyworksuccessfully.Finally yet importantly, I am thankful to Love Singhal, Anil Berry andOtherswhohelpedmedirectlyorindirectlyincompletingthisbook.HemantJain
TABLEOFCONTENTSTABLEOFCONTENTSCHAPTER0:HOWTOUSETHISBOOKWHATTHISBOOKISABOUTPREPARATIONPLANSSUMMARY
CHAPTER1:INTRODUCTION-PROGRAMMINGOVERVIEWINTRODUCTIONVARIABLEPOINTERSARRAYTWODIMENSIONALARRAYARRAYINTERVIEWQUESTIONSSTRUCTUREPOINTERTOSTRUCTUREDYNAMICMEMORYALLOCATIONFUNCTIONCONCEPTOFSTACKSYSTEMSTACKANDFUNCTIONCALLSPARAMETERPASSING,CALLBYVALUEPARAMETERPASSING,CALLBYREFERENCERECURSIVEFUNCTIONEXERCISES
CHAPTER2:ALGORITHMSANALYSISINTRODUCTIONASYMPTOTICANALYSISBIG-ONOTATIONOMEGA-ΩNOTATIONTHETA-ΘNOTATIONCOMPLEXITYANALYSISOFALGORITHMSTIMECOMPLEXITYORDERDERIVINGTHERUNTIMEFUNCTIONOFANALGORITHMTIMECOMPLEXITYEXAMPLESMASTERTHEOREMMODIFIEDMASTERTHEOREMEXERCISE
CHAPTER3:APPROACHTOSOLVEALGORITHMDESIGNPROBLEMSINTRODUCTIONCONSTRAINTSIDEAGENERATION
COMPLEXITIESCODINGTESTINGEXAMPLESUMMARY
CHAPTER4:ABSTRACTDATATYPEABSTRACTDATATYPE(ADT)DATA-STRUCTUREARRAYLINKEDLISTSTACKQUEUETREESBINARYTREEBINARYSEARCHTREES(BST)PRIORITYQUEUE(HEAP)HASH-TABLEDICTIONARY/SYMBOLTABLEGRAPHSGRAPHALGORITHMSSORTINGALGORITHMSCOUNTINGSORTENDNOTE
CHAPTER5:SEARCHINGINTRODUCTIONWHYSEARCHING?DIFFERENTSEARCHINGALGORITHMSLINEARSEARCH–UNSORTEDINPUTLINEARSEARCH–SORTEDBINARYSEARCHSTRINGSEARCHINGALGORITHMSHASHINGANDSYMBOLTABLESHOWSORTINGISUSEFULINSELECTIONALGORITHM?PROBLEMSINSEARCHINGEXERCISE
CHAPTER6:SORTINGINTRODUCTIONTYPEOFSORTINGBUBBLE-SORTMODIFIED(IMPROVED)BUBBLE-SORTINSERTION-SORTSELECTION-SORTMERGE-SORT
QUICK-SORTQUICKSELECTBUCKETSORTGENERALIZEDBUCKETSORTHEAP-SORTTREESORTINGEXTERNALSORT(EXTERNALMERGE-SORT)COMPARISONSOFTHEVARIOUSSORTINGALGORITHMS.SELECTIONOFBESTSORTINGALGORITHMEXERCISE
CHAPTER7:LINKEDLISTINTRODUCTIONLINKEDLISTTYPESOFLINKEDLISTSINGLYLINKEDLISTDOUBLYLINKEDLISTCIRCULARLINKEDLISTDOUBLYCIRCULARLISTEXERCISE
CHAPTER8:STACKINTRODUCTIONTHESTACKABSTRACTDATATYPESTACKUSINGARRAY(MACRO)STACKUSINGARRAY(DYNAMICMEMORY)STACKUSINGARRAY(GROWINGCAPACITYIMPLEMENTATION)STACKUSINGARRAY(GROWING-REDUCINGCAPACITYIMPLEMENTATION)STACKUSINGLINKEDLISTPROBLEMSINSTACKPROSANDCONSOFARRAYANDLINKEDLISTIMPLEMENTATIONOFSTACK.USESOFSTACKEXERCISE
CHAPTER9:QUEUEINTRODUCTIONTHEQUEUEABSTRACTDATATYPEQUEUEUSINGARRAYQUEUEUSINGLINKEDLISTPROBLEMSINQUEUEEXERCISE
CHAPTER10:TREEINTRODUCTIONTERMINOLOGYINTREEBINARYTREE
TYPESOFBINARYTREESPROBLEMSINBINARYTREEBINARYSEARCHTREE(BST)PROBLEMSINBINARYSEARCHTREE(BST)EXERCISE
CHAPTER11:PRIORITYQUEUEINTRODUCTIONTYPESOFHEAPHEAPADTOPERATIONSOPERATIONONHEAPHEAP-SORTUSESOFHEAPPROBLEMSINHEAPEXERCISE
CHAPTER12:HASH-TABLEINTRODUCTIONHASH-TABLEHASHINGWITHOPENADDRESSINGHASHINGWITHSEPARATE-CHAININGPROBLEMSINHASHINGEXERCISE
CHAPTER13:GRAPHSINTRODUCTIONGRAPHREPRESENTATIONADJACENCYMATRIXADJACENCYLISTGRAPHTRAVERSALSDEPTHFIRSTTRAVERSALBREADTHFIRSTTRAVERSALPROBLEMSINGRAPHDIRECTEDACYCLICGRAPHTOPOLOGICALSORTMINIMUMSPANNINGTREES(MST)SHORTESTPATHALGORITHMSINGRAPHEXERCISE
CHAPTER14:STRINGALGORITHMSINTRODUCTIONSTRINGMATCHINGDICTIONARY/SYMBOLTABLEPROBLEMSINSTRINGMEMMOVEFUNCTIONEXERCISE
CHAPTER15:ALGORITHMDESIGNTECHNIQUESINTRODUCTIONBRUTEFORCEALGORITHMGREEDYALGORITHMDIVIDE-AND-CONQUER,DECREASE-AND-CONQUERDYNAMICPROGRAMMINGREDUCTION/TRANSFORM-AND-CONQUERBACKTRACKINGBRANCH-AND-BOUNDA*ALGORITHMCONCLUSION
CHAPTER16:BRUTEFORCEALGORITHMINTRODUCTIONPROBLEMSINBRUTEFORCEALGORITHMCONCLUSION
CHAPTER17:GREEDYALGORITHMINTRODUCTIONPROBLEMSONGREEDYALGORITHM
CHAPTER18:DIVIDE-AND-CONQUER,DECREASE-AND-CONQUERINTRODUCTIONGENERALDIVIDE-AND-CONQUERRECURRENCEMASTERTHEOREMPROBLEMSONDIVIDE-AND-CONQUERALGORITHM
CHAPTER19:DYNAMICPROGRAMMINGINTRODUCTIONPROBLEMSONDYNAMICPROGRAMMINGALGORITHM
CHAPTER20:BACKTRACKINGANDBRANCH-AND-BOUNDINTRODUCTIONPROBLEMSONBACKTRACKINGALGORITHM
CHAPTER21:COMPLEXITYTHEORYANDNPCOMPLETENESSINTRODUCTIONDECISIONPROBLEMCOMPLEXITYCLASSESCLASSPPROBLEMSCLASSNPPROBLEMSCLASSCO-NPNP–HARD:NP–COMPLETEPROBLEMSREDUCTIONENDNOTE
CHAPTER22:INTERVIEWSTRATEGYINTRODUCTIONRESUMENONTECHNICALQUESTIONSTECHNICALQUESTIONS
CHAPTER23:SYSTEMDESIGNSYSTEMDESIGNSYSTEMDESIGNPROCESSSCALABILITYTHEORYDESIGNSIMPLIFIEDFACEBOOKDESIGNFACEBOOKFRIENDSSUGGESTIONFUNCTIONDESIGNASHORTENINGSERVICELIKEBITLYSTOCKQUERYSERVERDESIGNABASICSEARCHENGINEDATABASEDESIGNABASICSEARCHENGINECACHINGDUPLICATEINTEGERINMILLIONSOFDOCUMENTSZOMATOYOUTUBEDESIGNIRCTCALARMCLOCKDESIGNFORELEVATOROFABUILDINGVALETPARKINGSYSTEMOODESIGNFORAMCDONALDSSHOPOBJECTORIENTEDDESIGNFORARESTAURANTOBJECTORIENTEDDESIGNFORALIBRARYSYSTEMSUGGESTASHORTESTPATHEXERCISE
APPENDIXAPPENDIXA
CHAPTER0:HOWTOUSETHISBOOK
WhatthisbookisaboutThisbookisaboutusageofdatastructuresandalgorithmsincomputerprogramming.Datastructuresarethewaysinwhichdataisarrangedincomputersmemory.Algorithmsaresetofinstructionstosolvesomeproblembymanipulatingthesedatastructures.DesigninganefficientalgorithmtosolveacomputerscienceproblemisaskillofComputerprogrammer.TheskillwhichtechcompanieslikeGoogle,Amazon,Microsoft,Facebook,Adobeandmanyothersarelookingfor inan interview.Oncewearecomfortablewithaprogramminglanguage, thenextstep is tolearnhowtowriteefficientalgorithms.ThisbookassumesthatyouareaClanguagedeveloper.YouarenotanexpertinClanguage,butyouarewellfamiliarwithconceptsofpointers,functions,arraysandrecursion.Atthestartofthisbook,wewillberevisingtheClanguagefundamentalsthatwillbeusedthroughoutthisbook.Wewillbelookingintosomeoftheproblemsinarraysandrecursiontoo.TheninthecomingchapterwewillbelookingintoComplexityAnalysis.Followedbythevariousdatastructures and their algorithms.Will look into a linked list, stack, queue, trees, heap,Hash-Table andgraphs.Wewillalsobelookingintosorting,searchingtechniques.Moreover, we will be looking into analysis of various algorithm techniques, such as brute forcealgorithms, greedy algorithms, divide & conquer algorithms, dynamic programming, reduction &backtracking.Intheend,wewillbelookingintosystemdesignthatwillgiveasystematicapproachtosolvethedesignproblemsinanInterview.
PreparationPlansGiven the limited timeyouhavebeforeyournext interview, it is important tohaveasolidpreparationplan.Thepreparationplandependsuponthetimeandwhichcompaniesyouareplanningtotarget.Belowarethethree-preparationplanfor1Month,3Monthand5Monthdurations.
1MonthPreparationPlansBelowisalistoftopicsandapproximatetimeuserneedtotaketofinishthesetopics.Thesearethemostimportantchaptersthatmusttobepreparedbeforeappearingforaninterview.Thisplanshouldbeusedwhenyouhavealimitedtimebeforeaninterview.Thesechapterscover90%ofdata structures and algorithm interview questions. In this plan sincewe are reading about the variousADTinchapter4sowecanusethesedatatypeeasilywithoutknowingtheinternaldetailshowtheyareimplemented.Chapter24isforsystemdesign,youmustreadthischapterifyouarethreeormoreyearsofexperienced.Anyway,readingthischapterwillgivethereaderabroaderperspectiveofvariousdesigns.Time Chapters Explanation
Week1
Chapter 1: Introduction - ProgrammingOverviewChapter2:AlgorithmsAnalysisChapter 3: Approach To Solve AlgorithmDesignProblemsChapter4:AbstractDataType
You will get a basic understanding of how to findcomplexity of a solution. You will know how tohandlenewproblems.Youwill read about avarietyofdatatypesandtheiruses.
Week2Chapter5:SearchingChapter6:SortingChapter14:StringAlgorithms
Searching,SortingandStringalgorithmconsistsofamajorportionoftheinterviews.
Week3Chapter7:LinkedListChapter8:StackChapter9:Queue
Linkedlistsareoneofthefavouritesinaninterview.
Week4Chapter10:TreeChapter23:InterviewStrategyChapter24:SystemDesign
This portion youwill read about Trees and SystemDesign.You are good to go for interviews. Best ofluck.
3MonthPreparationPlanThisplanshouldbeusedwhenyouhavesome time toprepare foran interview.Thispreparationplanincludesnearlyeverythinginthisbookexceptvariousalgorithmtechniques.Algorithmproblemsthatarebasedondynamicprogramming,divide&conqueretc.WhichareaskedinvaryspecificcompanieslikeGoogle,Facebook,etc.Therefore,untilyouareplanningtofaceinterviewwiththemyoucanparkthesetopicsforsometimeandfocusontherestofthetopics.Again,samethingherewithsystemdesignproblems,themoreexperienceyouare,themoreimportantthischapterbecomes.However,ifyouareafresherfromcollege,thenalsoyoushouldreadthischapter.Time Chapters Explanation
Week1
Chapter 1: Introduction - ProgrammingOverviewChapter2:AlgorithmsAnalysisChapter 3: Approach To Solve AlgorithmDesignProblemsChapter4:AbstractDataType
Youwill get abasicunderstandingofhow to findcomplexity of a solution. You will know how tohandle new problems. You will read about avarietyofdatatypesandtheiruses.
Week 2 &Week3
Chapter5:SearchingChapter6:SortingChapter14:StringAlgorithms
Searching, sorting and string algorithmconsists ofamajorportionoftheinterviews.
Week 4 &Week5
Chapter7:LinkedListChapter8:StackChapter9:Queue
Linked lists are one of the favourites in aninterview.
Week 6 &Week7
Chapter10:TreeChapter11:Heap
This portion you will read about trees and heapdatastructures.
Week 8 &Week9
Chapter12:Hash-TableChapter13:Graphs
Hash-Table are used throughout this book invarious places but now it is time to understandhowHash-Tableareactuallyimplemented.Graphsare used to propose a solution many real lifeproblems.
Week10 Chapter23:InterviewStrategyChapter24:SystemDesign
Interview strategy and system design chapter arethefinalchaptersofthiscourse.
Week 11 &Week12 Revisionofthechapterslistedabove.
At this time, you need to revise all the chaptersthat we have seen in this book.Whatever is leftneeds to be completed and the exercise that maybeleftneedingtobesolvedinthisperiod.
5MonthPreparationPlanThispreparationplanismadeontopof3-monthplan.Inthisplan,thestudentsshouldlookforalgorithmdesign chapters. In addition, in the rest of the time they need to practice more and more fromwww.topcoder.com and other resources. If you are targeting google, Facebook, etc., then it is highlyrecommendedtojointopcoderandpracticeasmuchaspossible.Time Chapters Explanation
Week1Week2
Chapter 1: Introduction - ProgrammingOverviewChapter2:AlgorithmsAnalysisChapter3:ApproachToSolveAlgorithmDesignProblemsChapter4:AbstractDataType
You will get a basic understanding of how tofind complexity of a solution. You will knowhow to handle new problems. You will readaboutavarietyofdatatypesandtheiruses.
Week 3Week4Week5
Chapter5:SearchingChapter6:SortingChapter14:StringAlgorithms
Searching, sorting and string algorithm consistsofamajorportionoftheinterviews.
Week 6Week7Week8
Chapter7:LinkedListChapter8:StackChapter9:Queue
Linked lists are one of the favourites in aninterview.
Week9Week10
Chapter10:TreeChapter11:Heap Thisportionyouwillreadabouttrees
Week11Week12
Chapter12:Hash-TableChapter13:Graphs
Hash-Table are used throughout this book invarious places but now it is time to understandhow Hash-Table are actually implemented.
Graphsareusedtoproposeasolutionmanyreallifeproblems.
Week13Week14Week 15Week16
Chapter15:AlgorithmDesignTechniquesChapter16:BruteForceChapter17:GreedyAlgorithmChapter 18: Divide-And-Conquer, Decrease-And-ConquerChapter19:DynamicProgrammingChapter 20: Backtracking And Branch-And-BoundChapter 21: Complexity Theory And NpCompleteness
Thesechapterscontainvariousalgorithmstypesand their usage. Once the user is familiar withmostof this algorithm.Then thenext step is tostart solving topcoder problems fromhttps://www.topcoder.com/
Week 17Week18
Chapter22:InterviewStrategyChapter23:SystemDesign
Interviewstrategyandsystemdesignchapterarethefinalchaptersofthiscourse.
Week 19Week20
Revisionofthechapterslistedabove.
At this time,youneed to reviseall thechaptersthatwehaveseeninthisbook.Whateverisleftneedstobecompletedandtheexercisethatmaybeleftneedingtobesolvedinthisperiod.
SummaryTheseare fewpreparationplans thatcanbe followed tocomplete thisbook therebypreparing for theinterview.Itishighlyrecommendedthattheusershouldreadtheproblemstatementfirst,thenheshouldtrytosolvetheproblemsbyhimselfandthenonlyheshouldlookintothesolutiontofindtheapproachofthebookpracticingmoreandmoreproblemsopenourthinkingcapacityandyouwillbeabletohandlenew problems in an interview. System design is a topic that is not asked much from a fresher fromcollege, but as you gain experience its importance increase. We will recommend practicing all theproblems given in this book, then solve more and more problems from online resources likewww.topcoder.com,www.careercup.com,www.geekforgeek.cometc.
CHAPTER1:INTRODUCTION-PROGRAMMINGOVERVIEW
IntroductionThis chapter emphasizes on the fundamentals ofCProgramming language. Itwill talk about variables,pointers,recursion,arraysetc.
Variable"Variables"aresimplystoragelocationstoholddata.Foreveryvariable,somememoryisallocated.Thesizeof thismemorydependsonthetypeof thevariable.Forexample,2bytesareallocatedfor integertype,4bytesareallocatedforfloattype,etc.Program1.11.#include<stdio.h>2.3.intmain()4.5.intvar1,var2,var3;6.var1=100;7.var2=200;8.var3=var1+var2;9.printf("Adding%dand%dwillgive%d",var1,var2,var3);10.return0;11.Analysis:Line5:Memory isallocated forvariablesvar1,var2andvar3.Wheneverwedeclareavariable, thenmemoryisallocatedforstoringthevalueinthevariable.Inourexample,2bytesareallocatedforeachofthevariable.Line6&7:Value100isstoredinvariablevar1andvalue200isstoredinvariablevar2.Line8:Valueofvar1andvar2isaddedandstoredinvar3.Line9:Finally,thevalueofvar1,var2andvar3isprintedtoscreen.
PointersPointersarenothingmorethanvariablesthatstorememoryaddressesofanothervariableandcanbeusedtoaccess thevalue storedat thoseaddresses.Variousoperators suchas*,&,and [], enableus tousepointers.Address-Thenumericlocationofaplaceinmemory.Anaddressofamemoryisusedbythecomputertoidentifydatastoredinitquickly.Justasthepostaladdressofahouseisusedtoidentifyahousebythepostalworker.Dereference -Apointerstoresanaddressofa location in thememory.Toget thevaluestored in thataddress,weneed todereference thepointer,meaningweneed togo to that location andget thevaluestoredthere.Program1.21.intmain()2.3.intvar;4.int*ptr;5.var=10;6.ptr=&var;7.8.printf("Valuestoredatvariablevaris%d\n",var);9.printf("Valuestoredatvariablevaris%d\n",*ptr);10.11.printf("Theaddressofvariablevaris%p\n",&var);12.printf("Theaddressofvariablevaris%p\n",ptr);13.return0;14.Analysis:Line3:Anintegertypevariablevariscreated.Line4:Apointertointptriscreated.Line6:Addressofvarisstoredinptr.Line8&9:Valuestoredinvariablevarisprintedtoscreen.*Operatorisusedtogetthevaluestoredatthepointerlocation.Line11&12:Memoryaddressofvarisprintedtothescreen.&operatorisusedtogettheaddressofavariable.PointtoRemember:
1.Youcandefineapointerbyincludinga*beforethenameofthevariable.(Pointerdeclaration)2.Youcangetthevaluestoredataddressbyadding*beforethevariablename.(Pointeruse)3.Youcangettheaddressofavariablebyusing&operator.
ArrayAnarray is adata structureused to storemultipledataelementsof the samedata type.All thedata isstoredsequentially.Thevaluestoredatanyindexcanbeaccessedinconstanttime.Program1.31.intmain()2.3.intarr[10];4.for(inti=0;i<10;i++)5.6.arr[i]=i;7.8.printArray(arr,10);9.Analysis:Line3:Definesanarrayofintegers.Thearrayisofsize10-whichmeansthatitcanstore10integersinsideit.Line 6: Array elements are accessed using subscript operator []. Lowest subscript is 0 and highestsubscriptis(sizeofarray–1).Value0to9isstoredinthearrayatindex0to9.Line8:ArrayanditssizearepassedtoprintArray()function.Program1.41.voidprintArray(intarr[],intcount)2.3.printf("Valuesstoredinarrayare:");4.for(inti=0;i<count;i++)5.6.printf("[%d]",arr[i]);7.8.Analysis:Line1:ArrayvariablearranditsvariablecountarepassedasargumentstoprintArray()function.Line4-7:Finally,arrayvaluesareprintedtoscreenusingtheprintffunctioninaloop.PointtoRemember:1.Arrayindexalwaysstartsfrom0indexandhighestindexissize-1.2.Thesubscriptoperatorhashighestprecedenceifyouwritearr[2]++.Thenthevalueofarr[2]willbe
incremented.Eachelementofthearrayhasamemoryaddress.Thefollowingprogramwillprintthevaluestoredinanarrayandtheiraddress.
Program1.51.voidprintArrayAddress(intarr[],intcount)2.3.printf("Valuesstoredinarrayare:");4.for(inti=0;i<count;i++)5.6.printf("Data:[%d]hasAddress:[%p]\n",arr[i],arr+i);7.8.Analysis:Line 6: Value stored in an array is printed. The address of the various elements in the array is alsoprinted.PointtoRemember:Forarrayelements,consecutivememorylocationisallocated.Program1.61.voidprintArrayUsingPointer(intarr[],intcount)2.3.printf("Valuesstoredinarrayare:");4.int*ptr=arr;5.for(inti=0;i<count;i++)6.7.printf("Data:[%d]hasAddress:[%p]\n",*ptr,ptr);8.ptr++;9.10.Analysis:Line4:Apointertoanintiscreatedanditwillpointtothearray.Line7:valuestoredinpointerisprintedtoscree.Line8:Pointerisincremented.
TwoDimensionalArrayWecandefinetwodimensionalormultidimensionalarray.Itisanarrayofarray.Program1.71.intmain()2.3.intarr[4][2];4.intcount=0;5.for(inti=0;i<4;i++)6.for(intj=0;j<2;j++)7.arr[i][j]=count++;8.9.print2DArray((int**)arr,4,2);10.print2DArrayAddress((int**)arr,4,2);11.1.voidprint2DArray(int*arr[],introw,intcol)2.3.for(inti=0;i<row;i++)4.for(intj=0;j<col;j++)5.printf("[%d]",*(arr+i*col+j));6.7.1.voidprint2DArrayAddress(int*arr[],introw,intcol)2.3.for(inti=0;i<row;i++)4.for(intj=0;j<col;j++)5.printf("Value:%d,Address:%p\n",*(arr+i*col+j),(arr+i*col+j));6.Analysis:·Anarrayiscreatedwithdimension4x2.Thearraywillhave4rowsand2columns.·Valueisassignedtothearray·Finallythevaluestoredinarrayisprintedtoscreenandtheaddressusedtostorevaluesisprintedby
usingprint2DArray()andprint2DarrayAddress()function.Wecandefineapointerarraysimilartoanarrayofinteger.Thearrayelementwillstoreapointerinsideit.Program1.81.voidprintArray(int*arr[],intcount)2.3.int*ptr;4.for(inti=0;i<count;i++)
5.6.ptr=arr[i];7.printf("[%d]",*ptr);8.9.1.voidprintArrayAddress(int*arr[],intcount)2.3.int*ptr;4.for(inti=0;i<count;i++)5.6.ptr=arr[i];7.printf("Valueis:%d,Addressis:%p\n",*ptr,ptr);8.9.1.intmain()2.3.intone=1,two=2,three=3;4.int*arr[3];5.arr[0]=&one;6.arr[1]=&two;7.arr[2]=&three;8.printArray(arr,3);9.printArrayAddress(arr,3);10.Analysis:·Threevariables,one,twoandthreearedefined.·Pointerarrayarrisdefied.·Theaddressofone,twoandthreeisstoredinsidearrayarr.·printArray()andprintArrayAddress()functionsareusedtoprintvaluestoredinthearray.
ArrayInterviewQuestionsThefollowingsectionwilldiscussthevariousalgorithmsthatareapplicabletoarraysandwillfollowbylistofpracticeproblemswithsimilarapproaches.
SumArrayProgram:Writeafunctionthatwillreturnthesumofalltheelementsoftheintegerarraygivenarrayanditssizeasanargument.Program1.9intSumArray(intarr[],intsize)
inttotal=0;intindex=0;for(index=0;index<size;index++)total=total+arr[index];returntotal;
SequentialSearchWriteafunctionwhichwillsearchinanarraythatsomevalueispresentinthearrayornot.Program1.10:intSequentialSearch(intarr[],intsize,intvalue)
inti=0;for(i=0;i<size;i++)if(value==arr[i])returni;return-1;
Analysis:·Sincewehavenoideaaboutthedatastoredinarrayorifthedataisnotsortedthenwehavetosearch
thearrayinsequentialmanneronebyone.·Ifwefindthevalue,wearelookingforwereturnthatindex.·Else,wereturn-1index,aswedidnotfoundthevaluewearelookingfor.In the above example, the data are not sorted. If the data is sorted, a binary searchmay be done.Weexamine themiddlepositionat each step.Dependingupon thedata thatweare searching isgreateror
smallerthanthemiddlevalue.Wewillsearcheithertheleftortherightportionofthearray.Ateachstep,weareeliminatinghalfofthesearchspacetherebymakingthisalgorithmveryefficientthelinearsearch.
BinarySearchProgram1.11:Binarysearchinasortedarray./*BinarySearchAlgorithm–IterativeWay*/intBinarySearch(intarr[],intsize,intvalue)
intlow=0,mid;inthigh=size-1;while(low<=high)mid=low+(high-low)/2;/*Toavoidtheoverflow*/if(arr[mid]==value)returnmid;elseif(arr[mid]<value)low=mid+1;elsehigh=mid-1;return-1;
Analysis:·Sincewehavedatasortedinincreasing/decreasingorder,wecanapplymoreefficientbinarysearch.
Ateachstep,wereduceoursearchspacebyhalf.·Ateachstep,wecomparethemiddlevaluewiththevaluewearesearching.Ifmidvalueisequaltothe
valuewearesearchingforthenwereturnthemiddleindex.·Ifthevalueissmallerthanthemiddlevalue,wesearchthelefthalfofthearray.·Ifthevalueisgreaterthanthemiddlevaluethenwesearchtherighthalfofthearray.·Ifwefindthevaluewearelookingforthenitsindexisreturnedor-1isreturnedotherwise.
RotatinganArraybyKpositions.Forexample,anarray[10,20,30,40,50,60]rotateby2positionsto[30,40,50,60,10,20]Program1.12voidrotateArray(int*a,intn,intk)
reverseArray(a,k);reverseArray(&a[k],n-k);reverseArray(a,n);
voidreverseArray(int*a,intn)
for(inti=0,j=n-1;i<j;i++,j--)a[i]^=a[j]^=a[i]^=a[j];
1,2,3,4,5,6,7,8,9,10=>5,6,7,8,9,10,1,2,3,41,2,3,4,5,6,7,8,9,10=>4,3,2,1,10,9,8,7,6,5=>5,6,7,8,9,10,1,2,3,4Analysis:·Rotatingarrayisdoneintwopartstrick.Inthefirstpart,wefirstreverseelementsofarrayfirsthalfand
thensecondhalf.·Thenwereversethewholearraytherebycompletingthewholerotation.
Findthelargestsumcontiguoussubarray.Givenanarrayofpositiveandnegativeintegers,findacontiguoussubarraywhosesum(sumofelements)ismaximized.
Program1.13intmaxSubArraySum(inta[],intsize)
intmaxSoFar=0,maxEndingHere=0;for(inti=0;i<size;i++)maxEndingHere=maxEndingHere+a[i];if(maxEndingHere<0)maxEndingHere=0;if(maxSoFar<maxEndingHere)maxSoFar=maxEndingHere;returnmaxSoFar;
Analysis:·Maximumsubarrayinanarrayisfoundinasinglescan.Wekeeptrackofglobalmaximumsumsofar
andthemaximumsum,whichincludethecurrentelement.·Whenwefindglobalmaximumvaluesofarislessthanthemaximumvaluecontainingcurrentvaluewe
updatetheglobalmaximumvalue.·Finallyreturntheglobalmaximumvalue.
StructureStructuresareusedwhenwewanttoprocessdataofmultipledatatypesasasingleentity.Program1.14:DemonstratingStructure1.structcoord2.intx;3.inty;4.;5.intmain()6.7.structcoordpoint;8.point.x=10;9.point.y=10;10.printf("Xaxiscoordvalueis%d\n",point.x);11.printf("Yaxiscoordvalueis%d\n",point.y);12.printf("Sizeofstructureis%dbytes\n",sizeof(point));13.return0;14.Output:Xaxiscoordvalueis10Yaxiscoordvalueis10Sizeofstructureis8bytesAnalysis:Line1-4:Wehavedeclaredstructure"coord",whichcontaintwoelementsinsideit.Thetwoelementsxandycorrespondingtox-axisandy-axiscoordinates.Line8:Wehavedeclaredavariable"point"oftype"structcoord”.Line9-10:Wehaveassignedcoordinate(10,10),toxandyelementofthe"point".Variouselementsofastructureareassessedusingthedot(.)operator.Line11-12:Weareprinting thevalue stored in thex andy elements of thepoint that is of type structcoord.Line13:Weareprintingthesizeofstructpoint.Sincestructureconsistsofmorethanoneelement,thenthesizeofastructureisthesumofalltheelementsinsideit.
PointertostructurePointerscanbeusedtoaccessthevariouselementsofastructure.Thevariouselementsofastructureareaccessedbypointerusing->operator.Program1.15:Pointertostructure1.#include<stdio.h>2.structstudent3.introllNo;4.char*firstName;5.char*lastName;6.;7.intmain()8.9.inti=0;10.structstudentstud;11.structstudent*ptrStud;12.ptrStud=&stud;13.ptrStud->rollNo=1;14.ptrStud->firstName="john";15.ptrStud->lastName="smith";16. printf("Roll No: %d Student Name: %s %s ", ptrStud->rollNo, ptrStud->firstName, ptrStud->lastName);17.18.return0;19.Analysis:Line2-6:Wehavedeclaredastructstudentthatcontainrollno,firstandlastnameofastudent.Line12:Wehavedeclaredapointertostructstudent.Line13-15:PointerptrStudispointingtostud.WehaveusedptrStudtoassignavaluetostructstud.Wehaveused->operatortoaccessthevariouselementsofthestructurepointedbyptrStud.Note: If we have used the stud to assignwewould have used "." operator. The same structurewhenaccessedusingpointerweuseindirectionoperator"->".Line16:Wehavefinallyprintedallthevariouselementsofstructurevariablestud.Note:Inthesamewayyoucanuse->operatortoaccesselementsoftheUnion.
DynamicMemoryAllocationIn c language, dynamic memory is allocated using the malloc(), calloc() and realloc() functions. Thedynamicmemoryrequiredtobefreedusingthefree()function.
MallocfunctionDefinitionofthemallocfunctionisasbelow.void*malloc(size_tsize);Itallocatesamemoryblockoflength"size"bytesandreturnapointertotheblock.ItwillreturnNULLifthesystemdoesnothaveenoughmemory.TheCstandarddefinesvoid*isagenericpointerthatisrequiredtobecastedtotherequiredtype.MostCcompilersneedthiscasting.However,thelatestANSICstandarddoesnotrequire.E.g.int*p=(int*)malloc(sizeof(int));
CallocfunctionDefinitionofcallocfunctionisasbelow.void*calloc(size_tnum,size_tsize);Itallocatesamemoryblockoflength"num*size"bytesandreturnapointertotheblock.ItwillreturnNULLifthesystemdoesnothaveenoughmemory.Onethingmoreitdoesthat,itinitializeseverybytetozero.
ReallocfunctionDefinitionofreallocfunctionisasbelow.void*realloc(void*ptr,size_tnewSize);Itisusedtochangethememoryblocksizeofapreviouslyallocatedblockofmemorypointedbyptr.ItreturnsamemoryblockofthenewSize.Iftheblocksizeifincreased,thenthecontentoftheoldmemoryblockiscopiedtoanewlyallocatedregion.Ifthepointerreturnedbythefunctionisdifferentfromtheoldpointerptr.Thenptrwillnolongerpointtoavalidlocation.Soingeneralyoushouldnotuseptronceitispassedtorealloc()function.IfptrisNULL,reallocworkssameasmalloc().Note:Again,youneed tocast the returnvalueofmalloc/calloc/reallocbeforeusing it. int*i= (int*)malloc(size);
FreeFunctionThememory that is allocatedusingmalloc/calloc/reallocneed tobe freedusing a free() function.The
syntaxofthefree()functionisasbelow.voidfree(void*pointer);Apointer topreviously allocatedmemory ispassed to free() function.The free() functionwill put theallocatedmemoryblockbacktoheapsection.
FunctionFunctionsareusedtoprovidemodularitytotheprogram.Byusingfunction,wecandividecomplextasksintosmallermanageabletasks.Theuseofthefunctionisalsotoavoidduplicatecode.Forexample,wecandefineafunctionsum()whichtaketwointegersandreturntheirsum.Thenwecanusethisfunctionmultipletimeswheneverwewantsumoftwointegers.Program1.16:DemonstratingFunctionCalls1.#include<stdio.h>2./*functiondeclaration*/3.intsum(intnum1,intnum2);4.intmain()5.6./*localvariabledefinition*/7.intx=10;8.inty=20;9.intresult;10./*callingafunctiontofindsum*/11.result=sum(x,y);12.printf("Sumis:%d\n",result);13.return0;14.15./*functionreturningthesumoftwonumbers*/16.intsum(intnum1,intnum2)17.18./*localvariabledeclaration*/19.intresult;20.result=num1+num2;21.returnresult;22.Output:Sumis:30Analysis:Line3:functiondeclarationofsum()functionLine11:sumfunctioniscalledfromthismainbypassingvariablexandywithvalue10and20atthispointcontrolflowwillgotoLine16.Line16:variablespassedtosumfunctionarecopiedintonum1andnum2localvariables.Line20&21: thesumiscalculatedandsaved inavariable result.And the result is returned.Controlflowcomesbacktolinenumber11.Line11-12:returnvalueofthesumfunctionissavedinalocalvariableresultandprintedtothescreen.
ConceptofStackAstackisamemoryinwhichvaluesarestoredandretrievedin“lastinfirstout”manner.Dataisaddedtostackusingpushoperationanddataistakenoutofstackusingpopoperation.
1.Initiallythestackwasempty.Thenwehaveaddedvalue1tostackusingpush(1)operator.2.Similarly,push(2)andpush(3)3.Popoperationtakethetopofthestack.Stackdataisaddedanddeletedin“lastin,firstout”manner.4.Firstpop()operationwilltake3outofthestack.5.Similarly,otherpopoperationwilltake2then1outofthestack6.Intheend,thestackisemptywhenalltheelementsaretakenoutofthestack.
SystemstackandFunctionCallsWhenthefunctioniscalled,thecurrentexecutionisstoppedandthecontrolgoestothecalledfunction.Afterthecalledfunctionexits/returns,theexecutionresumesfromthepointatwhichtheexecutionwasstopped.Togettheexactpointatwhichexecutionshouldberesumed,theaddressofthenextinstructionisstoredinthestack.Whenthefunctioncallcompletetheaddressatthetopofthestackistakenout.Program1.171.voidfun2()2.3.printf("fun2line1\n");4.5.6.voidfun1()7.8.printf("fun1line1\n");9.fun2();10.printf("fun1line2\n");11.12.13.intmain()14.15.printf("mainline1\n");16.fun1();17.printf("mainline2\n");18.Output:mainline1fun1line1fun2line1fun1line2mainline2Analysis:Line13:Everyprogramstartswithmain()function.Line15:Thisisthefirststatementthatwillbeexecuted.Andwewillget“mainline1”asoutput.Line16:fun1()iscalled.Beforecontrolgoestofun1()thennextinstructionthatisaddressofline17isstoredinthesystemstack.Line6:Controlgoestofun1()function.Line8:Thisisthefirststatementinsidefun1(),thiswillprint“fun1line1”tooutput.Line9:fun2()iscalledfromfun1().Beforecontrolgoestofun2()addressofthenextinstructionthatisaddressofline10isaddedtothesystemstack.
Line1:Controlgoestofun2()function.Line3:“fun2line1”isprintedtoscreen.Line10:Whenfun2exits,controlcomeback to fun1.Moreover, theprogramreads thenext instructionfromthestack,andline10isexecuted.Andprint“fun1line2”toscreen.Line17:When fun1 exits, control comesback to themain function.Program reads thenext instructionfromthestack,linenumber17isexecuted,andfinally“mainline2”isprintedtoscreen.Pointstoremember:1.Functionsareimplementedusingastack.2.Whenafunctioniscalledtheaddressofthenextinstructionispushedintothestack.3.Whenafunctionisfinishedtheaddressoftheexecutionistakenoutofthestack.
Parameterpassing,CallbyvalueArgumentscanbepassedfromonefunctiontootherusingparameters.Bydefault,alltheparametersarepassedbyvalue.Thatmeansaseparatecopyiscreatedinsidethecalledfunctionandthevariableinthecallingfunctionremainsunchanged.Program1.181.voidincrement(intvar)2.3.var++;4.5.6.intmain()7.8.inti=10;9.printf("Valueofibeforeincrementis:%d\n",i);10.increment(i);11.printf("Valueofibeforeincrementis:%d\n",i);12.Output:Valueofibeforeincrementis:10Valueofibeforeincrementis:10Analysis:Line8:variable”i”isdeclaredandvalue10isinitializedtoit.Line9:valueif”i”isprinted.Line10:incrementfunctioniscalled.Whenafunctioniscalledthevalueoftheparameteriscopiedintoanothervariableofthecalledfunction.Flowofcontrolgoestolineno1.Line 3: value of var is incremented by 1. However, remember, it is just a copy inside the incrementfunction.Line11:Whenthefunctionexits,thevalueof”i”isstill10.Pointstoremember:1.Passbyvaluejustcreatesacopyofvariable.2.Passbyvalue,valuebeforeandafterthefunctioncallremainsame.
Parameterpassing,CallbyReferenceIfyouneedtochangethevalueoftheparameterinsidethefunction,thenyoushouldusecallbyreference.Clanguagebydefaultpassesbyvalue.Therefore,tomakeithappen,youneedtopasstheaddressofavariableandchangingthevalueofthevariableusingthisaddressinsidethecalledfunction.Program1.191.voidincrement(int*ptr)2.3.(*ptr)++;4.5.intmain()6.7.inti=10;8.printf("Valueofibeforeincrementis:%d\n",i);9.increment(&i);10.printf("Valueofibeforeincrementis:%d\n",i);11.Output:Valueofibeforeincrementis:10Valueofibeforeincrementis:11Analysis:Line9:Addressof“i” ispassed to thefunction increment.Function increment takesapointer to intasargument.Line3:Variableattheaddressptrisaccessedanditsvalueisincremented.Line10:Finally,incrementedvalueisprintedtoscreen.Pointstoremember:1.Callbyreferenceisimplementedindirectlybypassingtheaddressofavariabletothefunction.
RecursiveFunctionArecursivefunctionisafunctionthatcallsitself,directlyorindirectly.A recursive function consists of two parts: TerminationCondition andBody (which include recursiveexpansion).1.TerminationCondition:A recursive function always contains oneormore terminating condition.A
conditioninwhichrecursivefunctionisprocessingasimplecaseanddonotcallitself.2.Body(includingrecursiveexpansion):Themainlogicoftherecursivefunctioncontainedinthebody
ofthefunction.Italsocontainstherecursionexpansionstatementthatinturncallsthefunctionitself.Threeimportantpropertiesofrecursivealgorithmare:1)Arecursivealgorithmmusthaveaterminationcondition.2)Arecursivealgorithmmustchangeitsstate,andmovetowardstheterminationcondition.3)Arecursivealgorithmmustcallitself.Note:Thespeedofarecursiveprogramisslowerbecauseofstackoverheads.If thesametaskcanbedoneusingan iterativesolution(loops), thenweshouldpreferan iterativesolution(loops) inplaceofrecursiontoavoidstackoverhead.Note:Withoutterminationcondition,therecursivefunctionmayrunforeverandwillfinallyconsumeallthestackmemory.
FactorialProgram1.20:FactorialCalculation.N!=N*(N-1)….2*1.1.intfactorial(unsignedinti)2.3./*TerminationCondition*/4.if(i<=1)5.return1;6./*Body,RecursiveExpansion*/7.returni*factorial(i-1);8.Analysis:Eachtimefunctionfniscallingfn-1.TimeComplexityisO(N)
PrintBase10IntegersProgram1.211.voidprintInt(unsignedintnumber)2.3.chardigit=number%10+'0';4.if(number/=10)
5.printInt(number/10);6.printf("%c",digit);7.Analysis:Line3:Eachtimeremainderiscalculatedandstoreditscharequivalentindigit.Line4-5:ifthenumberisgreaterthan10thenthenumberdividedby10ispassedtoprintInt()function.Line6:Numberwillbeprintedwithhigherorderfirstthenthelowerorderdigits.TimeComplexityisO(N)
PrintBase16IntegersProgram1.22:Genericprinttosomespecificbasefunction.1.voidprintInt(unsignedintnumber,constintbase)2.3.char*conversion="0123456789ABCDEF";4.chardigit=number%base;5.if(number/=base)6.printInt(number,base);7.printf("%c",conversion[digit]);8.Analysis:Line1:basevalueisalsoprovidedalongwiththenumber.Line4:remainderofthenumberiscalculatedandstoredindigit.Line5-6:ifthenumberisgreaterthanbasethen,numberdividedbybaseispassedasanargumenttotheprintInt()functionrecursively.Line7:Numberwillbeprintedwithhigherorderfirstthenthelowerorderdigits.TimeComplexityisO(N)
IntegertoStringProgram1.231.char*intToStr(char*p,unsignedintnumber)2.3.chardigit=number%10+'0';4.if(number/=10)5.p=intToStr(p,number);6.*p++=digit;7.return(p);8.Analysis:Line1:characterbufferp ispassedasargumentandnumber ispassed,whichneed tobeconverted tostring.
Line3:leastsignificantdigitofthenumberisconvertedintocorrespondingcharacter.Line4-5: if thenumber isgreater than10 then,numberdividedby10 ispassedas anargument to theintToStr()functionrecursively.Line6:Thecharacterisstoredinthebufferphigherorderfirst,thenthelowerorderdigits.TimeComplexityisO(N)
TowerofHanoiTheTowerofHanoi(alsocalledtheTowerofBrahma)WearegiventhreerodsandNnumberofdisks,initiallyallthedisksareaddedtofirstrod(theleftmostone)indecreasingsizeorder.Theobjectiveistotransfertheentirestackofdisksfromfirsttowertothirdtower(therightmostone),movingonlyonediskatatimeandneveralargeroneontoasmaller.
Program1.241.voidtowerOfHanoi(intnum,charsrc,chardst,chartemp)2.3.if(num<1)4.return;5.6.towerOfHanoi(num-1,src,temp,dst);7.printf("\nMovedisk%dfrompeg%ctopeg%c",num,src,dst);8.towerOfHanoi(num-1,temp,dst,src);9.1.intmain()2.3.intnum=4;4.printf("ThesequenceofmovesinvolvedintheTowerofHanoiare:\n");5.towerOfHanoi(num,'A','C','B');6.return0;7.Analysis:TowerOfHanoiproblemifwewanttomoveNdisksfromsourcetodestination,thenwefirst
moveN-1disksfromsourcetotemp,thenmovethelowestNthdiskfromsourcetodestination.ThenwillmoveN-1disksfromtemptodestination.
Greatestcommondivisor(GCD)Program1.251.intGCD(intm,intn)2.3.if(m<n)4.return(GCD(n,m));5.if(m%n==0)6.return(n);7.return(GCD(n,m%n));8.Analysis:Euclid’salgorithmisusedtofindgcd.GCD(n,m)==GCD(m,nmodm)
FibonaccinumberProgram1.261.intfibonacci(intn)2.3.if(n<=1)4.returnn;5.returnfibonacci(n-1)+fibonacci(n-2);6.Analysis:Fibonaccinumberarecalculatedbyaddingsumoftheprevioustwonumber.Thereisaninefficiencyinthesolutionwewilllookbettersolutionincomingchapters.
AllpermutationsofanintegerarrayProgram1.271.voidprintArray(intarr[],intcount)2.3.printf("Valuesstoredinarrayare:");4.for(inti=0;i<count;i++)5.6.printf("%d",arr[i]);7.8.printf("\n");9.10.voidswap(int*arr,intx,inty)11.inttemp=arr[x];12.arr[x]=arr[y];13.arr[y]=temp;14.return;
15.16.voidpermutation(int*arr,inti,intlength)17.if(length==i)18.printArray(arr,length);19.return;20.21.intj=i;22.for(j=i;j<length;j++)23.swap(arr,i,j);24.permutation(arr,i+1,length);25.swap(arr,i,j);26.27.return;28.29.intmain()30.31.intarr[5];32.for(inti=0;i<5;i++)33.34.arr[i]=i;35.36.permutation(arr,0,5);37.Analysis:In permutation function at each recursive call number at index, “i” is swapped with all thenumbersthatarerightofit.Sincethenumberisswappedwithallthenumbersinitsrightonebyoneitwillproduceallthepermutationpossible.
BinarysearchusingrecursionProgram1.281./*BinarySearchAlgorithm–RecursiveWay*/2.intBinarySearchRecursive(intarr[],intlow,inthigh,intvalue)3.4.if(low>high)5.return-1;6.intmid=low+(high-low)/2;/*Toavoidtheoverflow*/7.if(arr[mid]==value)8.returnmid;9.elseif(arr[mid]<value)10.returnBinarySearchRecursive(arr,mid+1,high,value);11.else12.returnBinarySearchRecursive(arr,low,mid-1,value);13.
Analysis:Similar iterativesolutionwehadalreadyseen.Nowletus lookinto therecursivesolutionof thesameprobleminthissolutionalso,wearedivingthesearchspaceintohalfanddoingthesamewhatwehaddoneintheiterativesolution.
Exercises1.Findaverageofalltheelementsinanarray.2.Findthesumofalltheelementsofatwodimensionalarray.3.Findthelargestelementinthearray.4.Findthesmallestelementinthearray.5.Findthesecondlargestnumberinthearray.6.Printall themaxima’s inanarray. (Avalue isamaximumif thevaluebeforeandafter its indexare
smallerthanitisordoesnotexist.)Hint:a)Starttraversingarrayfromtheendandkeeptrackofthemaxelement.b)Ifweencounteranelement>max,printtheelementandupdatemax.
7.Printalternateelementsinanarray.8.Givenanarraywithvalue0or1,writeaprogramtosegregate0ontheleftsideand1ontherightside.9.Givenalistofintervals,mergealloverlappingintervals.
Input:[1,4],[3,6],[8,10]Output:[1,6],[8,10]
10.Writeafunctionthatwilltakeintervalsasinputandtakescareofoverlappingintervals.11.Reverse an array in-place. (You cannot use any additional array inotherwardsSpaceComplexity
shouldbeO(1).)Hint:Usetwovariable,startandend.Startsetto0andendsetto(n-1).Incrementstartanddecrementend.Swapthevaluesstoredatarr[start]andarr[end].Stopwhenstartisequaltoendorstartisgreaterthanend.
12.Givenanarrayof0sand1s.Weneedtosortitsothatallthe0sarebeforeallthe1s.Hint:Usetwovariable,startandend.Startsetto0andendsetto(n-1).Incrementstartanddecrementend.Swapthevaluesstoredatarr[start]andarr[end]onlywhenarr[start]==1andarr[end]==0.Stopwhenstartisequaltoendorstartisgreaterthanend.
13.Givenanarrayof0s,1sand2s.Weneedtosortitsothatallthe0sarebeforeallthe1sandallthe1s
arebefore2s.Hint:Sameasabovefirstthink0sand1sasonegroupandmoveallthe2sontherightside.Thendoasecondpassoverthearraytosort0sand1s.
14.Findtheduplicateelementsinanarrayofsizenwhereeachelementisintherange0ton-1.
Hint:Approach1:Compareeachelementwithalltheelementsofthearray(usingtwoloops)O(n2)solutionApproach2:MaintainaHash-Table.Setthehashvalueto1ifweencountertheelementforthefirsttime.Whenwesamevalueagainwecanseethatthehashvalueisalready1sowecanprintthatvalue.
O(n)solution,butadditionalspaceisrequired.Approach3:Wewillexploittheconstraint"everyelementisintherange0ton-1".Wecantakeanarrayarr[]ofsizenandsetalltheelementsto0.Wheneverwegetavaluesayval1.Wewillincrementthevalueatarr[var1]indexby1.Intheend,wecantraversethearrayarrandprinttherepeatedvalues.AdditionalSpaceComplexitywillbeO(n)whichwillbeless thanHash-Tableapproach.
15.Findthemaximumelementinasortedandrotatedarray.Complexity:O(logn)
Hint:Usebinarysearchalgorithm.16.Givenanarraywith'n'elements&avalue'x',findtwoelementsinthearraythatsumsto'x'.
Hint:Approach1:Sortthearray.Approach2:UsingaHash-Table.
17.Write a function to find the sum of every number in an int number. Example: input= 1984, output
shouldbe32(1+9+8+4).18.WriteafunctiontocomputeSum(N)=1+2+3+…+N.
CHAPTER2:ALGORITHMSANALYSIS
IntroductionComputer programmer learn by experience. We learn by seeing solved problems and solving newproblemsbyourselves.Studyingvariousproblem-solvingtechniquesandbyunderstandinghowdifferentalgorithmsaredesignedhelpsustosolvethenextproblemthatisgiventous.Byconsideringanumberofdifferentalgorithms,wecanbegintodeveloppatternsothatthenexttimeasimilarproblemarises,wearebetterabletosolveit.Whenanintervieweraskstodevelopaprograminaninterviewer,whatarethestepsthatanintervieweeshouldfollow.Wewillbetakingasystematicapproachtohandletheproblemandfinallyreachingtothesolution.
AlgorithmAnalgorithmisasetofstepstoaccomplishatask.Analgorithminacomputerprogramisasetofstepsappliedoverasetofinputtoproduceasetofoutput.Knowledgeofalgorithmhelpsustogetourdesiredresultfasterbyapplyingtherightalgorithm.Themostimportantpropertiesofanalgorithmare:1.Correctness:The algorithm should be correct. It should be able to process all the given inputs and
providecorrectoutput.2.Efficiency:Thealgorithmshouldbeefficientinsolvingproblems.Algorithmiccomplexityisdefinedashowfastaparticularalgorithmperforms.ComplexityisrepresentedbyfunctionT(n)-timeversustheinputsizen.
AsymptoticanalysisAsymptoticanalysisisusedtocomparetheefficiencyofalgorithmindependentlyofanyparticulardatasetorprogramminglanguage.Wearegenerallyinterestedintheorderofgrowthofsomealgorithmandnotinterestedintheexacttimerequiredforrunninganalgorithm.ThistimeisalsocalledAsymptotic-runningtime.
Big-ONotationDefinition:“f(n)isbig-Oofg(n)”orf(n)=O(g(n)),iftherearetwo+veconstantscandn0suchthatf(n)≤cg(n)foralln≥n0,Inotherwords,cg(n)isanupperboundforf(n)foralln≥n0Thefunctionf(n)growthisslowerthancg(n)
Example:n2+n=O(n2)
Omega-ΩNotationDefinition:“f(n)isomegaofg(n).”orf(n)=Ω(g(n))iftherearetwo+veconstantscandn0suchthatcg(n)≤f(n)foralln≥n0Inotherwords,cg(n)islowerboundforf(n)Functionf(n)growthisfasterthancg(n)
Findrelationshipoff(n)=ncandg(n)=cnF(n)=Ω(g(n))
Theta-ΘNotationDefinition:“f(n)isthetaofg(n).”orf(n)=Θ(g(n))iftherearethree+veconstantsc1,c2andn0suchthatc1g(n)≤f(n)≤c2g(n)foralln≥n0g(n)isanasymptoticallytightboundonf(n).Functionf(n)growsatthesamerateasg(n).
Example:n3+n2+n=Ɵ(n2)Example:n2+n=Ɵ(n2)Findrelationshipoff(n)=2n2+nandg(n)=n2f(n)=O(g(n))f(n)=Ɵ(g(n))f(n)=Ω(g(n))Note:-AsymptoticAnalysisisnotperfect,butthatisthebestwayavailableforanalysingalgorithms.Forexample, say thereare twosortingalgorithms first take f(n)=10000nlognand f(n)=n2 time.Theasymptoticanalysissaysthatthefirstalgorithmisbetter(asitignoresconstants)butactuallyforasmallset of data when n is small then 10000, the second algorithm will perform better. To consider thisdrawbackofasymptoticanalysiscaseanalysisofthealgorithmisintroduced.
Complexityanalysisofalgorithms1)WorstCase complexity: It is the complexityof solving theproblem for theworst input of sizen. Itprovidestheupperboundforthealgorithm.Thisisthemostcommonanalysisdone.2)AverageCasecomplexity:Itisthecomplexityofsolvingtheproblemonanaverage.Wecalculatethetimeforallthepossibleinputsandthentakeanaverageofit.3)BestCasecomplexity:Itisthecomplexityofsolvingtheproblemforthebestinputofsizen.
TimeComplexityOrderAlistofcommonlyoccurringalgorithmTimeComplexityinincreasingorder:Name NotationConstant O(1)Logarithmic O(logn)Linear O(n)N-LogN O(n.logn)Quadratic O(n2)Polynomial O(nc)cisaconstant&c>1Exponential O(cn)cisaconstant&c>1FactorialorN-power-N O(n!)orO(nn)
ConstantTime:O(1)Analgorithmissaidtoruninconstanttimeregardlessoftheinputsize.Examples:1.Accessingnthelementofanarray2.Pushandpopofastack.3.EnqueueandDequeueofaqueue.4.AccessinganelementofHash-Table.5.Bucketsort
LinearTime:O(n)Analgorithmissaidtoruninlineartimeiftheexecutiontimeofthealgorithmisdirectlyproportionaltotheinputsize.Examples:1.Arrayoperationslikesearchelement,findmin,findmaxetc.2.Linkedlistoperationsliketraversal,findmin,findmaxetc.Note:whenweneedtosee/traverseallthenodesofadata-structureforsometaskthencomplexityisnolessthanO(n)
LogarithmicTime:O(logn)Analgorithmissaidtoruninlogarithmictimeiftheexecutiontimeofthealgorithmisproportionaltothelogarithm of the input size. Each step of an algorithm, a significant portion of the input is pruned outwithouttraversingit.Example:1.BinarysearchNote:Wewillreadaboutthesealgorithmsinthisbook.
N-LogNTime:O(nlog(n))Analgorithmissaidtoruninlogarithmictimeiftheexecutiontimeofanalgorithmisproportionaltothe
productofinputsizeandlogarithmoftheinputsize.Example:1.Merge-Sort2.Quick-Sort(Averagecase)3.Heap-SortNote:Quicksortisaspecialkindofalgorithmtosortalistofnumbers.Itsworst-casecomplexityisO(n2)andaveragecasecomplexityisO(n.logn)
QuadraticTime:O(n2)Analgorithmissaidtoruninlogarithmictimeiftheexecutiontimeofanalgorithmisproportionaltothesquareoftheinputsize.Examples:1.Bubble-Sort2.Selection-Sort3.Insertion-Sort
DerivingtheRuntimeFunctionofanAlgorithm
Constants
Eachstatementtakesaconstanttimetorun.TimeComplexityisO(1)
Loops
The running time of a loop is a product of running time of the statement inside a loop and number ofiterationsintheloop.TimeComplexityisO(n)
NestedLoop
Therunningtimeofanestedloopisaproductofrunningtimeofthestatementsinsideloopmultipliedbyaproductofthesizeofalltheloops.TimeComplexityisO(nc)Wherecisanumberofloops.Fortwoloops,itwillbeO(n2)
Consecutive Statements
Justaddtherunningtimesofalltheconsecutivestatements
If-Else Statement
Considertherunningtimeofthelargerofifblockorelseblock.Moreover,ignoretheotherone.
Logarithmicstatement
Ifeachiterationtheinputsizeofdecreasesbyaconstantmultiplefactors.O(logn)
TimeComplexityExamplesExample1intfun(intn)
intm=0;for(inti=0;i<n;i++)m+=1;returnm;
TimeComplexity:O(n)Example2intfun(intn)
inti=0,j=0,m=0;for(i=0;i<n;i++)for(j=0;j<n;j++)m+=1;returnm;
TimeComplexity:O(n2)Example3intfun(intn)
inti=0,j=0,m=0;for(i=0;i<n;i++)for(j=0;j<i;j++)m+=1;returnm;
TimeComplexity:O(N+(N-1)+(N-2)+...)==O(N(N+1)/2)==O(n2)Example4intfun(intn)
inti=0,m=0;i=1;while(i<n)m+=1;i=i*2;
returnm;
Eachtimeproblemspaceisdividedintohalf.TimeComplexity:O(log(n))Example5intfun(intn)
inti=0,m=0;i=n;while(i>0)m+=1;i=i/2;returnm;
Sameasaboveeachtimeproblemspaceisdividedintohalf.TimeComplexity:O(log(n))Example6intfun(intn)
inti=0,j=0,k=0,m=0;i=n;for(i=0;i<n;i++)for(j=0;j<n;j++)for(k=0;k<n;k++)m+=1;returnm;
Outerloopwillrunfornnumberofiterations.Ineachiterationoftheouterloop,innerloopwillrunforniterationsoftheirown.Finalcomplexitywillben*n*nTimeComplexity:O(n3)Example7intfun(intn)
inti=0,j=0,k=0,m=0;i=n;for(i=0;i<n;i++)for(j=0;j<n;j++)m+=1;
for(i=0;i<n;i++)for(k=0;k<n;k++)m+=1;returnm;
These two groups of for loop are in consecutive so their complexity will add up to form the finalcomplexityoftheprogram.TimeComplexity:T(n)=O(n2)+O(n2)=O(n2)Example8intfun(intn)
inti=0,j=0,m=0;
for(i=0;i<n;i++)for(j=0;j<sqrt(n);j++)m+=1;returnm;
TimeComplexity:O(n*√n)=O(n3/2)Example9intfun(intn)
inti=0,j=0,m=0;
for(i=n;i>0;i/=2)for(j=0;j<i;j++)m+=1;returnm;
Eachtimeproblemspaceisdividedintohalf.TimeComplexity:O(log(n))Example10intfun(intn)
inti=0,j=0,m=0;
for(i=0;i<n;i++)for(j=i;j>0;j--)
m+=1;returnm;
O(N+(N-1)+(N-2)+...)=O(N(N+1)/2)//arithmeticprogression.TimeComplexity:O(n2)Example11intfun(intn)
inti=0,j=0,k=0,m=0;
for(i=0;i<n;i++)for(j=i;j<n;j++)for(k=j+1;k<n;k++)m+=1;
returnm;
TimeComplexity:O(n3)Example12intfun(intn)
inti=0,j=0,m=0;
for(i=0;i<n;i++)for(;j<n;j++)m+=1;returnm;
Thinkcarefullyonceagainbeforefindingasolution,jvalueisnotresetateachiteration.TimeComplexity:O(n)Example13intfun(intn)
inti=1,j=0,m=0;
for(i=1;i<=n;i*=2)for(j=0;j<=i;j++)m+=1;
returnm;Theinnerloopwillrunfor1,2,4,8,…ntimesinsuccessiveiterationoftheouterloop.TimeComplexity:T(n)=O(1+2+4+….+n/2+n)=O(n)
MasterTheoremThemastertheoremsolvesrecurrencerelationsoftheform:T(n)=aT(n/b)+f(n)Wherea≥1andb>1."n"isthesizeoftheproblem."a"isanumberofsubproblemintherecursion.“n/b”isthesizeofeachsub-problem."f(n)"isthecostofthedivisionoftheproblemintosubproblemormergeofresultsofsubproblemstogettheresult.Itispossibletodetermineanasymptotictightboundinthesethreecases:Case1:when )andconstantЄ>1,thenthefinalTimeComplexitywillbe:
Case2:when )andconstantk≥0,thenthefinalTimeComplexitywillbe:
)Case3:when andconstantЄ>1,thenthefinalTimeComplexitywillbe:T(n)=Ɵ(f(n))
Example14:TakeanexampleofMerge-Sort,T(n)=2T(n/2)+nSol:-Logba=log22=1
)Case2appliesand )T(n)=Ɵ(nlog(n))Example15:BinarySearchT(n)=T(n/2)+O(1)Sol:-Logba=log21=0
)
Case2appliesand )T(n)=Ɵ(log(n))Example16:BinarytreetraversalT(n)=2T(n/2)+O(1)Sol:-Logba=log22=1F(n)=1= )Case1appliesandT(n)=Ɵ(n)Example17:TakeanexampleT(n)=T(n/2)+n2Sol:-Logba=log22=1f(n)=n2=Case3appliesandT(n)=Ɵ(f(n))T(n)=O(n2)Example18:TakeanexampleT(n)=4T(n/2)+n2Sol:-Logba=log24=2f(n)=n2= )Case2appliesandT(n)= )T(n)=Ɵ(n2logn)
ModifiedMastertheoremThisisashortcuttosolvingthesameproblemeasilyandfast.IftherecurrencerelationisintheformofT(n)=aT(n/b)+dxs
Example19:T(n)=2T(n/2)+n2Sol:-r=log22s=2Case3:log22<sT(n)=Ɵ(f(n))=Ɵ(n2)
Example20:T(n)=T(n/2)+2nSol:-r=log21=0s=1Case3T(n)=Ɵ(n)Example21:T(n)=16T(n/4)+nSol:-r=2s=1Case1T(n)=Ɵ(n2)Example22:T(n)=2T(n/2)+nlognSol:-Thereislogninf(n)sousemastertheoremshortcutwillnotword.T(n)=nlog(n)= )
)=Ɵ(nlog(n))Example23:T(n)=2T(n/4)+n0.5Sol:-r=log42=0.5=s
Case2:)=Ɵ(n0.5log1.5n)
Example24:T(n)=2T(n/4)+n0.49Sol:-Case1:
=Ɵ(n0.5)Example25:T(n)=3T(n/3)+√nSol:-r=log33=1s=½Case1T(n)=Ɵ(n)Example26:T(n)=3T(n/4)+nlognSol:-Thereislogninf(n)soseeifmastertheorem.f(n)=nlogn= )Case3:T(n)=Ɵ(nlog(n))Example27:T(n)=3T(n/3)+n/2Sol:-r=1=sCase2:T(n)=Ɵ(nlog(n))
Exercise1.Trueorfalse
a)5n+10n2=O(n2)b)nlogn+4n=O(n)c)log(n2)+4log(logn)=O(logn)d)12n1/2+3=O(n2)e)3n+11n2+n20=O(2n)
2.Whatisthebest-caseruntimeComplexityofsearchinganarray?3.Whatistheaverage-caseruntimeComplexityofsearchinganarray?
CHAPTER3:APPROACHTOSOLVEALGORITHMDESIGNPROBLEMS
IntroductionKnowthetheoreticalknowledgeofthealgorithmisessential,butitisnotsufficient.Youneedtohaveasystematicapproachtosolveaproblem.Ourapproachisfundamentallydifferenttosolveanyalgorithmdesign question.Wewill follow a systematic five-step approach to reach to our solution.Master thisapproachandyouwillbebetterthanmostofthecandidatesininterviews.Fivestepsforsolvingalgorithmdesignquestionsare:1.Constraints2.IdeasGeneration3.Complexities4.Coding5.Testing
ConstraintsSolving a technical question is not just about knowing the algorithms and designing a good softwaresystem.The interviewerwants to knowyou approach towards any given problem.Many peoplemakemistakes as they do not ask clarifying questions about a given problem.They assumemany things andbeginworkingwiththat.Thereisdata,whichismissingthatyouneedtocollectfromyourinterviewerbeforebeginningtosolveaproblem.Inthisstep,youwillcapturealltheconstraintsabouttheproblem.Weshouldnevertrytosolveaproblemthatisnotcompletelydefined.Interviewquestionsarenotlikeexampaperwhereallthedetailsaboutaproblem arewell defined. In the interview, the interviewer actually expects you to ask questions andclarifytheproblem.Forexample:Whentheproblemstatementsaysthatwriteanalgorithmtosortnumbers.Thefirstinformationyouneedtocaptureiswhatkingofdata.LetussupposeinterviewerrespondwiththeanswerInteger.Thesecondinformationthatyouneedtoknowwhatisthesizeofdata.Youralgorithmdiffersiftheinputdatasizeif100integersor1billionintegers.BasicguidelinefortheConstraintsforanarrayofnumbers:1.Howmanynumbersofelementsinthearray?2.Whatistherangeofvalueineachelement?Whatistheminandmaxvalue?3.Whatisthekindofdataineachelementisitanintegerorafloatingpoint?4.Doesthearraycontainuniquedataornot?BasicguidelinefortheConstraintsforanarrayofstring:1.Howmanynumbersofelementsinthearray?2.Whatisthelengthofeachstring?Whatistheminandmaxlength?3.Doesthearraycontainuniquedataornot?BasicguidelinefortheConstraintsforaGraph1.Howmanynodesarethereinthegraph?2.Howmanyedgesarethereinthegraph?3.Isitaweightedgraph?Whatistherangeofweights?4.Isthegraphdirectedorundirected?5.Isthereisaloopinthegraph?6.Istherenegativesumloopinthegraph?7.Doesthegraphhaveself-loops?We have already seen this in graph chapter that depending upon the constraints the algorithm appliedchangesandsoisthecomplexityofthesolution.
IdeaGenerationWehavecoveredalotoftheoreticalknowledgeinthisbook.Itisimpossibletocoverallthequestionsasnewonesarecreatedeveryday.Therefore,weshouldknowhowtohandlenewproblems.Evenifyouknowthesolutionofaproblemaskedbytheinterviewerthenalsoyouneedtohaveadiscussionwiththeinterviewerandreachtothesolution.Youneedtoanalysetheproblemalsobecausetheinterviewermaymodifyaquestionalittlebitandtheapproachtosolveitwillvary.Well,howtosolveanewproblem?Thesolutiontothisproblemisthatyouneedtodoalotofpracticeandthemoreyoupracticethemoreyouwillbeabletosolveanynewquestion,whichcomeinfrontofyou.Whenyouhavesolvedenoughproblems,youwillbeabletoseeapatterninthequestionsandabletosolvenewproblemseasily.Followingisthestrategythatyouneedtofollowtosolveanunknownproblem:1.Trytosimplifythetaskinhand.2.Tryafewexamples3.Thinkofasuitabledata-structure.4.Thinkaboutsimilarproblemsyouhavealreadysolved.
TrytosimplifythetaskinhandLetuslookintothefollowingproblem:Husbandsandtheirwivesarestandinginrandominaline.TheyhavebeennumberedforhusbandsH1,H2,H3and soon.And their correspondingwiveshavenumberW1,W2,W3and soon.Youneed toarrangethemsothatH1willstandfirst,followedbyW1,thenH2followedbyW2andsoon.Atthefirstlook,itlooksdifficult,butitisasimpleproblem.Trytofindarelationofthefinalposition.P(Hi)=i*2-1P(Wi)=i*2TherestofthealgorithmweareleavingyoutodosomethinglikeInsertion-Sortandyouaredone.
TryafewexamplesInthesameaboveproblemifyouhavetrieditwithsomeexamplefor3husbandandwifepairthenyoumayhavereached to thesameformula thatwehaveshownin theprevioussection.Sometimethinkingsomemoreexamplestrytosolvetheproblemathand.
Thinkofasuitabledata-structureForsomeproblems,itisstraightforwardwhichdatastructureismostsuitable.Forexample,ifwehaveaproblemfindingmin/maxofsomegivenvalue,thenprobablyheapisthedatastructurewearelookingfor.Wehaveseenanumberofdatastructurethroughoutthisbook.Inaddition,wehavetofigureoutwhichdata-structurewillsuiteourneed.Letuslookintoaproblem:Wearegivenastreamofdataatanytimewecanbeaskedtotellthemedianvalueofthedataandmaybewecanbeaskedtopopmediandata.
Wecanthinkaboutsomesortoftree,maybebalancedtreewheretherootisthemedian.Waitbutitisnotsoeasytomakesurethetreeroottobeamedian.Aheapcangiveusminimumormaximumsowecannotgetthedesiredresultfromittoo.However,whatifweuse twoheaponemaxheap andonemin heap.The smaller valueswill go tomin heap and thebiggervalueswillgotomaxheap.Inaddition,wecankeepthecountofhowmanyelementsarethereintheheap.Therestofthealgorithmyoucanthinkyourself.For every new problem think about the data structure, you know and may be one of them or somecombinationofthemwillsolveyourproblem.ThinkaboutsimilarproblemsyouhavealreadysolvedLetussupposeyouaregiventwolinkedlistheadpointerandtheymeetatsomepointandneedtofindthepointof intersection.However, inplaceoftheendofboththelinkedlist tobeanullpointer thereisaloop.Youknowhowtofindintersectionpointoftwointersectinglinkedlist,youknowhowtofindifalinkedlist have a loop (three-pointer solution). Therefore, you can apply both of these solutions to find thesolutionoftheprobleminhand.
ComplexitiesSolving a problem is not just finding a correct solution. The solution should be fast and should havereasonablememory requirement.Youhavealready readaboutBig-Onotation in thepreviouschapters.You shouldbe able todoBig-Oanalysis. In caseyou think the solutionyouhaveprovided is not thatoptimalandthereissomemoreefficientsolution,thenthinkagainandtrytofigureoutthisinformation.MostinterviewersexpectthatyoushouldbeabletofindthetimeandSpaceComplexityofthealgorithms.YoushouldbeabletocomputethetimeandSpaceComplexityinstantly.Wheneveryouaresolvingsomeproblem,youshouldfindthecomplexityassociatedwithitfromthisyouwouldbeabletochoosethebestsolutions.Insomeproblemsthereissometrade-offsbetweenspaceandTimeComplexity,soyoushouldknowthesetrade-offs.Sometimetakingsomebitmorespacesavesalotoftimeandmakeyouralgorithmmuchfaster.
CodingAt this point, you have already captured all the constraints of the problem, proposed few solutions,evaluatedthecomplexitiesofthevarioussolutionsandpickedtheonesolutiontodofinalcoding.Neverever,jumpintocodingbeforediscussingconstraints,Ideagenerationandcomplexitywiththeinterviewer.WeareaccustomedtocodinginanIDElikevisualstudio.Somanypeoplestrugglewhenaskedtowritecodeonawhiteboardorsomeblanksheet.Therefore,weshouldhavealittlepracticetothecodingonasheetofpaper.Youshouldthinkbeforecodingbecausethereisnobackbuttoninsheetofpaper.Alwaystrytowritemodularcode.Smallfunctionsneedtobecreatedsothatthecodeiscleanandmanaged.Ifthere is a swap function so just use this function and tell the interviewer that youwill write it later.Everybodyknowsthatyoucanwriteswapcode.
TestingOncethecodeiswritten,youarenotdone.Itismostimportantthatyoutestyourcodewithseveralsmalltest cases. It shows that you understand the importance of testing. It also gives confidence to yourinterviewerthatyouarenotgoingtowriteabuggycode.Onceyouaredonewith,yourcodingitisagoodpracticethatyougothroughyourcodelinebylinewithsomesmalltestcase.Thisisjusttomakesureyourcodeisworkingasitissupposedtowork.Youshouldtestfewtestcases.Normaltestcases:Thesearethepositivetestcases,whichcontainthemostcommonscenario,andfocusisontheworkingofthebaselogicofthecode.Forexample,ifwe`aregoingtowritesomealgorithmforlinkedlist,thenthismaycontainwhatwillhappenwhenalinkedlistwith3or4nodesisgivenasinput.Thesetestcasesyoushouldalwaysruninyourheadbeforesayingthecodeisdone.Edgecases:Theseare the testcases,whicharegoing to test theboundariesof thecode.For thesamelinkedlistalgorithm,edgecasesmaybehowthecodebehaveswhenanemptylistispassedorjustonenodeispassed.Thesetestcasesyoushouldalwaysruninyourheadbeforesayingthecodeisdone.Edgecasesmayhelptomakeyourcodemorerobust.Justfewchecksneedtobeaddedtothecodetotakecareofthecondition.Loadtesting:Inthiskindoftest,yourcodewillbetestedwithahugedata.Thiswillallowustotestifyourcodeisslowortoomuchmemoryintensive.Alwaysfollowthesefivestepsnever jumptocodingbeforedoingconstraintanalysis, ideageneration,andComplexityAnalysis:.Atleastnever,missthetestingphase.
ExampleLetussupposetheintervieweraskyoutogiveabestsortingalgorithm.Some interviewee will directly jump to Quick-SortO(nlogn). Oops, mistake you need to ask manyquestionsbeforebeginningtosolvethisproblem.Questions1:Whatisthetypeofdata?Aretheyintegers?Answer:Yes,theyareintegers.Questions2:Howmuchdataarewegoingtosort?Answer:Maybethousands.Questions3:Whatexactlyisthisdataabout?Answer:Theystoreaperson’sageQuestions4:Whatkindofdata-structureusedtoholdthisdata?Answer:DataaregivenintheformofsomearrayQuestions5:Canwemodifythegivendata-structure?Inaddition,many,manymore…?Answer:No,youcannotmodifythedatastructureprovidedOkfromthefirstanswer,wewilldeducethatthedataisinteger.Thedataisnotsobigitjustcontainsafewthousandentries.Thethirdanswerisinterestingfromthiswededucethattherangeofdatais1-150.Dataisprovidedinanarray.Fromfifthsanswerwededucethatwehavetocreateourowndatastructureandwecannotmodifythearrayprovided.Sofinally,weconclude,wecanjustusebucketsorttosortthedata.Therangeisjust1-150soweneedjust151-capacityintegralarray.DataisunderthousandssowedonothavetoworryaboutdataoverflowandwegetthesolutioninlineartimeO(N).Note:Wewillreadsortinginthecomingchapters.
SummaryAtthispoint,youknowtheprocessofhandlingnewproblemsverywell.Inthecomingchapterwewillbelookingintoalotofvariousdatastructureandtheproblemstheysolve.Ahugenumberofproblemsaresolved in thisbook.However, is recommendedso first try tosolve thembyyourself, then lookfor thesolution.Alwaysthinkaboutthecomplexityoftheproblem.Intheinterviewinteractionisthekeytogetproblemdescribedcompletelyanddiscussyourapproachwiththeinterviewer.
CHAPTER4:ABSTRACTDATATYPE
Abstractdatatype(ADT)Anabstractdatatype(ADT)isalogicaldescriptionofhowweviewthedataandtheoperationsthatareallowedonit.ADTisdefinedasauserpointofviewofadatatype.ADTconcernsaboutthepossiblevaluesofthedataandwhatareinterfaceexposedbyit.ADTdoesnotconcernabouttheactualimplementationofthedatastructure.Forexample,auserwantstostoresomeintegersandfindameanofit.Doesnottalkabouthowexactlyitwillbeimplemented.
Data-StructureDatastructuresareconcreterepresentationsofdataandaredefinedasaprogrammerpointofview.Data-structurerepresentshowdatawillbestoredinmemory.Alldata-structureshavetheirownprosandcons.Dependingupontheproblemathand,wepickadata-structurethatisbestsuitedforit.Forexample,wecanstoredatainanarray,alinked-list,stack,queue,tree,etc.
Array
Array represent a collection of multiple elements of the same datatype. Arrays are fixed size datastructure,thesizeofthisdatastructuremustbeknownatcompiletimeandcannotbechangedafterthat.Arraysarethemostcommondatastructureusedtostoredata.Aswecannotchangethesizeofanarray,wegenerallydeclarelargesizearraytohandleanyfuturedataexpansion.Thisendsupincreatinglargesizearray,wheremostofthespaceisunused.
ArrayADTOperationsBelowistheAPIofarray:Insert(x,k):addsanelementatkthpositionIfwewant to just store value k at index x, andwedo not care about the value already stored in thatlocationthanthisoperationisdoneinO(1)constanttime.However, if we care about the previous values stored in an array, so insertions and deletions of anelementistimeconsumingaswehavetoshiftotherelementsbyonepositionrespectively.Inthiscase,insertionanddeletiontakeO(n)time.Delete(k):deleteelementatkthpositionIfwewanttomarkthatthereisnovaluestoredatindexk,thisoperationisdoneinO(1)constanttime.However, if we care about the other values stored in an array, so deletions of an element are timeconsumingaswehave to shift other elementsbyoneposition respectively. In this case,deletion takesO(n)time.FindKth(k):findelementatpositionkThebiggestadvantageofthearrayisthatifweknowindexofanelementthenitcanbeaccessedinO(1)TimeComplexity.WejustneedtoreturnA[k].Find(x):findpositionofelementIfthearrayisunsorted,thenfind(x)anelementwilltakeO(n)TimeComplexity.Ifthearrayissorted,thenfind(x)isfastusingbinarysearchandwilltakeO(logn)TimeComplexity.PrintList():displayalltheelementsinthelistPrintListJustrunthroughthearrayandprintoneelementatatime.Usesoneloop.LineartimeO(n).IsEmpty():checkifnumberofelementsarezeroSearchingifthereisnoelementstoredinarrayalsotakelineartimeO(n).
LinkedList
Linkedlistsaredynamicdatastructureandmemoryisallocatedatruntime.Theconceptoflinkedlistisnottostoredatacontiguously.Uselinksthatpointtothenextelements.Performance wise linked lists are slower than arrays because there is no direct access to linked listelements.Thelinkedlistisausefuldatastructurewhenwedonotknowthenumberofelementstobestoredaheadoftime.Therearemanyflavoursoflinkedlist:linear,circular,doubly,anddoublycircular.
LinkedListADTOperationsBelowistheAPIofLinkedlist.Insert(k):addsktothestartofthelistInsertanelementat thestartof the list. Justcreateanewelementandmovepointers.So that thisnewelementbecomesthenewelementofthelist.ThisoperationwilltakeO(1)constanttime.Delete():deleteelementatthestartofthelistDeleteanelementatthestartofthelist.Wejustneedtomoveonepointer.ThisoperationwillalsotakeO(1)constanttime.PrintList():displayalltheelementsofthelist.Startwiththefirstelementandthenfollowthepointers.ThisoperationwilltakeO(N)time.Find(k):findthepositionofelementwithvaluekStartwiththefirstelementandfollowthepointeruntilwegetthevaluewearelookingfororreachtheendofthelist.ThisoperationwilltakeO(N)time.Note:binarysearchdoesnotworkonlinkedlists.FindKth(k):findelementatpositionkStartfromthefirstelementandfollowthelinksuntilyoureachthekthelement.ThisoperationwilltakeO(N)time.IsEmpty():checkifthenumberofelementsinthelistarezero.JustchecktheheadpointerofthelistitshouldbeNull.Whichmeanslistisempty.ThisoperationwilltakeO(1)time.
Stack
StackisaspecialkindofdatastructurethatfollowsLast-In-First-Out(LIFO)strategy.Thismeansthattheelementthatisaddedtostacklastwillbethefirsttoberemoved.Thevariousapplicationsofstackare:1.Recursion:recursivecallsarepreformedusingsystemstack.2.Postfixevaluationofexpression.3.Backtracking4.Depth-firstsearchoftreesandgraphs.5.Convertingadecimalnumberintoabinarynumberetc.
StackADTOperationsPush(k):AddsanewitemtothetopofthestackPop():Removeanelementfromthetopofthestackandreturnitsvalue.Top():ReturnsthevalueoftheelementatthetopofthestackSize():ReturnsthenumberofelementsinthestackIsEmpty():determineswhetherthestackisempty.Itreturns1ifthestackisemptyorreturn0.Note:AlltheaboveStackoperationsareimplementedinO(1)TimeComplexity.
Queue
AqueueisaFirst-In-First-Out(FIFO)kindofdatastructure.Theelementthatisaddedtothequeuefirstwillbethefirsttoberemovedfromthequeueandsoon.Queuehasthefollowingapplicationuses:1.Accesstosharedresources(e.g.,printer)2.Multiprogramming3.Messagequeue
QueueADTOperations:Enqueue():Addanewelementtothebackofthequeue.Dequeue():removeanelementfromthefrontofthequeueandreturnitsvalue.Front():returnthevalueoftheelementatthefrontofthequeue.Size():returnsthenumberofelementsinsidethequeue.IsEmpty():returns1ifthequeueisemptyotherwisereturn0Note:AlltheaboveQueueoperationsareimplementedinO(1)TimeComplexity.
TreesTreeisahierarchicaldatastructure.Thetopelementofa treeiscalledtherootof thetree.Except therootelement,everyelementinatreehasaparentelement,andzeroormorechildelements.Thetreeisthemostusefuldatastructurewhenyouhavehierarchicalinformationtostore.Therearemanytypesoftrees,forexample,binary-tree,Red-blacktree,AVLtree,etc.
BinaryTreeAbinarytreeisatypeoftreeinwhicheachnodehasatmosttwochildren(0,1or2)whicharereferredasleftchildandrightchild.
BinarySearchTrees(BST)
Abinarysearchtree(BST)isabinarytreeonwhichnodesareorderedinthefollowingway:1.Thekeyintheleftsubtreeislessthanthekeyinitsparentnode.2.Thekeyintherightsubtreeisgreaterorequalthekeyinitsparentnode.
BinarySearchTreeADTOperationsInsert(k):Insertanelementkintothetree.Delete(k):Deleteanelementkfromthetree.Search(k):Searchaparticularvaluekintothetreeifitispresentornot.FindMax():Findthemaximumvaluestoredinthetree.FindMin():Findtheminimumvaluestoredinthetree.TheaverageTimeComplexityofall theaboveoperationsonabinarysearchtree isO(logn), thecasewhenthetreeisbalanced.Theworst-caseTimeComplexitywillbeO(n)whenthetreeisskewed.Abinarytreeisskewedwhentreeisnotbalanced.Therearetwotypesofskewedtree.1.RightSkewedbinarytree:Abinarytreeinwhicheachnodeishavingeitheronlyarightchildorno
childatall.2.LeftSkewedbinarytree:Abinarytreeinwhicheachnodeishavingeitheronlyaleftchildornochild
atall.
BalancedBinarysearchtreeTherearefewbinarysearchtree,whichalwayskeepsthemselvesbalanced.MostimportantamongthemareRed-BlackTree(RB-Tree)andAVLtree.Thestandardtemplatelibrary(STL)isimplementedusingthisRed-BlackTree(RB-Tree).
PriorityQueue(Heap)
Priorityqueueisimplementedusingabinaryheapdatastructure.Inaheap,therecordsarestoredinanarraysothateachkeyislargerthanitstwochildrenkeys.Eachnodeintheheapfollowthesamerulethattheparentvalueisgreaterthanitschildren.Therearetwotypesoftheheapdatastructure:1.Maxheap:eachnodeshouldbegreaterthanorequaltoeachofitschildren.2.Minheap:eachnodeshouldbesmallerthanorequaltoeachofitschildren.Aheapisausefuldatastructurewhenyouwanttogetmax/minonebyonefromdata.Heap-Sortusesmaxheaptosortdatainincreasing/decreasingorder.
HeapADTOperationsInsert()-Addinganewelementtotheheap.TheTimeComplexityofthisoperationisO(log(n))Extract()-Extractingmaxformaxheapcase(orminforminheapcase).TheTimeComplexityofthisoperationisO(log(n))Heapify()–Toconvertalistofnumbersinanarrayintoaheap.ThisoperationhasaTimeComplexityO(n)Delete()–Deleteanelementfromtheheap.TheTimeComplexityofthisoperationisO(log(n))
Hash-Table
AHash-Table isadatastructure thatmapskeys tovalues.Eachpositionof theHash-Table iscalledaslot.TheHash-Tableusesahash function tocalculatean indexofanarrayof slots.Weuse theHash-Tablewhenthenumberofkeysactuallystoredissmallrelativelytothenumberofpossiblekeys.Theprocessofstoringobjectsusingahashfunctionisasfollows:1.CreateanarrayofsizeMtostoreobjects,thisarrayiscalledHash-Table.2.Findahashcodeofanobjectbypassingitthroughthehashfunction.3.TakemoduleofhashcodebythesizeofHash-Tabletogettheindexofthetablewhereobjectswillbestored.4.Finallystoretheseobjectsinthedesignatedindex.TheprocessofsearchingobjectsinHash-Tableusingahashfunctionisasfollows:1.Findahashcodeoftheobjectwearesearchingforbypassingitthroughthehashfunction.2.Takemoduleofhashcodeby thesizeofHash-Table toget the indexof the tablewhereobjectsarestored.3.Finally,retrievetheobjectfromthedesignatedindex.
Hash-TableAbstractDataType(ADT)ADTofHash-Tablecontainsthefollowingfunctions:Insert(x):Addobjectxtothedataset.Delete(x):Deleteobjectxfromthedataset.Search(x):Searchobjectxindataset.TheHash-Tableisausefuldatastructureforimplementingdictionary.TheaveragetimetosearchforanelementinaHash-TableisO(1).AHashTablegeneralizesthenotionofanarray.
Dictionary/SymbolTableAsymboltableisamappingbetweenastring(key)andavalue,whichcanbeofanydatatype.Avaluecanbeanintegersuchasoccurrencecount,dictionarymeaningofawordandsoon.
BinarySearchTree(BST)forStringsBinarySearchTree(BST)isthesimplestwaytoimplementsymboltable.Simplestringcomparefunctioncan be used to compare two strings. If all the keys are random, and the tree is balanced. Then on anaveragekeylookupcanbedoneinlogarithmictime.
Hash-TableTheHash-Table is another data structure,which can be used for symbol table implementation.BelowHash-Tablediagram,wecanseethenameofthatpersonistakenasthekey,andtheirmeaningisthevalueofthesearch.Thefirstkeyisconvertedintoahashcodebypassingittoappropriatehashfunction.InsidehashfunctionthesizeofHash-Tableisalsopassed,whichisusedtofindtheactualindexwherevalueswillbestored.Finally,thevaluethatismeaningofnameisstoredintheHash-Table,oryoucanstoreareferencetothestringwhichstoremeaningcanbestoredintotheHash-Table.
Hash-Tablehasanexcellentlookupofconstanttime.Let us supposewewant to implement autocomplete the box feature ofGoogle search.When you typesomestringtosearchingooglesearch,itproposesomecompletestringevenbeforeyouhavedonetyping.BSTcannotsolvethisproblemasrelatedstringscanbeinbothrightandleftsubtree.TheHash-Tableisalsonotsuitedfor this job.OnecannotperformapartialmatchorrangequeryonaHash-Table.Hash function transforms string to a number.Moreover, a good hash functionwill give adistributedhashcodeevenforpartialstringandthereisnowaytorelatetwostringsinaHash-Table.Trie and Ternary Search tree are a special kind of tree, which solves partialmatch, and range queryproblemwell.
TrieTrieisatree,inwhichwestoreonlyonecharacterateachnode.Thisfinalkeyvaluepairisstoredintheleaves.EachnodehasKchildren,oneforeachpossiblecharacter.Forsimplicitypurpose,letusconsiderthatthecharactersetis26,correspondstodifferentcharactersofEnglishalphabets.Trie is an efficient data structure. Using Trie, we can search the key in O(M) time.WhereM is themaximumstringlength.Trieisalsosuitableforsolvingpartialmatchandrangequeryproblems.
TernarySearchTrie/TernarySearchTreeTrieshavingaverygoodsearchperformanceofO(M)whereMisthemaximumsizeofthesearchstring.However,trieshavingveryhighspacerequirement.EverynodeTriecontainreferencestomultiplenodes,each reference corresponds to possible characters of the key. To avoid this high space requirementTernarySearchTrie(TST)isused.
A TST avoid the heavy space requirement of the traditional Trie while still keeping many of itsadvantages. In aTST, eachnode contains a character, an endofkey indicator, and threepointers.Thethree pointers are corresponding to current char hold by the node(equal), characters less than andcharactergreaterthan.TheTimeComplexityofternarysearchtreeoperationisproportionaltotheheightoftheternarysearchtree.Intheworstcase,weneedtotraverseupto3timesthatmanylinks.However,thiscaseisrare.Therefore,TSTisaverygoodsolutionforimplementingSymbolTable,Partialmatchandrangequery.
Graphs
Agraphisadatastructurewhichrepresentsanetwork,thatconnectsacollectionofnodescalledvertices,andthereconnections,callededges.Anedgecanbeseenasapathbetweentwonodes.Theseedgescanbeeitherdirectedorundirected.Ifapathisdirectedthenyoucanmoveonlyinonedirection,whileinanundirectedpathyoucanmoveinboththedirections.
GraphAlgorithms
Depth-FirstSearch(DFS)TheDFSalgorithmwestartfromstartingpointandgointodepthofgraphuntilwereachadeadendandthenmoveup toparentnode(Backtrack). InDFS,weusestack toget thenextvertex tostartasearch.Alternatively,wecanuserecursion(systemstack)todothesame.
Breadth-FirstSearch(BFS)InBFS algorithm, a graph is traversed in layer-by-layer fashion. The graph is traversed closer to thestartingpoint.ThequeueisusedtoimplementBFS.
SortingAlgorithms
Sortingistheprocessofplacingelementsfromacollectionintoascendingordescendingorder.Sortingarrangesdataelementsinordersothatsearchingbecomeeasier.TherearegoodsortingfunctionsavailablewhichdoessortinginO(nlogn)time,sointhisbookwhenweneedsortingwewillusesort()functionandwillassumethatthesortingisdoneinO(nlogn)time.
CountingSortCountingsortisthesimplestandmostefficienttypeofsorting.Countingsorthasastrictrequirementofapredefinedrangeofdata.Like,sorthowmanypeopleareinwhichagegroup.Weknowthattheageofpeoplecanvarybetween1and130.
Ifweknowtherangeofinput,thensortingcanbedoneusingcountinginO(n+k).
EndnoteThis chapter have provided a brief introduction of the various data structures, algorithms and theircomplexities.Inthecomingchapterswewilllookintoallthesedatastructureindetails.Ifyouknowtheinterface of the various data structures, then you can use them while solving other problems withoutknowingtheinternaldetailshowtheyareimplemented.
CHAPTER5:SEARCHING
IntroductionInComputerScience,Searchingisthealgorithmicprocessoffindingaparticulariteminacollectionofitems.Theitemmaybeakeywordinafile,arecordinadatabase,anodeinatreeoravalueinanarrayetc.
WhySearching?Imagineyouareinalibrarywithmillionsofbooks.Youwanttogetaspecificbookwithspecifictitle.Howwill you find?Youwill just start searching by initial letter of the book title. Then you continuematchingwithawholebooktitleuntilyoufindyourdesiredbook.(Bydoingthissmallheuristicyouhavereducedthesearchspacebyafactorof26,considerwehaveanequalnumberofbookswhosetitlebeginwithparticularchar.)Similarly,computerstores lotsof informationand to retrieve this informationefficiently,weneedveryefficientsearchingalgorithms.Tomakesearchingefficient,wekeepthedatainsomeproperorder.Therearecertainwaysoforganizingthedata.Ifyoukeepthedatainproperorder,itiseasytosearchrequiredelement.Forexample,Sortingisoneoftheprocessformakingdataorganized.
DifferentSearchingAlgorithms·LinearSearch–UnsortedInput·LinearSearch–SortedInput·BinarySearch(SortedInput)·StringSearch:Tries,SuffixTrees,TernarySearch.·HashingandSymbolTables
LinearSearch–UnsortedInputWhenelementsofanarrayarenotorderedorsortedandwewant tosearchforaparticularvalue,weneedtoscanthefullarrayunlesswefindthedesiredvalue.Thiskindofalgorithmknownasunorderedlinear search.Themajor problemwith this algorithm is less performance or highTimeComplexity inworstcase.Example5.1intlinearSearchUnsorted(intarr[],intsize,intvalue)
inti=0;for(i=0;i<size;i++)if(value==arr[i])returni;return-1;
TimeComplexity:O(n).Asweneedtotraversethecompletearrayinworstcase.Worstcaseiswhenyourdesiredelementisatthelastpositionofthearray.Here,‘n’isthesizeofthearray.SpaceComplexity:O(1).Noextramemoryisusedtoallocatethearray.
LinearSearch–SortedIf elements of the array are sorted either in increasing order or in decreasing order, searching for adesiredelementwillbemuchmoreefficientthanunorderedlinearsearch.Inmanycases,wedonotneedtotraversethecompletearray.Followingexampleexplainswhenyouencounteragreaterelementfromthe increasing sorted array, you stop searching further. This is how this algorithm saves the time andimprovestheperformance.Example5.2intlinearSearchSorted(intarr[],intsize,intvalue)
inti=0;for(i=0;i<size;i++)if(value==arr[i])returni;elseif(value<arr[i])return-1;return-1;
TimeComplexity:O(n).Asweneedtotraversethecompletearrayinworstcase.Worstcaseiswhenyour desired element is at the last position of the sorted array. However, in the average case thisalgorithmismoreefficienteventhoughthegrowthrateissameasunsorted.SpaceComplexity:O(1).Noextramemoryisusedtoallocatethearray.
BinarySearchHowdowesearchawordinadictionary?Ingeneral,wegotosomeapproximatepage(mostlymiddle)andstartsearchingfromthatpoint. Ifweseethewordthatwearesearchingissamethenwearedonewiththesearch.Else,ifweseethepageisbeforetheselectedpages,thenapplythesameprocedureforthefirsthalfotherwisetothesecondhalf.BinarySearchalsoworksinthesameway.Wegetthemiddlepointfromthesortedarrayandstartcomparingwiththedesiredvalue.Note:Binarysearchrequiresthearraytobesortedotherwisebinarysearchcannotbeapplied.Example5.3/*BinarySearchAlgorithm–IterativeWay*/intBinarysearch(intarr[],intsize,intvalue)
intlow=0;inthigh=size-1;intmid;while(low<=high)mid=low+(high-low)/2;/*Toavoidtheoverflow*/if(arr[mid]==value)returnmid;elseif(arr[mid]<value)low=mid+1;elsehigh=mid-1;return-1;
TimeComplexity:O(logn).Wealwaystakehalfinputandthrowingouttheotherhalf.SotherecurrencerelationforbinarysearchisT(n)=T(n/2)+c.Usingmastertheorem(divideandconquer),wegetT(n)=O(logn)SpaceComplexity:O(1)Example5.4/*BinarySearchAlgorithm–RecursiveWay*/intBinarySearchRecursive(intarr[],intlow,inthigh,intvalue)
if(low>high)return-1;intmid=low+(high-low)/2;/*Toavoidtheoverflow*/if(arr[mid]==value)returnmid;elseif(arr[mid]<value)
returnBinarySearchRecursive(arr,mid+1,high,value);elsereturnBinarySearchRecursive(arr,low,mid-1,value);
TimeComplexity:O(logn).SpaceComplexity:O(logn)forsystemstackinrecursion
StringSearchingAlgorithmsReferStringchapter.
HashingandSymbolTablesReferHash-Tablechapter.
HowsortingisusefulinSelectionAlgorithm?Selectionproblemscanbeconvertedtosortingproblems.Oncethearrayissorted,itiseasytofindtheminimum/maximum(ordesiredelement)fromthesortedarray.Themethod‘SortingandthenSelecting’isinefficientforselectingasingleelement,butitisefficientwhenmanyselectionsneedtobemadefromthearray. It is because only one initial expensive sort is needed, followed by many cheap selectionoperations.Forexample,ifwewanttogetthemaximumelementfromanarray.Aftersortingthearray,wecansimplyreturnthelastelementfromthearray.Whatifwewanttogetsecondmaximum.Now,wedonothavetosort thearrayagainandwecanreturn thesecondlastelementfromthesortedarray.Similarly,wecanreturnthekthmaximumelementbyjustonescanofthesortedlist.Sowiththeabovediscussion,sortingisusedtoimprovetheperformance.IngeneralthismethodrequiresO(nlogn)(forsorting)time.Withtheinitialsorting,wecanansweranyqueryinonescan,O(n).
ProblemsinSearching
PrintDuplicatesinArrayGivenanarrayofnnumbers,printtheduplicateelementsinthearray.Firstapproach:Exhaustive searchorBrute force, for eachelement in array find if there is someotherelementwiththesamevalue.Thisisdoneusingtwoforloop,firstlooptoselecttheelementandsecondlooptofinditsduplicateentry.Example5.5voidprintRepeating(intarr[],intsize)
inti,j;printf("Repeatingelementsare");for(i=0;i<size;i++)for(j=i+1;j<size;j++)if(arr[i]==arr[j])printf("%d",arr[i]);
TheTimeComplexityisO(n2)andSpaceComplexityisO(1)Secondapproach:Sorting,Sortalltheelementsinthearrayandafterthisinasinglescan,wecanfindtheduplicates.Example5.6voidprintRepeating(intarr[],intsize)
inti;Sort(arr,size);printf("Repeatingelementsare");for(i=1;i<size;i++)if(arr[i]==arr[i-1])printf("%d",arr[i]);
SortingalgorithmstakeO(n.log(n))timeandsinglescantakeO(n)time.TheTimeComplexityofanalgorithmisO(n.log(n)) andSpaceComplexityisO(1)Thirdapproach:Hash-Table,usingHash-Table,wecankeeptrackoftheelementswehavealreadyseenandwecanfindtheduplicatesinjustonescan.Example5.7
voidprintRepeating(intarr[],intsize)
HashTableh;inti;printf("Repeatingelementsare");for(i=0;i<size;i++)if(findValue(h,arr[i]))printf("%d",arr[i]);elseaddValue(h,arr[i]);
Hash-TableinsertandfindtakeconstanttimeO(1)sothetotalTimeComplexityofthealgorithmisO(n)time.SpaceComplexityisalsoO(n)Forthapproach:Counting,thisapproachisonlypossibleifweknowtherangeoftheinput.Ifweknowthat,theelementsinthearrayareintherange0ton-1.Wecanreserveandarrayoflengthnandwhenweseeanelementwecanincreaseitscount.Injustonesinglescan,weknowtheduplicates.Ifweknowtherangeoftheelements,thenthisisthefastestwaytofindtheduplicates.Example5.8voidprintRepeating(intarr[],intsize)
int*count=(int*)calloc(sizeof(int),size);inti;printf("Repeatingelementsare");for(i=0;i<size;i++)if(count[arr[i]]==1)printf("%d",arr[i]);elsecount[arr[i]]++;
Counting approach just uses an array so insert and find take constant time O(1) so the total TimeComplexityofthealgorithmisO(n)time.SpaceComplexityforcreatingcountarrayisalsoO(n)
Findmax,appearingelementinanarrayGivenanarrayofnnumbers,findtheelement,whichappearsmaximumnumberoftimes.Firstapproach:Exhaustive search orBrute force, for each element in array find howmany times this
particularvalueappears inarray.Keep trackof themaxCountandwhensomeelementcount isgreaterthanmaxCountthenupdatethemaxCount.Thisisdoneusingtwoforloop,firstlooptoselecttheelementandsecondlooptocounttheoccurrenceofthatelement.TheTimeComplexityisO(n2),andSpaceComplexityisO(1)Example5.9intgetMax(intarr[],intsize)
inti,j;intmax=arr[0],count=1,maxCount=1;
for(i=0;i<size;i++)count=1;for(j=i+1;j<size;j++)if(arr[i]==arr[j])count++;if(count>maxCount)max=arr[i];maxCount=count;returnmax;
Secondapproach:Sorting,Sortalltheelementsinthearrayandafterthisinasinglescan,wecanfindthecounts.SortingalgorithmstakeO(n.log(n))timeandsinglescantakeO(n)time.TheTimeComplexityofanalgorithmisO(n.log(n)) andSpaceComplexityisO(1)Example5.10intgetMax(intarr[],intsize)
intmax=arr[0],maxCount=1,curr=arr[0],currCount=1;inti;Sort(arr,size);for(i=1;i<size;i++)if(arr[i]==arr[i-1])currCount++;elsecurrCount=1;curr=arr[i];
if(currCount>maxCount)maxCount=currCount;max=curr;returnmax;
Thirdapproach:Counting,Thisapproachisonlypossibleifweknowtherangeoftheinput.Ifweknowthat,theelementsinthearrayareintherange0ton-1.Wecanreserveandarrayoflengthnandwhenweseeanelementwecanincreaseitscount.Injustonesinglescan,weknowtheduplicates.Ifweknowtherangeoftheelements,thenthisisthefastestwaytofindthemaxcount.Counting approach just uses array so to increase count take constant time O(1) so the total TimeComplexityofthealgorithmisO(n)time.SpaceComplexityforcreatingcountarrayisalsoO(n)Example5.11#defineN100intgetMax(intarr[],intsize)
intmax=arr[0],maxCount=1;int*count=(int*)calloc(sizeof(int),N);inti;
for(i=0;i<size;i++)count[arr[i]]++;if(count[arr[i]]>maxCount)maxCount=count[arr[i]];max=arr[i];returnmax;
Majoritye lementinanArrayGivenanarrayofnelements.Findthemajorityelement,whichappearsmorethann/2times.Return0incasethereisnomajorityelement.Firstapproach:Exhaustive search orBrute force, for each element in array find howmany times thisparticularvalueappears inarray.Keep trackof themaxCountandwhensomeelementcount isgreaterthanmaxCountthenupdatethemaxCount.Thisisdoneusingtwoforloop,firstlooptoselecttheelement
andsecondlooptocounttheoccurrenceofthatelement.Oncewehave the final,maxCountwe can see if it is greater than n/2, if it is greater thanwehave amajorityifnotwedonothaveanymajority.TheTimeComplexityisO(n2)+O(1)=O(n2)andSpaceComplexityisO(1)Example5.12intgetMajority(intarr[],intsize)
inti,j;intmax=0,count=0,maxCount=0;for(i=0;i<size;i++)for(j=i+1;j<size;j++)if(arr[i]==arr[j])count++;if(count>maxCount)max=arr[i];maxCount=count;if(maxCount>size/2)returnmax;elsereturn0;
Secondapproach:Sorting,Sortalltheelementsinthearray.Ifthereisamajoritythanthemiddleelementattheindexn/2mustbethemajoritynumber.Sojustsinglescancanbeusedtofinditscountandseeifthemajorityisthereornot.SortingalgorithmstakeO(n.logn)timeandsinglescantakeO(n)time.TheTimeComplexityofanalgorithmisO(n.logn) andSpaceComplexityisO(1)Example5.13intgetMajority(intarr[],intsize)
intmajIndex=size/2,count=1;inti;intcandidate;Sort(arr,size);candidate=arr[majIndex];count=0;for(i=0;i<size;i++)
if(arr[i]==candidate)count++;if(count>size/2)returnarr[majIndex];elsereturn0;
Thirdapproach: This is a cancelation approach (Moore’sVotingAlgorithm), if all the elements standagainstthemajorityandeachelementiscancelledwithoneelementofmajorityifthereismajoritythenmajorityprevails.·Setthefirstelementofthearrayasmajoritycandidateandinitializethecounttobe1.·Startscanningthearray.
oIfwegetsomeelementwhosevaluesameasamajoritycandidate,thenweincreasethecount.oIfwegetanelementwhosevalueisdifferentfromthemajoritycandidate,thenwedecrementthe
count.oIfcountbecome0,thatmeanswehaveanewmajoritycandidate.Makethecurrentcandidateas
majoritycandidateandresetcountto1.oAttheend,wewillhavetheonlyprobablemajoritycandidate.
·Nowscanthroughthearrayonceagaintoseeifthatcandidatewefoundabovehaveappearedmorethann/2times.
Countingapproachjustscansthrowarraytwotimes.TheTimeComplexityofthealgorithmisO(n)time.SpaceComplexityforcreatingcountarrayisalsoO(1)Example5.14intgetMajority(intarr[],intsize)
intmajIndex=0,count=1;inti;intcandidate;for(i=1;i<size;i++)if(arr[majIndex]==arr[i])count++;elsecount--;if(count==0)majIndex=i;count=1;candidate=arr[majIndex];count=0;
for(i=0;i<size;i++)if(arr[i]==candidate)count++;if(count>size/2)returnarr[majIndex];elsereturn0;
FindthemissingnumberinanArrayGivenanarrayofn-1elements,whichareintherangeof1ton.Therearenoduplicatesinthearray.Oneoftheintegerismissing.Findthemissingelement.Firstapproach:ExhaustivesearchorBruteforce,foreachvalueintherange1ton,findifthereissomeelementinarraywhichhavethesamevalue.Thisisdoneusingtwoforloop,firstlooptoselectvalueintherange1tonandthesecondlooptofindifthiselementisinthearrayornot.TheTimeComplexityisO(n2)andSpaceComplexityisO(1)Example5.15intfindMissingNumber(intarr[],intsize)
inti,j,found=0;for(i=1;i<=size;i++)found=0;for(j=0;j<size;j++)if(arr[j]==i)found=1;break;if(found==0)returni;
Secondapproach:Sorting,Sortalltheelementsinthearrayandafterthisinasinglescan,wecanfindtheduplicates.SortingalgorithmstakeO(n.logn)timeandsinglescantakeO(n)time.TheTimeComplexityofanalgorithmisO(n.logn)andSpaceComplexityisO(1)
Thirdapproach:Hash-Table,usingHash-Table,wecankeeptrackoftheelementswehavealreadyseenandwecanfindthemissingelementinjustonescan.Hash-TableinsertandfindtakeconstanttimeO(1)sothetotalTimeComplexityofthealgorithmisO(n)time.SpaceComplexityisalsoO(n)Forthapproach:Counting,weknowtherangeoftheinputsocountingwillwork.Asweknowthat, theelementsinthearrayareintherange0ton-1.Wecanreserveandarrayoflengthnandwhenweseeanelementwecanincreaseitscount.Injustonesinglescan,weknowthemissingelement.Counting approach just uses an array so insert and find take constant time O(1) so the total TimeComplexityofthealgorithmisO(n)time.SpaceComplexityforcreatingcountarrayisalsoO(n)Fifthapproach:Youareallowedtomodifythegiveninputarray.Modifythegiveninputarrayinsuchawaythatinthenextscanyoucanfindthemissingelement.Whenyouscanthroughthearray.Whenatindex“index”,thevaluestoredinthearraywillbearr[index]soadd thenumber“n+1” toarr[arr[ index]].Alwaysread thevaluefromthearrayusingareminderoperator“%”.Whenyouscanthearrayforthefirsttimeandmodifiedallthevalues,thenonesinglescanyou can see if there is some value in the arraywhich is smaller than “n+1” that index is themissingnumber.Inthisapproach,thearrayisscannedtwotimesandtheTimeComplexityofthisalgorithmisO(n).SpaceComplexityisO(1)Sixthapproach:Summationformulatofindthesumofnnumbersfrom1ton.Subtractthevaluesstoredinthearrayandyouwillhaveyourmissingnumber.TheTimeComplexityofthisalgorithmisO(n).SpaceComplexityisO(1)Seventhapproach:XORapproachtofindthesumofnnumbersfrom1ton.XORthevaluesstoredinthearrayandyouwillhaveyourmissingnumber.TheTimeComplexityofthisalgorithmisO(n).SpaceComplexityisO(1).Example5.16intfindMissingNumber(intarr[],intsize,intrange)
inti;intxorSum=0;//gettheXORofallthenumbersfrom1torangefor(i=1;i<=range;i++)xorSum^=i;
//loopthroughthearrayandgettheXORofelementsfor(i=0;i<size;i++)
xorSum^=arr[i];returnxorSum;
Note: Same problem can be asked in many forms (sometimes you have to do the xor of the rangesometimeyoudonot):1.Therearenumbers in the rangeof1-noutofwhichall appears single timebutone that appear two
times.2.All theelements in the range1-nareappearing16 timesandoneelementappear17 times.Find the
elementthatappears17times.
FindPairinanArrayGivenanarrayofnnumbers,findtwoelementssuchthattheirsumisequalto“value”Firstapproach:Exhaustive searchorBrute force, for eachelement in array find if there is someotherelement, which sum up to the desired value. This is done using two for loop, first loop to select theelementandsecondlooptofindanotherelement.TheTimeComplexityisO(n2) andSpaceComplexityisO(1)Example5.17intFindPair(int*arr,intsize,intvalue)
inti,j;for(i=0;i<size;i++)for(j=i+1;j<size;j++)if((arr[i]+arr[j])==value)printf("Thepairis%d,%d",arr[i],arr[j]);return1;return0;
Secondapproach:Sorting,Stepsareasfollows:1.Sortalltheelementsinthearray.2.Taketwovariablefirstandsecond.Variablefirst=0andsecond=size-13.Computesum=arr[first]+arr[second]4.Ifthesumisequaltothedesiredvalue,thenwehavethesolution5.Ifthesumislessthanthedesiredvalue,thenwewillincreasefirst6.Ifthesumisgreaterthanthedesiredvalue,thenwewilldecreasethesecond7.Werepeattheaboveprocesstillwegetthedesiredpairorwegetfirst>=second(don’thaveapair)SortingalgorithmstakeO(n.logn)timeandsinglescantakeO(n)time.
TheTimeComplexityofanalgorithmisO(n.logn)andSpaceComplexityisO(1)Example5.18intFindPair(int*arr,intsize,intvalue)
intfirst=0,second=size-1;intcurr;Sort(arr,size);
while(first<second)curr=arr[first]+arr[second];if(curr==value)printf("Thepairis%d,%d",arr[first],arr[second]);return1;elseif(curr<value)first++;elsesecond--;return0;
Thirdapproach:Hash-Table,usingHash-Table,wecankeeptrackoftheelementswehavealreadyseenandwecanfindthepairinjustonescan.1.Foreachelement,insertthevalueinHashtable.Letsaycurrentvalueisarr[index]2.Ifthevalue-arr[index]isalreadyinaHashtable.3.Ifvalue-arr[index]isintheHashtablethenwehavethedesiredpair.4.Else,proceedtothenextentryinthearray.Hash-TableinsertandfindtakeconstanttimeO(1)sothetotalTimeComplexityofthealgorithmisO(n)time.SpaceComplexityisalsoO(n)Example5.19intFindPair(int*arr,intsize,intvalue)
HashTableh;inti;for(i=0;i<size;i++)if(findValue(h,value-arr[i]))printf("Thepairis%d,%d",arr[i],value-arr[i]);
return1;addValue(h,arr[i]);return0;
Forthapproach:Counting,Thisapproachisonlypossibleifweknowtherangeoftheinput.Ifweknowthat,theelementsinthearrayareintherange0ton-1.Wecanreserveandarrayoflengthnandwhenweseeanelementwecanincreaseitscount.InplaceoftheHashtableintheaboveapproach,wewillusethisarrayandwillfindoutthepair.Counting approach just uses an array so insert and find take constant time O(1) so the total TimeComplexityofthealgorithmisO(n)time.SpaceComplexityforcreatingcountarrayisalsoO(n)
FindthePairintwoArraysGiventwoarrayXandY.Findapairofelements(xi,yi)suchthatxi∈Xandyi∈Ywherexi+yi=value.Firstapproach:ExhaustivesearchorBrute force, loop throughelementxiofXandsee ifyoucanfind(value–xi)inY.Twoforloop.TheTimeComplexityisO(n2)andSpaceComplexityisO(1)Secondapproach:Sorting,SortalltheelementsinthesecondarrayY.ForeachelementifXyoucanseeifthatelementisthereinYbyusingbinarysearch.SortingalgorithmstakeO(m.logm)andsearchingwilltakeO(n.logm) time.TheTimeComplexityofanalgorithmisO(n.logm)orO(m.logm)andSpaceComplexityisO(1)Thirdapproach:Sorting,Stepsareasfollows:1.SorttheelementsofbothXandYinincreasingorder.2.TakethesumofthesmallestelementofXandthelargestelementofY.3.Ifthesumisequaltovalue,wegotourpair.4.Ifthesumissmallerthanvalue,takenextelementofX5.Ifthesumisgreaterthanvalue,takethepreviouselementofYSortingalgorithmstakeO(n.logn)+O(m.logm)forsortingandsearchingwilltakeO(n+m) time.TheTimeComplexityofanalgorithmisO(n.logn)SpaceComplexityisO(1)Forthapproach:Hash-Table,Stepsareasfollows:1.ScanthroughalltheelementsinthearrayYandinsertthemintoHashtable.2.NowscanthroughalltheelementsofarrayX,letussupposethecurrentelementisxiseeifyoucan
find(value-xi)intheHashtable.3.Ifyoufindthevalue,yougotyourpair.4.Ifnot,thengotothenextvalueinthearrayX.
Hash-TableinsertandfindtakeconstanttimeO(1)sothetotalTimeComplexityofthealgorithmisO(n)time.SpaceComplexityisalsoO(n)Fifthapproach: Counting, This approach is only possible ifwe know the range of the input. Same asHashtableimplementationjustuseasimplearrayinplaceofHashtableandyouaredone.Counting approach just uses an array so insert and find take constant time O(1) so the total TimeComplexityofthealgorithmisO(n)time.SpaceComplexityforcreatingcountarrayisalsoO(n).
TwoelementswhosesumisclosesttozeroGivenanArrayof integers,both+veand-ve.Youneed tofind the twoelementssuch that theirsumisclosesttozero.Firstapproach:ExhaustivesearchorBruteforce,foreachelementinarrayfindtheotherelementwhosevaluewhenaddedwillgiveminimumabsolutevalue.Thisisdoneusingtwoforloop,firstlooptoselecttheelementandsecondlooptofindtheelementthatshouldbeaddedtoitsothattheabsoluteofthesumwillbeminimumorclosetozero.TheTimeComplexityisO(n2)andSpaceComplexityisO(1)Example5.20voidminAbsSumPair(intarr[],intsize)
intl,r,minSum,sum,minFirst,minSecond;
/*Arrayshouldhaveatleasttwoelements*/if(size<2)printf("InvalidInput");return;
/*Initializationofvalues*/minFirst=0;minSecond=1;minSum=abs(arr[0]+arr[1]);
for(l=0;l<size-1;l++)for(r=l+1;r<size;r++)sum=abs(arr[l]+arr[r]);if(sum<minSum)
minSum=sum;minFirst=l;minSecond=r;printf("Thetwoelementswithminimumsumare%d&%d",arr[minFirst],arr[minSecond]);
Secondapproach:SortingStepsareasfollows:1.Sortalltheelementsinthearray.2.TaketwovariablefirstIndex=0andsecondIndex=size-13.Computesum=arr[firstIndex]+arr[secondIndex]4.Ifthesumisequaltothe0thenwehavethesolution5.Ifthesumislessthanthe0thenwewillincreasefirst6.Ifthesumisgreaterthanthe0thenwewilldecreasethesecond7.Werepeattheaboveprocess3to6,tillwegetthedesiredpairorwegetfirst>=secondExample5.21voidminAbsSumPair(intarr[],intsize)
intl,r,minSum,sum,minFirst,minSecond;/*Arrayshouldhaveatleasttwoelements*/if(size<2)printf("InvalidInput");return;Sort(arr,size);/*Initializationofvalues*/minFirst=0;minSecond=size-1;minSum=abs(arr[minFirst]+arr[minSecond]);for(l=0,r=size-1;l<r;)sum=(arr[l]+arr[r]);if(abs(sum)<minSum)minSum=abs(sum);minFirst=l;minSecond=r;
if(sum<0)l++;elseif(sum>0)r++;elsebreak;printf("Thetwoelementswithminimumsumare%d&%d",arr[minFirst],arr[minSecond]);
FindmaximainabitonicarrayAbitonicarraycomprisesofan increasingsequenceof integers immediately followedbyadecreasingsequenceofintegers.Sincetheelementsaresortedinsomeorder,weshouldgoforalgorithmsimilartobinarysearch.Thestepsareasfollows:1.Taketwovariableforstoringstartandendindex.Variablestart=0andend=size-12.Findthemiddleelementofthearray.3.Seeifthemiddleelementisthemaxima.Ifyes,returnthemiddleelement.4.Alternatively,Ifthemiddleelementinincreasingpart,thenweneedtolookforinmid+1andend.5.Alternatively,ifthemiddleelementisinthedecreasingpart,thenweneedtolookinthestartandmid-
1.6.Repeatstep2to5untilwegetthemaxima.Example5.22intSearchBotinicArrayMax(intarr[],intsize)
intstart=0,end=size-1,mid;intmaximaFound=0;if(size<3)printf("error");return0;while(start<=end)mid=(start+end)/2;if(arr[mid-1]<arr[mid]&&arr[mid+1]<arr[mid])//maximamaximaFound=1;break;elseif(arr[mid-1]<arr[mid]&&arr[mid]<arr[mid+1])//increasingstart=mid+1;
elseif(arr[mid-1]>arr[mid]&&arr[mid]>arr[mid+1])//decreasingend=mid-1;elsebreak;if(maximaFound==0)printf("error");return0;returnarr[mid];
SearchelementinabitonicarrayAbitonicarraycomprisesofan increasingsequenceof integers immediately followedbyadecreasingsequenceofintegers.Tosearchanelementinabitonicarray:1.Findtheindexormaximumelementinthearray.Byfindingtheendofincreasingpartofthearray,using
modifiedbinarysearch.2.Oncewehavethemaximumelement,searchthegivenvalueinincreasingpartofthearrayusingbinary
search.3.Ifthevalueisnotfoundinincreasingpart,searchthesamevalueindecreasingpartofthearrayusing
binarysearch.Example5.23intSearchBitonicArray(intarr[],intsize,intkey)
intmax=FindMaxBitonicArray(arr,size);
intk=BinarySearch(arr,0,max,key,true);if(k!=-1)returnk;elsereturnBinarySearch(arr,max+1,size-1,key,false);
intFindMaxBitonicArray(intarr[],intsize)
intstart=0,end=size-1,mid;if(size<3)
printf("error");return0;while(start<=end)mid=(start+end)/2;if(arr[mid-1]<arr[mid]&&arr[mid+1]<arr[mid])//maximareturnmid;elseif(arr[mid-1]<arr[mid]&&arr[mid]<arr[mid+1])//increasingstart=mid+1;elseif(arr[mid-1]>arr[mid]&&arr[mid]>arr[mid+1])//increasingend=mid-1;elsebreak;printf("error");return0;
intBinarySearch(intarr[],intstart,intend,intkey,intisInc)
intmid;if(end<start)return-1;mid=(start+end)/2;if(key==arr[mid])returnmid;if(isInc&&key<arr[mid]||!isInc&&key>arr[mid])returnBinarySearch(arr,start,mid-1,key,isInc);elsereturnBinarySearch(arr,mid+1,end,key,isInc);
OccurrencecountsinsortedArrayGivenasortedarrayarr[]findthenumberofoccurrencesofanumber.Firstapproach:Bruteforce,Traversethearrayandinlineartimewewillgettheoccurrencecountofthenumber.Thisisdoneusingoneloop.TheTimeComplexityisO(n)andSpaceComplexityisO(1).Example5.24intfindKeyCount(intarr[],intsize,intkey)
inti,count=0;for(i=0;i<size;i++)if(arr[i]==key)count++;returncount;
Secondapproach:Sincewehavesortedarray,weshouldthinkaboutsomebinarysearch.1.First,weshouldfindthefirstoccurrenceofthekey.2.Thenweshouldfindthelastoccurrenceofthekey.3.Takethedifferenceofthesetwovaluesandyouwillhavethesolution.Example5.25intfindKeyCount(intarr[],intsize,intkey)
intfirstIndex,lastIndex;firstIndex=findFirstIndex(arr,0,size-1,key);lastIndex=findLastIndex(arr,0,size-1,key);return(lastIndex-firstIndex+1);
intfindFirstIndex(intarr[],intstart,intend,intkey)
intmid;if(end<start)return-1;
mid=(start+end)/2;
if(key==arr[mid]&&(mid==start||arr[mid-1]!=key))
returnmid;if(key<=arr[mid])//<=isusthenumber.tinsortedarray.returnfindFirstIndex(arr,start,mid-1,key);elsereturnfindFirstIndex(arr,mid+1,end,key);
intfindLastIndex(intarr[],intstart,intend,intkey)
intmid;if(end<start)return-1;
mid=(start+end)/2;
if(key==arr[mid]&&(mid==end||arr[mid+1]!=key))returnmid;
if(key<arr[mid])//<returnfindLastIndex(arr,start,mid-1,key);elsereturnfindLastIndex(arr,mid+1,end,key);
SeparateevenandoddnumbersinArrayGiven an array of even and odd numbers, write a program to separate even numbers from the oddnumbers.Firstapproach:allocateaseparatearray,thenscanthroughthegivenarray,andfillevennumbersfromthestartandoddnumbersfromtheend.Secondapproach:Algorithmisasfollows.1.Initializethetwovariableleftandright.Variableleft=0andright=size-1.2.Keepincreasingtheleftindexuntiltheelementatthatindexiseven.3.Keepdecreasingtherightindexuntiltheelementatthatindexisodd.4.Swapthenumberatleftandrightindex.
5.Repeatsteps2to4untilleftislessthanright.Example5.26voidswap(int*first,int*second)
inttemp=*first;*first=*second;*second=temp;
voidseperateEvenAndOdd(intarr[],intsize)
intleft=0,right=size-1;
while(left<right)if(arr[left]%2==0)left++;elseif(arr[right]%2==1)right--;elseswap(&arr[left],&arr[right]);left++;right--;
Stockpurchase-sellproblemGivenanarray,whosenthelementisthepriceofthestockonnthday.Youareaskedtobuyonceandsellonce,onwhatdateyouwillbebuyingandatwhatdateyouwillbesellingtogetmaximumprofit.
OrGivenanarrayofnumbers,youneedtomaximizethedifferencebetweentwonumbers,suchthatyoucansubtractthenumber,whichappearbeforeformthenumberthatappearafterit.Firstapproach:Bruteforce,foreachelementinarrayfindifthereissomeotherelementwhosedifferenceismaximum.Thisisdoneusingtwoforloop,firstlooptoselect,buydateindexandthesecondlooptofinditssellingdateentry.TheTimeComplexityisO(n2)andSpaceComplexityisO(1)
Secondapproach:Anothercleversolutionistokeeptrackofthesmallestvalueseensofarfromthestart.Ateachpoint,wecanfindthedifferenceandkeeptrackofthemaximumprofit.Thisisalinearsolution.TheTimeComplexityof thealgorithmisO(n) time.SpaceComplexity forcreatingcountarray isalsoO(1)Example5.27voidmaxProfit(intstocks[],intsize)
intbuy=0,sell=0;intcurMin=0;intcurrProfit=0;intmaxProfit=0;inti;
for(i=0;i<size;i++)if(stocks[i]<stocks[curMin])curMin=i;
currProfit=stocks[i]-stocks[curMin];
if(currProfit>maxProfit)buy=curMin;sell=i;maxProfit=currProfit;printf("\nPurchasedayis-%datprice%d",buy,stocks[buy]);printf("\nSelldayis-%datprice%d",sell,stocks[sell]);
FindamedianofanarrayGivenanarrayofnumbersofsizen,ifalltheelementsofthearrayaresortedthenfindtheelement,whichlieattheindexn/2.Firstapproach:Sortthearrayandreturntheelementinthemiddle.SortingalgorithmstakeO(n.logn).TheTimeComplexityofanalgorithmisO(n.logn)andSpaceComplexityisO(1)Example5.28
intgetMedian(intarr[],intsize)
sort(arr,size);returnarr[size/2];
Second approach: Use QuickSelect algorithm. This algorithm we will look into the next chapter. InQuickSortalgorithmjustskiptherecursivecallthatwedonotneed.TheaverageTimeComplexityofthisalgorithmwillbeO(1)
Findmedianoftwosortedarrays.Firstapproach:Keeptrackof theindexofboththearray,saytheindexareiand j.keep increasing theindexofthearraywhicheverhaveasmallervalue.Useacountertokeeptrackoftheelementsthatwehavealreadytraced.TheTimeComplexityofanalgorithmisO(n)andSpaceComplexityisO(1)Example5.29intfindMedian(intarrFirst[],intsizeFirst,intarrSecond[],intsizeSecond)
//cealingfunction.intmedianIndex=((sizeFirst+sizeSecond)+(sizeFirst+sizeSecond)%2)/2;inti=0,j=0;intcount=0;while(count<medianIndex-1)if(i<sizeFirst-1&&arrFirst[i]<arrSecond[j])i++;elsej++;
count++;
if(arrFirst[i]<arrSecond[j])returnarrFirst[i];elsereturnarrSecond[j];
Findkthelementoftwosortedarrays.FindkthSmallestElementintheUnionofTwoSortedArrays
Example5.301.#defineERROR-999;2.intmin(inta,intb)3.4.returna>b?b:a;5.6.intfind_kth(intfirst[],intsecond[],intsizeFirst,intsizeSecond,intk)7.8.if(sizeFirst+sizeSecond<k)9.returnERROR;10.11.if(sizeFirst==0)12.returnsecond[k-1];13.14.if(sizeSecond==0)15.returnfirst[k-1];16.17.if(k==1)18.returnmin(first[0],second[0]);19.20./*Nowdivideandconquer*/21.inti=min(sizeFirst,k/2);22.intj=min(sizeSecond,k/2);23.24.if(first[i-1]>second[j-1])25.returnfind_kth(first,second+j,i,sizeSecond-j,k-j);26.else27.returnfind_kth(first+i,second,sizeFirst-i,j,k-i);28.
Search01ArrayGivenanarrayof0’sand1’s.Allthe0’scomebefore1’s.Writeanalgorithmtofindtheindexofthefirst1.
OrYou are given an array which contains either 0 or 1, and they are in sorted order Ex. a[ ] = 1,1,1,1,0,0,0Howwillyoucountnoof1`sand0's?Firstapproach:BinarySearch,sincethearrayissortedusingbinarysearchtofindthedesiredindex.TheTimeComplexityofanalgorithmisO(logn)andSpaceComplexityisO(1)Example5.31intBinarySearch01Wrapper(intarr[],intsize)
if(size==1&&arr[0]==1)return0;
returnBinarySearch01(arr,0,size-1);
intBinarySearch01(intarr[],intstart,intend)
intmid;if(end<start)return-1;
mid=(start+end)/2;
if(1==arr[mid]&&0==arr[mid-1])returnmid;
if(0==arr[mid])returnBinarySearch01(arr,mid+1,end);elsereturnBinarySearch01(arr,start,mid-1);
SearchinsortedrotatedArrayGivenasortedarrayofnintegerswhichisrotatedanunknownnumberoftimes.Findanelementinthearray.Firstapproach:Sincethearrayissorted,wecanusemodifiedbinarysearchtofindtheelement.TheTimeComplexityofanalgorithmisO(logn)andSpaceComplexityisO(1)Example5.32intBinarySearchRotateArray(intarr[],intstart,intend,intkey)
intmid;if(end<start)return-1;
mid=(start+end)/2;
if(key==arr[mid])
returnmid;if(arr[mid]>arr[start])if(arr[start]<=key&&key<arr[mid])returnBinarySearchRotateArray(arr,start,mid-1,key);elsereturnBinarySearchRotateArray(arr,mid+1,end,key);elseif(arr[mid]<key&&key<=arr[end])returnBinarySearchRotateArray(arr,mid+1,end,key);elsereturnBinarySearchRotateArray(arr,start,mid-1,key);
intBinarySearchRotateArrayWrapper(intarr[],intsize,intkey)
returnBinarySearchRotateArray(arr,0,size-1,key);
FirstRepeatedelementinthearrayGivenanunsortedarrayofnelements,findthefirstelement,whichisrepeated.Firstapproach:Exhaustive searchorBrute force, for eachelement in array find if there is someotherelementwiththesamevalue.Thisisdoneusingtwoforloop,firstlooptoselecttheelementandsecondlooptofinditsduplicateentry.TheTimeComplexityisO(n2)andSpaceComplexityisO(1)Example5.33intFirstRepeated(int*arr,intsize)
inti,j;for(i=0;i<size;i++)
for(j=i+1;j<size;j++)if(arr[i]==arr[j])returnarr[i];
return0;
Secondapproach:Hash-Table,usingHash-Table,wecankeeptrackofthenumberoftimesaparticularelement came in the array. First scan just populate the Hashtable. In the second, scan just look theoccurrenceof theelements in theHashtable. Ifoccurrence ismoreforsomeelement, thenwehaveoursolutionandthefirstrepeatedelement.Hash-TableinsertandfindtakeconstanttimeO(1)sothetotalTimeComplexityofthealgorithmisO(n)time.SpaceComplexityisalsoO(n)formaintaininghash.
TransformArrayHowwouldyouswapelementsofanarraylike[a1a2a3a4b1b2b3b4]toconvertitinto[a1b1a2b2a3b3a4b4]?
Approach:·Firstswapelementsinthemiddlepair·Nextswapelementsinthemiddletwopairs·Nextswapelementsinthemiddlethreepairs·Iteraten-1steps.
Ex:withn=4.a1a2a3a4b1b2b3b4a1a2a3b1a4b2b3b4a1a2b1a3b2a4b3b4a1b1a2b2a3b3a4b4Example5.34voidtransformArrayAB1(intarr[],intsize)
intN=size/2,i,j;for(i=1;i<N;i++)for(j=0;j<i;j++)swap(&arr[N-i+2*j],&arr[N-i+2*j+1]);
voidswap(int*a,int*b)
intt=*a;*a=*b;*b=t;
Find2ndlargestnumberinanarraywithminimumcomparisonsSupposeyouaregivenanunsortedarrayofndistinctelements.Howwillyouidentifythesecondlargestelementwithminimumnumberofcomparisons?First approach: Find the largest element in the array. Then replace the last element with the largestelement.Thensearchthesecondlargestelementinttheremainingn-1elements.Thetotalnumberofcomparisonsis:(n-1)+(n-2)
Secondapproach:Sortthearrayandthengivethe(n-1)element.Thisapproachisstillmoreinefficient.Third approach: Using priority queue / Heap. This approach we will look into heap chapter. UsebuildHeap() function tobuildheapfromthearray.This isdone inncomparisons.Arr[0] is the largestnumber,andthegreateramongarr[1]andarr[2]isthesecondlargest.Thetotalnumberofcomparisonsare:(n-1)+1=n
CheckiftwoarraysarepermutationofeachotherGiventwointegerarrays.Youhavetocheckwhethertheyarepermutationofeachother.Firstapproach:Sorting,Sortall theelementsofboth thearraysandCompareeachelementofboth thearraysfrombeginningtoend.Ifthereisnomismatch,returntrue.Otherwise,false.SortingalgorithmstakeO(n.logn)timeandcomparisontakeO(n)time.TheTimeComplexityofanalgorithmisO(n.logn)andSpaceComplexityisO(1)Example5.35intcheckPermutation(intarray1[],intsize1,intarray2[],intsize2)
if(size1!=size2)return0;
sort(array1,size1);sort(array2,size2);
for(inti=0;i<size1;i++)if(array1[i]!=array2[i])return0;return1;
Secondapproach:Hash-Table(Assumption:Noduplicates).Stepsare:1.CreateaHash-Tableforalltheelementsofthefirstarray.2.TraversetheotherarrayfrombeginningtotheendandsearchforeachelementintheHash-Table.3.IfalltheelementsarefoundintheHash-Table,returntrueotherwisereturnfalse.Hash-TableinsertandfindtakeconstanttimeO(1)sothetotalTimeComplexityofthealgorithmisO(n)time.SpaceComplexityisalsoO(n)TimeComplexity=O(n)(ForcreationofHash-Tableandlook-up),SpaceComplexity=O(n)(ForcreationofHash-Table).
Example5.36intcheckPermutation(intarray1[],intsize1,intarray2[],intsize2)
inti;if(size1!=size2)return0;Hashtableh;
for(i=0;i<size1;i++)insert(h,array1[i]);
for(i=0;i<size2;i++)if(!containsValue(h,array2[i]))returnfalse;returntrue;
RemoveduplicatesinanintegerarrayFirstapproach:SortingStepsareasfollows:1.Sortthearray.2. Take two pointers.A subarraywill be createdwith all unique elements starting from 0 to the first
pointer(Thefirstpointerpointstothelastindexofthesubarray).Thesecondpointeriteratesthroughthearrayfrom1totheend.Uniquenumberswillbecopiedfromthesecondpointerlocationtofirstpointerlocationandthesameelementsareignored.
TimeComplexitycalculation:Timetosortthearray=O(nlogn).Timetoremoveduplicates=O(n).OverallTimeComplexity=O(nlogn).NoadditionalspaceisrequiredsoSpaceComplexityisO(1).
Example5.37intremoveDuplicates(intarray[],intsize)
intj=0;inti;if(size==0)return0;sort(array,size);
for(i=1;i<size;i++)if(array[i]!=array[j])
j++;array[j]=array[i];returnj+1;
Searchingforanelementina2-dsortedarrayGivena2dimensionalarray.Eachrowandcolumnaresortedinascendingorder.Howwouldyoufindanelementinit?Thealgorithmworksas:1.Startwithelementatlastcolumnandfirstrow2.Iftheelementisthevaluewearelookingfor,returntrue.3. If theelement isgreater than thevalueweare lookingfor,go to theelementatpreviouscolumnbut
samerow.4.Iftheelementislessthanthevaluewearelookingfor,gototheelementatnextrowbutsamecolumn.5.Returnfalse,iftheelementisnotfoundafterreachingtheelementofthelastrowofthefirstcolumn.
Conditionrow<r&&column>=0isfalse.Runningtime=O(N).Example5.38intFindElementIn2DArray(int*arr[],intr,intc,intvalue)
introw=0;intcolumn=c-1;
while(row<r&&column>=0)if(arr[row][column]==value)return1;elseif(arr[row][column]>value)column--;elserow++;return0;
Exercise1.Given an array of n elements, find the first repeated element.Which of the followingmethodswill
workforus(andwhichofthemethodwillnotworkforus).Ifamethodwork,thenimplementsit.·Bruteforceexhaustivesearch.·UseHash-Tabletokeepanindexoftheelementsandusethesecondscantofindtheelement.·Sortingtheelements.·Ifweknowtherangeoftheelementthenwecanusecountingtechnique.
Hint:Whenorderinwhichelementsappearininputisimportant,wecannotusesorting.
2.Givenanarrayofn elements,write analgorithm to find three elements in anarraywhose sum is a
givenvalue.
Hint:Trytodothisproblemusingabruteforceapproach.Thentrytoapplythesortingapproachalongwithabruteforceapproach.TheTimeComplexitywillbeO(n2)
3. Given an array of –ve and +ve numbers, write a program to separate –ve numbers from the +ve
numbers.4.Givenanarrayof1’sand0’s,writeaprogramtoseparate0’sfrom1’s.
Hint:QuickSelect,counting5.Givenanarrayof0’s,1’sand2’s,writeaprogramtoseparate0’s,1’sand2’s.6.Givenanarraywhoseelementsismonotonicallyincreasingwithbothnegativeandpositivenumbers.
Writeanalgorithmtofindthepointatwhichlistbecomespositive.7.Givenasortedarray,findagivennumber.Iffoundreturntheindexifnot,findtheindexofthatnumber
ifitisinsertedintothearray.8.Findmaxinsortedrotatedarray.9.Findmininthesortedrotatedarray.
CHAPTER6:SORTING
IntroductionSorting is the process of placing elements from a collection into ascending or descending order. Forexample,whenweplaycards,sortcards,accordingtotheirvaluesothatwecanfindtherequiredcardeasily.Whenwegotosomelibrary,thebooksarearrangedaccordingtostreams(Algorithm,Operatingsystems,Networkingetc.).Sortingarrangesdataelementsinordersothatsearchingbecomeeasier.Whenbooksarearrangedinproperindexingorder,thenitiseasytofindabookwearelookingfor.ThischapterdiscussesalgorithmsforsortingasetofN items.Understandingsortingalgorithmsare thefirststeptowardsunderstandingalgorithmanalysis.Manysortingalgorithmsaredevelopedandanalysed.A sorting algorithm likeBubble-Sort, Insertion-Sort and Selection-Sort are easy to implement and aresuitableforthesmallinputset.However,forlargedatasettheyareslow.A sorting algorithm like Merge-Sort, Quick-Sort and Heap-Sort are some of the algorithms that aresuitableforsortinglargedataset.However,theyareoverkillifwewanttosortthesmalldataset.Somealgorithm,whichissuitablewhenwehavesomerangeinformationoninputdata.Some other algorithm is there to sort a huge data set that cannot be stored inmemory completely, forwhichexternalsortingtechniqueisdeveloped.Beforewestartadiscussionofthevariousalgorithmsonebyone.First,weshouldlookatcomparisonfunctionthatisusedtocomparetwovalues.Lessfunctionwillreturn1ifvalue1islessthanvalue2otherwise,itwillreturn0.intless(intvalue1,intvalue2)
returnvalue1<value2;Morefunctionwillreturn1ifvalue1ismorethanvalue2otherwiseitwillreturn0.intmore(intvalue1,intvalue2)
returnvalue1>value2;The value in various sorting algorithms is compared using one of the above functions and it will beswappeddependinguponthereturnvalueofthesefunctions.Ifmore()comparisonfunctionisused,thensortedoutputwillbeincreasinginorderandifless()isusedthanresultingoutputwillbeindescendingorder.
TypeofSortingInternalSorting:Alltheelementscanbereadintomemoryatthesametimeandsortingisperformedinmemory.
1.Selection-Sort2.Insertion-Sort3.Bubble-Sort4.Quick-Sort
ExternalSorting:Inthis,thedatasetissobigthatitisimpossibletoloadthewholedatasetintomemorysosortingisdoneinchunks.
1.Merge-SortThreethingstoconsiderinchoosing,sortingalgorithmsforapplication:
1.Numberofelementsinlist2.Anumberofdifferentordersoflistrequired3.Theamountoftimerequiredtomovethedataornotmovethedata
Bubble-SortBubble-Sortistheslowestalgorithmforsorting,butitisheavilyused,asitiseasytoimplement.InBubble-Sort,wecompareeachpairofadjacentvalues.Wewanttosortvaluesinincreasingordersoifthesecondvalueislessthanthefirstvaluethenweswapthesetwovalues.Otherwise,wewillgotothenextpair.Thus,smallervaluesbubbletothestartof thearray.WewillhaveNnumberofpassestoget thearraycompletelysorted.Afterthefirstpass,thelargestvaluewillbeintherightmostposition.
Example6.11.voidBubbleSort(int*arr,intsize)2.3.inti,j,temp;4.for(i=0;i<(size-1);i++)5.6.for(j=0;j<size-i-1;j++)7.8.if(more(arr[j],arr[j+1]))9.10./*Swapping*/11.temp=arr[j];12.arr[j]=arr[j+1];13.arr[j+1]=temp;14.15.16.17.Analysis:Line4:Theouterforloopsrepresentsthenumberofswapsthataredoneforcomparisonofdata.Line 6: The inner loop is actually used to do the comparison of data. At the end of each inner loop
iteration,thelargestvalueismovedtotheendofthearray.Inthefirstiterationthelargestvalue,intheseconditerationthesecondlargestandsoon.Line 8: more() function is used for comparison whichmeans when the value of the first argument isgreaterthanthevalueofthesecondargumentthenperformaswap.Bythiswearesortinginincreasingorderifwehave,theless()functioninplaceofmore()thanwewillgetdecreasingordersorting.Havealookintomore()functionincaseyouforgotintmore(intvalue1,intvalue2)
returnvalue1>value2;ComplexityAnalysis:Eachtimetheinnerloopexecutefor(n-1),(n-2),(n-3)…(n-1)+(n-2)+(n-3)+.....+3+2+1=n(n-1)/2Worstcaseperformance O(n2)Averagecaseperformance O(n2)SpaceComplexity O(1)asweneedonlyonetempvariableStableSorting Yes
Modified(improved)Bubble-SortWhenthereisnomoreswapinonepassoftheouterloop.Itindicatesthatalltheelementsarealreadyinordersoweshouldstopsorting.ThissortingimprovementinBubble-Sortisextremelyusefulwhenweknowthat,exceptfewelementsrestofthearrayisalreadysorted.Example6.21.voidBubbleSort(int*arr,intsize)2.3.inti,j,temp,swapped=1;4.for(i=0;i<(size-1)&&swapped;i++)5.6.swapped=0;7.for(j=0;j<size-i-1;j++)8.9.if(more(arr[j],arr[j+1]))10.11./*Swapping*/12.temp=arr[j];13.arr[j]=arr[j+1];14.arr[j+1]=temp;15.swapped=1;16.17.18.19.Byapplyingthisimprovement,bestcaseofthisalgorithm,whenanarrayisnearlysorted,isimproved.BestcaseisO(n)ComplexityAnalysis:Worstcaseperformance O(n2)Averagecaseperformance O(n2)SpaceComplexity O(1)Adaptive:Whenarrayisnearlysorted O(n)StableSorting Yes
Insertion-Sort
Insertion-SortTimeComplexityisO(n2)whichissameasBubble-Sortbutperformabitbetterthanit.Itisthewaywearrangeourplayingcards.Wekeepasortedsubarray.Eachvalueisinsertedintoitsproperpositioninthesortedsub-arrayintheleftofit.
Example6.31.voidinsertion(int*arr,intsize)2.3.inttemp,j;4.for(inti=1;i<size;i++)5.6.temp=arr[i];7.for(j=i;j>0&&more(arr[j-1],temp));j--)8.9.arr[j]=arr[j-1];10.11.arr[j]=temp;12.13.Analysis:Line4:Theouterloopisusedtopickthevaluewewanttoinsertintothesortedleftarray.
Line6:Thevaluewewanttoinsertwehavepickedandsavedinatempvariable.Line7-10:Thesearethelines,whichareimplementingtheinnerloopanddoingthecomparisonusingthemore()function.Thevaluesareshiftedtotherightuntilwefindtheproperpositionofthetempvalueforwhichwearedoingthisiteration.Line11:Thisistheline,whichweareactuallyplacingthetempvalueintotheproperposition.Line4-12:Ineachiterationoftheouterloop,thelengthofthesortedarrayincreasebyone.Whenweexittheouterloop,thewholearrayissorted.ComplexityAnalysis:WorstcaseTimeComplexity O(n2)BestcaseTimeComplexity O(n)AveragecaseTimeComplexity O(n2)SpaceComplexity O(1)Stablesorting Yes
Selection-SortSelection-Sortsearchesthewholeunsortedarrayandputthelargestvalueattheendofit.Thisalgorithmis having the same Time Complexity, but performs better than both bubble and Insertion-Sort as lessnumberofcomparisonsrequired.ThesortedarrayiscreatedbackwardinSelection-Sort.
Example6.4:1.voidSelectionSort(int*arr,intsize)2.3.inti,j,max,temp;4.for(i=0;i<size-1;i++)5.6.max=0;7.for(j=1;j<size-1-i;j++)8.9.if(arr[j]>arr[max])10.11.max=j;12.13.14.temp=arr[size-1-i];15.arr[size-1-i]=arr[max];16.arr[max]=temp;17.18.Analysis:Line4-17: It is theouter loop,which isused topick the largestvaluefromtheunsortedarray. Ineachiteration,thelargestvaluewillbeplacedattheendofthearray.Line6:Theindexofthemaxvalueisalwayssettothebeginningofthearrayandwewilliteratethroughthearraytofindtheproperindex.Line7-10:Thesearetheinnerlooplinestofindtheproperindexofthemaximum.
Line14-17:Thisisthefinalreplacementofthemaximumvaluetotheproperlocation.Thesortedarrayiscreatedbackward.ComplexityAnalysis:WorstCaseTimeComplexity O(n2)BestCaseTimeComplexity O(n2)AveragecaseTimeComplexity O(n2)SpaceComplexity O(1)StableSorting NoThesamealgorithmcanbeimplementedbycreatingthesortedarrayinthefrontofthearray.Example6.5:1.voidSelectionSort(int*arr,intsize)2.3.inti,j,min,temp;4.for(i=0;i<size-1;i++)5.6.min=i;7.for(j=i+1;j<size;j++)8.9.if(arr[j]<arr[min])10.11.min=j;12.13.14.temp=arr[i];15.arr[i]=arr[min];16.arr[min]=temp;17.18.
Merge-Sort
Example6.6:1.#include<stdio.h>23.voidmerge(int*arr,int*tempArray,intlowerIndex,intmiddleIndex,intupperIndex)4.5.intlowerStart=lowerIndex;6.intlowerStop=middleIndex;7.intupperStart=middleIndex+1;8.intupperStop=upperIndex;9.intcount=lowerIndex;10.while(lowerStart<=lowerStop&&upperStart<=upperStop)11.12.if(arr[lowerStart]<arr[upperStart])13.tempArray[count++]=arr[lowerStart++];14.else15.tempArray[count++]=arr[upperStart++];16.17.while(lowerStart<=lowerStop)18.19.tempArray[count++]=arr[lowerStart++];20.21.while(upperStart<=upperStop)22.23.tempArray[count++]=arr[upperStart++];24.25.for(inti=lowerIndex;i<=upperIndex;i++)26.arr[i]=tempArray[i];27.
1.voidmergeSrt(int*arr,int*tempArray,intlowerIndex,intupperIndex)2.3.if(lowerIndex>=upperIndex)4.return;5.intmiddleIndex=(lowerIndex+upperIndex)/2;6.mergeSrt(arr,tempArray,lowerIndex,middleIndex);7.mergeSrt(arr,tempArray,middleIndex+1,upperIndex);8.merge(arr,tempArray,lowerIndex,middleIndex,upperIndex);9.1.voidmergeSort(int*arr,intsize)2.3.int*tempArray=(int*)malloc(size*sizeof(int));4.mergeSrt(arr,tempArray,0,size-1);5.1.voidprintArray(int*arr,intsize)2.3.for(inti=0;i<size;i++)4.printf("%d",arr[i]);5.1.intmain()2.3.intarr[10]=3,4,2,1,6,5,7,8,1,1;4.mergeSort(arr,10);5.printArray(arr,10);6.·TheTimeComplexityofMerge-SortisO(nlogn)inall3cases(best,averageandworst)asMerge-Sort
alwaysdividesthearrayintotwohalvesandtakelineartimetomergetwohalves.·Itrequirestheequalamountofadditionalspaceastheunsortedlist.Hence,itisnotatallrecommended
forsearchinglargeunsortedlists.·ItisthebestSortingtechniqueforsortingLinkedLists.ComplexityAnalysis:WorstCaseTimeComplexity O(nlogn)BestCaseTimeComplexity O(nlogn)AverageTimeComplexity O(nlogn)SpaceComplexity O(n)StableSorting Yes
Quick-Sort
Example6.7:1.intmain()2.3.intarr[10]=4,5,3,2,6,7,1,8,9,10;4.printArray(arr,sizeof(arr)/sizeof(int));5.quickSort(arr,sizeof(arr)/sizeof(int));6.printArray(arr,sizeof(arr)/sizeof(int));7.1.voidswap(intarr[],intfirst,intsecond)2.3.inttemp=arr[first];4.arr[first]=arr[second];5.arr[second]=temp;
6.1.voidprintArray(intarr[],intsize)2.3.for(inti=0;i<size;i++)4.printf("%d",arr[i]);5.printf("\n");6.1.voidquickSort(intarr[],intsize)2.3.quickSortUtil(arr,0,size-1);4.1.voidquickSortUtil(intarr[],intlower,intupper)2.3.if(upper<=lower)4.return;5.6.intpivot=arr[lower];7.8.intstart=lower;9.intstop=upper;10.11.while(lower<upper)12.13.while(arr[lower]<=pivot)14.15.lower++;16.17.while(arr[upper]>pivot)18.19.upper--;20.21.if(lower<upper)22.23.swap(arr,upper,lower);24.25.26.swap(arr,upper,start);//upperisthepivotposition27.28.quickSortUtil(arr,start,upper-1);//pivot-1istheupperforleftsubarray.29.quickSortUtil(arr,upper+1,stop);//pivot+1isthelowerforrightsubarray.30.·ThespacerequiredbyQuick-Sortisveryless,onlyO(nlogn)additionalspaceisrequired.
·Quicksortisnotastablesortingtechnique,soitmightchangetheoccurrenceoftwosimilarelementsinthelistwhilesorting.
ComplexityAnalysis:WorstCaseTimeComplexity O(n2)BestCaseTimeComplexity O(nlogn)AverageTimeComplexity O(nlogn)SpaceComplexity O(nlogn)StableSorting No
QuickSelectQuickselectisverysimilartoQuick-Sortinplaceofsortingthewholearraywejustignoretheone-halfofthearrayateachstepofQuick-Sortandjustfocusontheregionofarrayonwhichweareinterested.Example6.8:1.voidquickSelect(intarr[],intlower,intupper,intk)2.3.if(upper<=lower)4.return;5.6.intpivot=arr[lower];7.8.intstart=lower;9.intstop=upper;10.11.while(lower<upper)12.13.while(arr[lower]<=pivot)14.15.lower++;16.17.while(arr[upper]>pivot)18.19.upper--;20.21.if(lower<upper)22.23.swap(arr,upper,lower);24.25.26.27.swap(arr,upper,start);//upperisthepivotposition28.if(k<upper)29.quickSelect(arr,start,upper-1,k);//pivot-1istheupperforleftsubarray.30.if(k>upper)31.quickSelect(arr,upper+1,stop,k);//pivot+1isthelowerforrightsubarray.32.1.intquickSelect(int*a,intcount,intindex)2.3.quickSelect(a,0,count-1,index-1);4.returna[index-1];5.ComplexityAnalysis:
WorstCaseTimeComplexity O(n2)BestCaseTimeComplexity O(logn)AverageTimeComplexity O(logn)SpaceComplexity O(nlogn)
BucketSortBucket sort is the simplest andmost efficient typeof sorting.Bucket sort has a strict requirementof apredefinedrangeofdata.Like,sorthowmanypeopleareinwhichagegroup.Weknowthattheageofpeoplecanvarybetween1and130.
Example6.9:1.voidBucketSort(intarray[],intn,intrange)2.3.inti,j;4.int*count=(int*)malloc(range*sizeof(int));5.6.for(i=0;i<range;i++)7.8.count[i]=0;9.10.11.for(i=0;i<n;i++)12.13.count[array[i]]++;14.15.16.j=0;17.18.for(i=0;i<range;i++)19.20.for(;count[i]>0;(count[i])--)21.22.array[j++]=i;23.24.25free(count);26Analysis:Line4:Wehavecreatedacountarraytostorecounts.Line6-9:countarrayelementsareinitializedtozero.
Line11-14:Indexcorrespondingtoinputarrayisincremented.Line:18-24:Finally,theinformationstoredincountarrayissavedinthearray.ComplexityAnalysis:Datastructure ArrayWorstcaseperformance O(n+k)Averagecaseperformance O(n+k)WorstcaseSpaceComplexity O(k)Wherek-isnumberofdistinctelements.n–isthetotalnumberofelementsinarray.
GeneralizedBucketSortTherearecaseswhentheelementfallingintoabucketarenotuniquebutareinthesamerange.Whenwewanttosortanindexofaname,wecanusethepointerbuckettostorenames.
The buckets are already sorted and the elements inside each bucket can be kept sorted by using anInsertion-Sortalgorithm.Weareleavingthisgeneralizedbucketsortimplementationtothereaderofthisbook. The similar data structure will be defined in the coming chapter of Hash-Table using separatechaining.
Heap-SortHeap-SortwehavealreadystudiedintheHeapchapter.ComplexityAnalysis:Datastructure ArrayWorstcaseperformance O(nlogn)Averagecaseperformance O(nlogn)WorstcaseSpaceComplexity O(1)
TreeSortingIn-order traversalof thebinarysearchtreecanalsobeseenasasortingalgorithm.Wewillsee this inbinarysearchtreesectionoftreechapter.ComplexityAnalysis:WorstCaseTimeComplexity O(n2)BestCaseTimeComplexity O(nlogn)AverageTimeComplexity O(nlogn)SpaceComplexity O(n)StableSorting Yes
ExternalSort(ExternalMerge-Sort)Whendataneedtobesortedishuge.Moreover,itisnotpossibletoloaditcompletelyinmemory(RAM)forsuchadatasetweuseexternalsorting.Specificdata issortedusingexternalMerge-Sortalgorithm.Firstdataarepickedinchunksanditissortedinmemory.Thenthissorteddataiswrittenbacktodisk.Wholedata are sorted in chunksusingMerge-Sort.Nowweneed to combine these sorted chunks intofinalsorteddata.Thenwecreatequeuesforthedata,whichwillreadfromthesortedchunks.Eachchunkwillhaveitsownqueue.Wewillpopfromthisqueueandthesequeuesareresponsibleforreadingfromthesortedchunks.LetussupposewehaveKdifferentchunksofsorteddataeachoflengthM.ThethirdstepisusingaMin-Heap,whichwilltakeinputdatafromeachofthisqueue.Itwilltakeoneelement from each queue. The minimum value is taken from the Heap and added to the final sortedelementoutput.ThenqueuefromwhichthisminelementisinsertedintheheapwillagainpoppedandonemoreelementfromthatqueueisaddedtotheHeap.Finally,whenthedataisexhaustedfromsomequeuethatqueueisremovedfromtheinputlist.Finally,wewillgetasorteddatacameoutfromtheheap.Wecanoptimizethisprocessfurtherbyaddinganoutputbuffer,whichwillstoredatacomingoutofHeapandwilldoalimitednumberofthewriteoperationinthefinalDiskspace.
Note:Noonewillbeaskingtoimplementexternalsortinginaninterview,butitisgoodtoknowaboutit.
Comparisonsofthevarioussortingalgorithms.
Sort AverageTime BestTime WorstTime Space Stability
Bubble-Sort O(n2) O(n2) O(n2) O(1) Stable
ModifiedBubble-Sort O(n2) O(n) O(n2) O(1) Stable
Selection-Sort O(n2) O(n2) O(n2) O(1) Unstable
Insertion-Sort O(n2) O(n) O(n2) O(1) Stable
Heap-Sort O(n*log(n)) O(n*log(n)) O(n*log(n)) O(1) Unstable
Merge-Sort O(n*log(n)) O(n*log(n)) O(n*log(n)) O(n) Stable
Quick-Sort O(n*log(n)) O(n*log(n)) O(n2)O(n)worstcaseO(log(n))averagecase Unstable
BucketSort O(nk) O(nk) O(nk) O(nk) Stable
SelectionofBestSortingAlgorithmNosortingalgorithmisperfect.Eachofthemhastheirownprosandcons.Letusreadonebyone:Quick-Sort:Whenyoudonotneedastablesortandaveragecaseperformancemattersmorethanworst-case performance.Whendata is random,weprefer theQuick-Sort.Average caseTimeComplexity ofQuick-Sort isO(nlogn)andworst-caseTimeComplexity isO(n2).SpaceComplexityofQuick-Sort isO(logn)auxiliarystorage,whichisstackspaceusedinrecursion.Merge-Sort:When you need a stable sort andTimeComplexity ofO(nlogn),Merge-Sort is used. Ingeneral,Merge-Sort is slower than Quick-Sort because of lot of copy happening in themerge phase.TherearetwousesofMerge-SortwhenwewanttomergetwosortedlinkedlistsandMerge-Sortisusedinexternalsorting.Heap-Sort:Whenyoudonotneedastablesortandyoucaremoreaboutworst-caseperformance thanaverage case performance. It has guaranteed to beO(nlogn) time complexity, and usesO(1) auxiliaryspace,meaningthatyouwillnotunpredictablyrunoutofmemoryonverylargeinputs.Insertion-Sort:Whenweneedastablesort,WhenNisguaranteedtobesmall,includingasthebasecaseofaQuick-SortorMerge-Sort.Worst-caseTimeComplexityisO(n2),ithasaverysmallconstant,soforsmaller input size it performs better thanMerge-Sort orQuick-Sort. It is also usefulwhen the data isalreadypre-sortedinthiscaseitsbestcaserunningtimeisO(N).Bubble-Sort:Whereweknow thedata is verynearly sorted.Sayonly twoelements areout of place.Theninonepass,BubbleSortwillmakethedatasortedandinthesecondpass,itwillseeeverythingissortedandthenexit.Onlytakes2passesofthearray.Selection-Sort:BestWorstAverageCaserunningtimeallO(n2).Itisonlyusefulwhenyouwanttodosomethingquick.Theycanbeusedwhenyouarejustdoingsomeprototyping.Counting-Sort:Whenyouaresortingdatawithinalimitedrange.Radix-Sort:Whenlog(N)issignificantlylargerthanK,whereKisthenumberofradixdigits.Bucket-Sort:Whenyourinputismoreorlessuniformlydistributed.Note:Astablesortisonethathasguaranteednottoreorderelementswithidenticalkeys.
Exercise1.Givenatextfile,printthewordswiththeirfrequency.Nowprintthekthwordintermoffrequency.
Hint:·Firstapproachmaybeyoucanusethesortingandreturnthekthelement.·Secondapproach:Youcanusethekthelementquickselectalgorithm.·Thirdapproach:YoucanuseHashtableorTrietokeeptrackofthefrequency.UseHeaptogetthe
Kthelement2.GivenK input streams of number in sorted order.You need tomake a single output stream,which
containsall theelementsof theKstreamsinsortedorder.The inputstreamssupportReadNumber()operationandoutputstreamsupportWriteNumber()operation.Hint:·ReadthefirstnumberfromalltheKinputstreamsandaddthemtoaPriorityQueue.(Nodesshould
keeptrackoftheinputstream)·DequeueoneelementatatimefromPQ,Putthiselementvaluetotheoutputstream,Readtheinput
streamnumberandfromthesameinputstreamaddanotherelementtoPQ.·Ifthestreamisempty,justcontinue·RepeatuntilPQisempty.
3.GivenK sorted arrays of fixed lengthM.Also, given a final output array of lengthM*K.Give an
efficientalgorithmtomergeallthearraysintothefinalarray,withoutusinganyextraspace.Hint:youcanusetheendofthefinalarraytomakePQ.
4.Howwillyousort1PBnumbers?1PB=1000TB.5.Whatwillbethecomplexityoftheabovesolution?6.Anyotherimprovementonquestion3solutionifthenumberofcoresiseight.7.GivenanintegerarraythatsupportthreefunctionfindMin,findMax,findMedian.Sortthearray.8.Givenapileofpatient filesofHigh,midand lowpriority.Sort these files such thathigherpriority
comesfirst,thenmidandlastlowpriority.Hint:Bucketsort.
9.WriteprosandconsofHeap-Sort,Merge-SortandQuick-Sort.10.Givenarotated-sortedarrayofNintegers.(Thearraywassortedthenitwasrotatedsomearbitrary
numberoftimes.)Ifalltheelementsinthearraywereuniquethefindtheindexofsomevalue.Hint:Modifiedbinarysearch
11. In the problem9,what if there are repetitions allowed and you need to find the index of the first
occurrenceoftheelementintherotated-sortedarray.
12.Mergetwosortedarraysintoasinglesortedarray.Hint:UsemergemethodofMerge-Sort.
13.Givenanarraycontain0’sand1’s,sortthearraysuchthatallthe0’scomebefore1’s.14.GivenanarrayofEnglishcharacters,sortthearrayinlineartime.15.Writeamethodtosortanarrayofstringssothatalltheanagramsarenexttoeachother.
·Loopthroughthearray.·Foreachword,sortthecharactersandaddittothehashmapwithkeysassortedwordandvalueas
theoriginalword.Attheendoftheloop,youwillgetallanagramsasthevaluetoakey(whichissortedbyitsconstituentchars).
·Iterateoverthehashmap,printallvaluesofakeytogetherandthenmovetothenextkey.SpaceComplexity:O(n),TimeComplexity:O(n)
CHAPTER7:LINKEDLIST
IntroductionLetussupposewehaveanarraythatcontainsfollowingfiveelements1,2,4,5,6.Wewanttoinsertanelementwithvalue“3”inbetween“2”and“4”.Inthearray,wecannotdosoeasily.Weneedtocreateanotherarraythatislongenoughtostorethecurrentvaluesandonemorespacefor“3”.Thenweneedtocopy these elements in the new space. This copy operation is inefficient. To remove this fixed lengthconstraintlinkedlistisused.
LinkedListThelinkedlistisalistofitems,callednodes.Nodeshavetwoparts,valuepartandlinkpart.Valuepartisused tostores thedata.Thevaluepartof thenodecanbeeitherabasicdata-type likean integerorsomeotherdata-typelikestructure.Thelinkpartisapointer,whichisusedtostoreaddressesofthenextelementinthelist.
TypesofLinkedlistTherearedifferenttypesoflinkedlists.Themaindifferenceamongthemishowtheirnodesrefertoeachother.
SinglyLinkedListEachnode(Exceptthelastnode)hasareferencetothenextnodeinthelinkedlist.Thelinkportionofnodecontainstheaddressofthenextnode.ThelinkportionofthelastnodecontainsthevalueNULL.
DoublyLinkedlistThenodeinthistypeoflinkedlisthasreferencetobothpreviousandthenextnodeinthelist.
CircularLinkedListThistypeissimilartothesinglylinkedlistexceptthatthelastelementpointstothefirstnodeofthelist.Thelinkportionofthelastnodecontainstheaddressofthefirstnode.
Thevariouspartsoflinkedlist1.Head:Headisapointerthatholdstheaddressofthefirstnodeinthelinkedlist.2.Nodes:Itemsinthelinkedlistarecallednodes.3.Value:Thedatathatisstoredineachnodeofthelinkedlist.4.Link:Linkpartofthenodeisusedtostoretheaddressofthenode.
a.Wewilluse“next”and“prev”tostoreaddressofnextorpreviousnode.
SinglyLinkedList
LookatNodeinthisexample,itsvaluepartisoftypeint(itcanbeofsomeotherdata-type).Thelinkisnamedasnextinthebelowstructure.WehavetypedeftheNode*toNodePtrsothatourcodelooksclean.structNode
intvalue;Node*next;
;typedefNode*NodePtr;
Note:Forasinglylinked,weshouldalwaystestthesethreetestcasesbeforesayingthatthecodeisgoodtogo.Thisonenodeandzeronodecaseisusedtocatchboundarycases.Itisalwaystotakecareofthesecasesbeforesubmittingcodetothereviewer.
·Zeroelement/Emptylinkedlist.·Oneelement/Justsinglenodecase.·Generalcase.
Note:Anyprogramthatislikelytochangetheheadpointeristobepassedasadoublepointer.Basic operation of a linked list requires traversing a linked list. The various operations that we canperformonlinkedlists,manyoftheseoperationsrequirelisttraversal:
·Insertanelementinthelist,thisoperationisusedtocreatealinkedlist.·Printvariouselementsofthelist.·Searchanelementinthelist.·Deleteanelementfromthelist.·Reversealinkedlist.
YoucannotuseHeadtotraversealinkedlistbecauseifweusethehead,thenwelosethenodesofthelist.Wehavetouseanotherpointervariableofsamedata-typeasthehead.
InsertelementinlinkedlistAnelementcanbeinsertedintoalinkedlistinvariousorders.Someoftheexamplecasesarementionedbelow:
1.Insertionofanelementatthestartoflinkedlist
2.Insertionofanelementattheendoflinkedlist3.Insertionofanelementatthe2ndpositioninlinkedlist4.Insertelementinsortedorderinlinkedlist
Example7.11.structNode2.intvalue;3.Node*next;4.;5.typedefNode*NodePtr;1.intInsertNode(NodePtr*ptrHead,intvalue)2.3.printf("InsertNode:%d",value);4.NodePtrtempPtr=(NodePtr)malloc(sizeof(Node));5.if(!tempPtr)6.return-1;7.tempPtr->value=value;8.tempPtr->next=*ptrHead;9.*ptrHead=tempPtr;10.return1;11.
Analysis:Line1:DoublepointerptrHeadispassedtothefunctionasargument,aswewanttoassignthenewnodetotheheadofthelinkedlist.Line3:Valuepassedasargumentisprintedtostandardoutput.Line4:MemoryisallocatedforthenewnodeofthelistandispointedbytrmpPtr.Line5-6:Hereitischeckedifthesystemisabletoallocatememoryifmalloc()succeededinallocatingmemory it returns theaddressof thatmemory location.Moreover, ifmalloc() fails toallocatememory,thenitreturnsNULL.Line7:ThevaluepassedasargumentisstoredinthememorypointedbytempPtrLine8:Thenewnodenextpointerwillpointtotheheadoftheoriginallist.Line9:Theheadof theoriginal listwillnowstartpointing to tempPtr.Therebyaddinganodeat thebeginningofthelinkedlistisdone.
TraversingLinkedListExample7.2:Printvariouselementsofalinkedlist1.voidPrintList(NodePtrhead)2.3.while(head)4.5.printf("value%d\n",head->value);6.head=head->next;7.8.Analysis:Line1:Thisfunctiontakestheheadofthelistasinputargument.Line3:Onthisline,wearecheckingiftheheadisnotNULL.Iftheheadisnotnullthenwhileblockwillexecute.Line5:Itprintsvaluestoredasthevalueofthecurrentnode.Line6:Atthisline,weareincrementingtheheadpointer,sothatitwillpointtothenextelementofthelinkedlist.
Completecodeforlistcreationandprintingthelist.Example7.3:1.intmain()2.3.NodePtrhead=NULL;4.intarr[5]=1,2,3,4,5;5.inti;6.for(i=0;i<5;i++)7.8.InsertNode(&head,arr[i]);
9.10.return0;11.Analysis:Line3:Headpointerof the list is createdand it is assigned thevalueNULL.Headpointing toNULLmeansthelistisempty.Line4:Inthis,wehavecreatedanarrayof5elements.Theseelementswillbesortedinthelist.Line6-9:ValuestoredinarrayarestoredinlistbycallingInsertNodefunction.
InsertanelementattheendoflinkedlistgivenHeadpointerExample7.6:InsertanelementattheendoflinkedlistgivenHeadpointer1.intInsertAtEnd(NodePtr*ptrHead,intvalue)2.3.printf("insertelement%d\n",value);4.NodePtrhead=*ptrHead;5.NodePtrtempNode=(NodePtr)malloc(sizeof(Node));6.if(!tempNode)7.return-1;8.tempNode->value=value;9.tempNode->next=NULL;10.if(head==NULL)11.12.tempNode->next=*ptrHead;13.*ptrHead=tempNode;14.return1;15.16.while(head->next!=NULL)17.18.head=head->next;19.20.tempNode->next=head->next;21.head->next=tempNode;22.return1;23.Analysis:Line3-14:Newnodeiscreatedandthevalueisstoredinsideit.Ifthelistisempty,thenitwillbepointedbytheheadpointerptrHead.Line16-19:Willtraverseuntiltheendofthelist.Line20-21:Finally,newnodeisaddedtotheendofthelist.Note:Thisoperationisun-efficientaseachtimeyouwanttoinsertanelementyouhavetotraversetotheendofthelist.Therefore,thecomplexityofcreationofthelistisn2.Sohowtomakeitefficientwehave
tokeep trackof the last elementbykeepinga tail pointer.Therefore, if it is required to always insertelementattheendoflinkedlist,sothatwewillkeeptrackofthetailpointeralso.
InsertionofanelementattheendInsertionofanelementattheendoflinkedlist(givenheadpointer,andtailpointer)
Example7.5:1.intInsertNode(NodePtr*ptrHead,NodePtr*ptrTail,intvalue)2.3.printf("InsertNode::%d",value);4.NodePtrtempPtr=(NodePtr)malloc(sizeof(Node));5.if(!tempPtr)6.return-1;7.tempPtr->value=value;8.tempPtr->next=NULL;
9.if(*ptrHead==NULL)10.11.*ptrTail=*ptrHead=tempPtr;12.13.else14.15.NodePtrtail=*ptrTail;16.tail->next=tempPtr;17.*ptrTail=tempPtr;18.19.return1;20.Analysis:Line4-8:NewlinkedlistnodeiscreatedandthevaluestoredinsideitanditsnextpointstoNULL.Line9-12:Ifthelinkedlistisempty.ThenthenewnodewillbepointedbyptrTailandptrHead.Line15-17:Iflinkedlistisnotempty,thenlistptrHeadwillnotchange.OnlytheptrTailwillbechangedanditwillpointtothisnewnode.
SortedInsertInsertanelementinsortedorderinlinkedlistgivenHeadpointer
Example7.4:
1.intSortedInsert(NodePtr*ptrHead,intvalue)2.3.NodePtrcurr=*ptrHead;4.NodePtrtempNode=(NodePtr)malloc(sizeof(Node));5.printf("Insertelement%d\n",value);6.if(!tempNode)7.return-1;8.tempNode->value=value;9.tempNode->next=NULL;10.if(curr==NULL||curr->value>value)11.12.tempNode->next=*ptrHead;13.*ptrHead=tempNode;14.return1;15.16.while(curr->next!=NULL&&17.curr->next->value<value)18.19.curr=curr->next;20.21.tempNode->next=curr->next;22.curr->next=tempNode;23.return1;24.Analysis:Line3:Headofthelistisstoredincurrpointer.Line4-9:Anewemptynodeofthelinkedlistiscreated.Itisinitializedbystoringanargumentvalueintoitsvalue.Nextofthenodewillpointtonull.Line10-15:Itchecksifthelistwasemptyorifthevaluestoredinthefirstnodeisgreaterthanthecurrentvalue.Thenthisnewcreatednodewillbeaddedtothestartofthelist.Line16-20:Weiteratethroughthelisttofindtheproperpositiontoinsertthenode.Line21-23:Finally,thenodewillbeaddedtothelist.
SearchElementinaLinked-ListSearchelementinlinkedlist.Givenaheadpointerandvalue.Returns1ifvaluefoundinlistelsereturns0.Searchinasinglelinkedlistcanbeonlydoneinonedirection.Sinceallelementsinthelisthasreferencetothenextiteminthelist.Therefore,traversaloflinkedlistislinearinnature.Example7.7:1.intSearchList(NodePtrhead,intvalue)2.
3.while(head)4.5.If(head->value==value)6.7.printf(“Thevalueisfound”);8.return1;9.10.head=head->next;11.12.return0;13.Analysis:Line1:Wedonotneedtomodifythelistsopointertotheheadispassed(nodoublepointerneeded).Line3:Whileloopwilliteratethroughthelistandwill.Line5:Valueofeachelementoflistiscomparedwiththegivenvalueiffound,then“Thevalueisfound”willbeprintedtothescreen.Inaddition,functionwillreturn1.Line12:Ifthevalueisnotfound,then0willbereturnedfromthefunction.
Deleteelementfromthelinkedlist
DeleteFirstelementinalinkedlist.Example7.8:1.voidDeleteFirstNodes(NodePtr*ptrHead)2.
3.printf("\nDeleteFirstNode\n");4.NodePtrtempNode=*ptrHead;5.if(tempNode==NULL)6.return;7.*ptrHead=tempNode->next;8.free(tempNode);9.Analysis:·Weneedtofindthesecondelementofthelistandassignitasheadofthelinkedlist.·Ifthelisthasatleastoneelement,itsfirstelementwillbedeletedandtheheadofthelistwillpointto
thesecondelement.
Deletenodefromthelinkedlistgivenitsvalue.
Example7.9:1.voidDeleteNode(NodePtr*ptrHead,intdelValue)2.3.printf("\nDeleteNode\n");4.NodePtrcurrNode=*ptrHead;5.NodePtrnextNode;
6.7.if(currNode->value==delValue)/*firstnode*/8.9.*ptrHead=currNode->next;10.free(currNode);11.return;12.13.14.while(currNode!=NULL)15.16.nextNode=currNode->next;17.if(nextNode&&nextNode->value==delValue)18.19.currNode->next=nextNode->next;20.free(nextNode);21.return;22.23.else24.25.currNode=nextNode;26.27.28.Analysis:Line7-12:If thevaluestoredinfirstnodeisthevaluethatneedtodeleted,sotheheadofthelistwillchangesothiscaseishandledseparately.Else,innoothercaseheadoflinkedlistwillchange.Line14-27:Wetraversethelinklist inalooptofindthenodethatneedtobedeleted.WearekeepingnextNodeasthenextofcurrNode.WealwayscomparenextNodevaluewiththedelValueandcurrNode’snextpointerwillpointtothenextNode’snextpointer.ThennextNodeisthenodethatneedstobedeletedwhichisfinallyfreeatline20.
Deletealltheoccurrenceofparticularvalueinlinkedlist.Example7.10:1.voidDeleteNodes(NodePtr*ptrHead,intdelValue)2.3.printf("\nDeleteNode\n");4.NodePtrcurrNode=*ptrHead;5.NodePtrnextNode;6.NodePtrdelNode;7.8.while(currNode!=NULL&&currNode->value==delValue)/*firstnode*/
9.10.*ptrHead=currNode->next;11.delNode=currNode;12.currNode=currNode->next;13.free(delNode);14.15.16.while(currNode!=NULL)17.18.nextNode=currNode->next;19.if(nextNode&&nextNode->value==delValue)20.21.currNode->next=nextNode->next;22.free(nextNode);23.24.else25.26.currNode=nextNode;27.28.29.Analysis:Line8-14:Whileloopwilldeleteallthenodesthatareatthefrontofthelist,whichhavevaluedequaltodelValue.Line 16-28: In this while loop, we will be deleting all the nodes that are having value equal to thedelValue.
DeletenodefromthelinkedlistgivenitspointerExample7.11:1.voidDeleteNodePtr(NodePtr*ptrHead,NodePtrptrDel)2.3.printf("\ndeleteNode\n");4.NodePtrcurrNode=*ptrHead;5.NodePtrnextNode;6.7.if(ptrDel==NULL)8.return;9.10.if(currNode==ptrDel)//firstnode11.12.*ptrHead=currNode->next;13.free(currNode);14.
15.while(currNode!=NULL)16.17.nextNode=currNode->next;18.if(nextNode==ptrDel)//nodetobedeleated19.20.currNode->next=nextNode->next;21.free(nextNode);return;22.23.else24.25.currNode=nextNode;26.27.28.Analysis:Line10-14:Ifthenodethatneedtobedeletedisthefirstnode,thentheheadofthelinkedlistwillchangeandwillpointtothenextelementofthelinkedlist.Line15-27:Inthiswhile,loopthenodethatissupposedtobedeletedwillberemovedfromthelist.
DeleteasinglelinkedlistDeletealltheelementsofalinkedlist,givenapointertoheadoflinkedlist.Example7.12:1.voidDeleteList(NodePtr*ptrHead)2.3.printf("\nDeleteList\n");4.NodePtrdeleteMe=*ptrHead;5.NodePtrnextNode;6.while(deleteMe!=NULL)7.8.nextNode=deleteMe->next;9.free(deleteMe);10.deleteMe=nextNode;11.12.*ptrHead=NULL;13.Analysis:Line6-11:Inthiswhileloopeachnodeisdeletedonebyone.Untilwholelistisdeleted.Line12:InthislineheadoflinkedlistisassignedthevalueNULLtherebymakingthelistempty.
Reversealinkedlist.ReverseasinglylinkedListiterativelyusingthreePointers
Example7.13:1.voidReverseList(NodePtr*ptrHead)2.3.NodePtrcurrNode=*ptrHead;4.NodePtrprevNode;5.NodePtrnextNode;6.if(!currNode)7.8.return;9.10.if(!currNode->next)11.12.return;13.14.prevNode=currNode;15.currNode=currNode->next;16.prevNode->next=NULL;17.while(tempNode)18.19.nextNode=currNode->next;20.currNode->next=prevNode;21.prevNode=currNode;22.currNode=nextNode;23.24.25.*ptrHead=prevNode;26.Analysis:Line17-24:Thelistisiterated.MakenextNodeequaltothenextnodeofthecurrNode.MakecurrNodenode’s next will point to prevNode. Then iterate the list bymaking prevNode point to currNode andcurrNodepointtonextNode.
RecursivelyReverseasinglylinkedListExample7.14:RecursivelyReversesinglylinkedListArgumentsarecurrentnodeanditsnextvalue.1.NodePtrreverseRecurseUtil(NodePtrcurrentNode,NodePtrnextNode)2.3.NodePtrret;4.if(!currentNode)5.returnNULL;6.7.if(!currentNode->next)
8.9.currentNode->next=nextNode;10.returncurrentNode;11.12.13.ret=reverseRecurseUtil(currentNode->next,currentNode);14.currentNode->next=nextNode;15.16.returnret;17.18.19.voidreverseRecurse(NodePtr*ptrHead)20.21.*ptrHead=reverseRecurseUtil(*ptrHead,NULL);22.Analysis:Line19-22:reverseRecursefunctionwillcallareverseRecurseUtilfunctiontoreversethelistandthepointerreturnedbythereverseRecurseUtilwillbetheheadofthereversedlist.Line9&14:thecurrentnodewillpointtothenextNodethatispreviousnodeoftheoldlist.Note:Alinkedlistcanbereversedusingtwoapproachestheoneapproachisbyusingthreepointers.TheSecondapproachisusingrecursionbotharelinearsolution,butthree-pointersolutionismoreefficient.
RemoveduplicatesfromthelinkedlistRemove duplicate values from the linked list. The linked list is sorted and it contains some duplicatevalues, you need to remove those duplicate values. (You can create the required linked list usingSortedInsert()function)Example7.15:1.voidRemoveDuplicate(NodePtrhead)2.3.NodePtrdeleteMe;4.while(head)5.6.if((head->next)&&head->value==head->next->value)7.8.deleteMe=head->next;9.head->next=deleteMe->next;10.free(deleteMe);11.12.else13.14.head=head->next;15.
16.17.Analysis:Line4-16:While loop isused to traverse the list.Whenever there isanodewhosevalue isequaltothenextnode’svalue,thatnodewillberemovedfromthelistanddeletethatnode.
CopyListReversedCopythecontentoflinkedlist inanotherlinkedlist inreverseorder.If theoriginallinkedlistcontainselementsinorder1,2,3,4,thenewlistshouldcontaintheelementsinorder4,3,2,1.Example7.16:1.voidCopyListReversed(NodePtrhead,NodePtr*ptrHead2)2.3.printf("copylist");4.NodePtrtempNode=NULL;5.NodePtrtempNode2=NULL;6.while(head)7.8.tempNode2=(NodePtr)malloc(sizeof(Node));9.tempNode2->value=head->value;10.tempNode2->next=tempNode;11.tempNode=tempNode2;12.head=head->next;13.14.*ptrHead2=tempNode;15.Analysis: Traverse the list and add the node’s value to the new list. Since the list is traversed in theforwarddirectionandeachnode’svalueisaddedtoanotherlistsotheformedlistisreverseofthegivenlist.
CopythecontentofgivenlinkedlistintoanotherlinkedlistCopythecontentofgivenlinkedlistintoanotherlinkedlist.Iftheoriginallinkedlistcontainselementsinorder1,2,3,4,thenewlistshouldcontaintheelementsinorder1,2,3,4.Example7.17:1.voidCopyList(NodePtrhead,NodePtr*ptrHead2)2.3.printf("copylist");4.NodePtrheadNode=NULL;5.NodePtrtailNode=NULL;6.NodePtrtempNode=NULL;7.
8.if(head==NULL)9.return;10.headNode=(NodePtr)malloc(sizeof(Node));11.tailNode=headNode;12.headNode->value=head->value;13.headNode->next=NULL;14.head=head->next;15.16.while(head)17.18.tempNode=(NodePtr)malloc(sizeof(Node));19.tempNode->value=head->value;20.tempNode->next=NULL;21.tailNode->next=tempNode;22.tailNode=tailNode->next;23.head=head->next;24.25.*ptrHead2=headNode;26.Analysis:Traversethelistandaddthenode’svaluetonewlist,butthistimealwaysattheendofthelist.Sincethelististraversedintheforwarddirectionandeachnode’svalueisaddedtotheendofanotherlist.Therefore,theformedlistissameasthegivenlist.
CompareListExample7.18:1.intcompareList(NodePtrhead1,NodePtrhead2)2.3.printf("comparelist");4.if(head1==NULL&&head2==NULL)5.return1;6.elseif((head1==NULL)||(head2==NULL)||(head1->value!=head2->value))7.return0;8.else9.returncompareList(head1->next,head2->next);10.Analysis:Line4-5:List iscomparedrecursively.Moreover, ifwereach theendof the listandboth the listsarenull.Thenboththelistsareequalandsoreturn1.Line6-7:List is compared recursively. If eitheroneof the list is emptyor thevalueofcorrespondingnodesisunequal,thenthisfunctionwillreturn0.Line9:Recursivelycallscomparelistfunctionforthenextnodeofthecurrentnodes.
CompareListExample7.19:1.intcompareList2(NodePtrhead1,NodePtrhead2)2.3.while(head1!=NULL&&head2!=NULL)4.5.if(head1->value!=head2->value)6.return0;7.head1=head1->next;8.head2=head2->next;9.10.if(head1==head2)11.12.return1;13.14.else15.16.return0;17.18.Analysis:Line3-9:Boththelistsaretraverseduntilonelistisemptyortherevalueofthecorrespondingnodeisunequal.Line11-17:Ifboth the listarereachedto theend,whichmeansboth thelistareequalsoreturn1elsereturn0.
FindLengthExample7.20:Findthelengthofgivenlinkedlist.1.intfindLength(NodePtrhead)2.3.intcount=0;4.while(head)5.6.count++;7.head=head->next;8.9.returncount;10.Analysis:Lengthoflinkedlistisfoundbytraversingthelistuntilwereachtheendoflist.
NthNodefromBeginning
Example7.21:1.NodePtrnthNodeFromBegining(NodePtrhead,intindex)2.3.intcount=0;4.while(head&&count<index-1)5.6.count++;7.head=head->next;8.9.10.if(head)11.returnhead;12.else13.returnNULL;14.Analysis:NthnodecanbefoundbytraversingthelistN-1numberoftimeandthenreturnthenodeifitisnotNULLelsereturnNULL.
NthNodefromEndExample7.22:1.NodePtrnthNodeFromEnd(NodePtrhead,intindex)2.3.intsize=findLength(head);4.intstartIndex;5.if(size&&size<index)6.7.printf("listdoesnothave%elements",index);8.returnNULL;9.10.startIndex=size-index+1;11.returnnthNodeFromBegining(head,startIndex);12.Analysis:First, find the lengthof list, thennthnode fromendwill be (length–nth+1)node from thebeginning.Example7.23:1.NodePtrnthNodeFromEnd(NodePtrhead,intindex)2.3.intcount=0;4.NodePtrtemp=NULL;5.NodePtrcurr=head;
6.while(curr&&count<index-1)7.8.count++;9.curr=curr->next;10.11.12.if(!curr)13.returnNULL;14.15.temp=head;16.17.while(curr)18.19.temp=temp->next;20.curr=curr->next;21.22.returntemp;23.Analysis:SecondapproachistousetwopointersoneisNsteps/nodesaheadoftheotherwhenforwardpointerreachtheendofthelistthenthebackwardpointerwillpointtothedesirednode.Example7.241.NodePtrnthNodeFromEnd(NodePtrhead,intindex)2.3.staticintcount=0;4.NodePtrretval;5.6.if(!head)7.returnNULL;8.9.retval=nthNodeFromEnd3(head->next,index);10.if(retval)11.returnretval;12.13.count++;14.if(count==index)15.returnhead;16.else17.returnNULL;18.
LoopDetect1.Traversethroughthelist.
2.IfthecurrentnodeisnotpresentintheHash-TabletheninsertitintotheHash-Table.3.Ifthecurrentnodeisalreadyinthehashtablethenwehavealoop.
LoopDetectWehavetofindifthereisaloopinthelinkedlist.Therearetwowaystofindifthereisaloopinalinkedlist.Onewayiscalled“Slowpointerandfastpointerapproach(SPFP)”theotheriscalled“Reverselistapproach”.Bothapproachesarelinearinnature,butstillinSPFPapproach,wedonotrequiretomodifythelinkedlistsoitispreferred.
Findifthereisaloopinalinkedlist.Ifthereisaloop,thenreturn1ifnot,thenreturn0.Useslowpointerfastpointerapproach.
Example7.25:1.intLoopDetect(NodePtrhead)2.3.printf("loopdetect");4.NodePtrslowPtr;5.NodePtrfastPtr;6.slowPtr=fastPtr=head;7.8.while(fastPtr->next&&fastPtr->next->next)9.10.slowPtr=slowPtr->next;11.fastPtr=fastPtr->next->next;
12.if(slowPtr==fastPtr)13.14.return1;15.16.17.return0;18.Analysis:Line8-16:Thelististraversedwithtwopointers,oneisslowpointerandanotherisfastpointer.Slowpointer alwaysmovesone-step.Fastpointer alwaysmoves two steps. If there isno loop, thencontrolwillcomeoutofwhileloop.Soreturn0.Line12-15:Ifthereisaloop,thentherecameapointinaloopwherethefastpointerwillcomeandtrytopassslowpointer.Whenthispointarrives,wecometoknowthatthereisaloopinthelist.Soreturn1.
ReverseListLoopDetectFindifthereisaloopinalinkedlist.Ifthereisaloop,thenreturn1ifnot,thenreturn0.Usereverselistapproach.Example7.26:1.intReverseListLoopDetect(NodePtrhead)2.3.NodePtr*ptrHead=&head;4.NodePtrhead2=head;5.reverseList(ptrHead);6.if(*ptrHead==head2)7.8.reverseList(ptrHead);9.return1;10.11.else12.13.reverseList(ptrHead);14.return0;15.16.Analysis:Line5:reversethelistatthisline.Line6:comparethereversedlistheadpointertothecurrentlistheadpointer.Line8-9:Iftheheadofreversedlistandtheoriginallist,aresamethenreversethelistbackandreturn1.Line13-14:Iftheheadofthereversedlistandtheoriginallistarenotsamethenreversethelistbackandreturn0.Whichmeansthereisnoloop.
LoopTypeDetectFindifthereisaloopinalinkedlist.Ifthereisnoloop,thenreturn0,ifthereisloopreturn1,ifthelistiscircularthen2.Useslowpointerfastpointerapproach.Example7.27:1.intLoopTypeDetect(NodePtrconsthead)2.3.printf("loopdetect");4.NodePtrslowPtr;5.NodePtrfastPtr;6.7.slowPtr=fastPtr=head;8.while(fastPtr->next&&fastPtr->next->next)9.10.11.slowPtr=slowPtr->next;12.fastPtr=fastPtr->next->next;13.14.if(head==fastPtr->next||head==fastPtr->next->next)15.16.return2;17.18.if(slowPtr==fastPtr)19.20.return1;21.22.23.return0;24.Analysis:ThisprogramissameastheloopdetectprogramexceptLine14-17.Line 14-17: If fast pointer reaches to the head of the list, then this means that there is a loop at thebeginningofthelist.
RemoveLoopExample7.28:1.voidRemoveLoop(NodePtr*ptrHead)2.3.intloopLength;4.NodePtrslowPtr,fastPtr,head;5.slowPtr=fastPtr=head=*ptrHead;
6.NodePtrloopNode=NULL;7.while(fastPtr->next&&fastPtr->next->next)8.9.fastPtr=fastPtr->next->next;10.slowPtr=slowPtr->next;11.12.if(fastPtr==slowPtr||fastPtr->next==slowPtr)13.14.loopNode=slowPtr;15.break;16.17.18.if(loopNode)19.20.NodePtrtemp=loopNode->next;21.loopLength=1;22.while(temp!=loopNode)23.24.loopLength++;25.temp=temp->next;26.27.temp=head;28.NodePtrbreakNode=head;29.30.for(inti=1;i<loopLength;i++)31.32.breakNode=breakNode->next;33.34.35.while(temp!=breakNode->next)36.37.temp=temp->next;38.breakNode=breakNode->next;39.40.breakNode->next=NULL;41.42.Analysis:Loopthroughthelistbytwopointer,onefastpointerandoneslowpointer.Fastpointerjumpstwonodesat a time and slowpointer jump one node at a time.The pointwhere these two pointer intersect is apointerintheloop.Thensincewehaveapointerinaloop,wecanfindthelengthoftheloop.TakeapointerbreakNodethatpoint to theheadof the linkedlist. Increment the listbylooplengthand
then increment theheadandbreakNodepointeruntilbothareequal.WhenyouhaveabreakNode thenassignitsnexttoNULL.Thereisonemoresolutionpossibletofindthebreakpoint.Thepointwhereslowpointerandfastpointermeets in loop,call it looppoint.Startonepointer fromheadandother fromlooppoint incrementbothone-stepata time theywillmeetat the looppoint.Mark thenextof looppointnullandyouaredone.(Thiscanbeprovedmathematically)
FindIntersection
Example7.29:1.NodePtrfindIntersection(NodePtrhead,NodePtrhead2)2.3.intl1=0;4.intl2=0;5.NodePtrtempHead=head;6.NodePtrtempHead2=head2;7.8.while(tempHead)9.10.l1++;11.tempHead=tempHead->next;12.13.while(tempHead2)14.15.l2++;16.tempHead2=tempHead2->next;17.18.19.intdiff;20.if(l1<12)21.22.NodePtrtemp=head;23.head=head2;24.head2=temp;25.diff=l2-l1;26.27.else28.
29.diff=l1-l2;30.31.32.for(;diff>0;diff--)33.34.head=head->next;35.36.while(head!=head2)37.38.head=head->next;39.head2=head2->next;40.41.42.returnhead;43.Analysis:Findlengthofboththelists.Findthedifferenceoflengthofboththelists.Incrementthelongerlistbydiffsteps,thenincrementboththelistsandgettheintersectionpoint.
DoublyLinkedListInaDoublyLinkedlist,therearetwopointersineachnode.Thesepointersarecalledprevandnext.Theprevpointerofthenodewillpointtothenodebeforeitandthenextpointerwillpointtothenodenexttothegivennode.
LetuslookatNodeinthisexamplethevalueisoftypeint,butitcanbeofsomeotherdata-type.Thetwolinkpointersareprevandnext.WehavetypedeftheNode*toNodePtrsothatourcodelooksclean.structNode
intvalue;structNode*next;structNode*prev;
;typedefNode*NodePtr;
Searchinasinglelinkedlistcanbeonlydoneinonedirection.Sinceallelementsinthelisthasreferencetothenextiteminthelist.Therefore,traversaloflinkedlistislinearinnature.Inadoublylinkedlist,wekeeptrackofbothheadofthelinkedlistandtailoflinkedlist.Note:Foradoublylinkedlist,fewcasesneedtokeepinmindwhilecoding:
·Zeroelementcase(headandtailbothcanbemodified)·Onlyelementcase(headandtailbothcanbemodified)·Firstelement(headcanbemodified)·Generalcase·Thelastelement(tailcanbemodified)
Note:Anyprogramthatislikelytochangeheadpointerortailpointeristobepassedasadoublepointer,whichispointingtoheadortailpointer.
TraversingLinkedList
Basic operation of a linked list requires traversing a linked list. The various operations that we canperformonlinkedlists,manyoftheseoperationsrequirelisttraversal:
·Insertanelementinthelist,thisoperationisusedtocreatealinkedlist.·Printvariouselementsofthelist.·Searchanelementinthelist.·Deleteanelementfromthelist.·Reversealinkedlist.
YoucannotuseHeadtotraversealinkedlistbecauseifweusethehead,thenwelosethenodesofthelist.Wehavetouseanotherpointervariableofsamedata-typeasthehead.Thelisttraversalcodeismentionednext.NodePtrcurrent=head;while(current!=NULL)
/*Processcurrentnode*/current=current->next;
Foranylinkedlistthereareonlythreecaseszeroelement,oneelement,andgenerallyAnyprogramthatislikelytochangetheheadpointeristobepassedadoublepointerFordoublylinkedlist,wehaveafewmorethings1.NULLvalues(headandtailbothcanbemodified)2.Onlyelement(headandtailbothcanbemodified)3.Firstelement(headcanbemodified)4.Generalcase5.Lastelement(tailcanbemodified)
1.structNode2.intvalue;3.structNode*next;4.structNode*prev;5.;6.typedefNode*NodePtr;
InsertNodeExample7.30:1./*INSERTVALUEINFRONT*/2.intinsertNode(NodePtr*ptrHead,NodePtr*ptrTail,intvalue)3.4.printf("\nINSERTNODE\n");5.NodePtrtemp=(NodePtr)malloc(sizeof(Node));6.if(!temp)7.return0;8.NodePtrhead=*ptrHead;9.if(!head)10.11.temp->value=value;12.temp->next=NULL;13.temp->prev=NULL;14.*ptrTail=temp;15.*ptrHead=temp;16.17.else18.19.temp->value=value;20.temp->next=head;21.temp->prev=NULL;22.head->prev=temp;23.*ptrHead=temp;24.25.return1;26.Analysis:Insertindoublelinkedlistissameasinsertinasinglylinkedlist.CreateanodeassignNULLtoprevpointerofthenode.Thenpointnexttothestartofthelist.IfthelistisemptythenptrTailwillpoint
tothenode.
SortedInsertDecreasing
Example7.31:1.//SORTEDINSERTDECREASING2.intsortedInsert(NodePtr*ptrHead,NodePtr*ptrTail,intvalue)3.4.printf("\nsortedinsert\n");5.NodePtrtemp=(NodePtr)malloc(sizeof(Node));6.if(!temp)7.return0;8.temp->value=value;9.NodePtrhead=*ptrHead;10.if(!head)//firstelement11.12.temp->next=NULL;13.temp->prev=NULL;14.*ptrHead=temp;15.*ptrTail=temp;16.return1;17.18.if(head->value<=value)//atthebegining19.
20.temp->next=head;21.temp->prev=NULL;22.head->prev=temp;23.*ptrHead=temp;24.return1;25.26.while(head->next&&head->next->value>value)//treversal27.28.head=head->next;29.30.if(!head->next)//attheend31.32.*ptrTail=temp;33.temp->next=NULL;34.temp->prev=head;35.head->next=temp;36.37.else///allother38.39.temp->next=head->next;40.temp->prev=head;41.head->next=temp;42.temp->next->prev=temp;43.44.return1;45.Analysis:Findtheproperlocationofthenodeandaddittothelist.Managenextandprevpointerofthenodesothatlistalwaysremaindoublelinkedlist.Example7.32:1./*2.PrintAsinglylinkedlist3.*/4.voidprintList(NodePtrhead)5.6.printf("Listis::");7.while(head!=NULL)8.9.printf("%d",head->value);10.head=head->next;11.12.printf("\n");13.
Analysis:Printlistissameasjusttraversingthelistandprintingthevaluesofthenodes.
ReverseadoublylinkedListiterativelyExample7.33:1./*2.ReverseadoublylinkedListiteratively3.*/4.5.voidreverseList(Node**ptrHead,Node**ptrTail)6.7.NodePtrhead=*ptrHead;8.NodePtrtempNode;9.while(head)10.11.tempNode=head->next;12.head->next=head->prev;13.head->prev=tempNode;14.if(!head->prev)15.16.*ptrTail=*ptrHead;17.*ptrHead=head;18.return;19.20.head=head->prev;21.22.return;23.Analysis:Traversethelistandswapnextandprevpointerofeachnode.WhenyouhavereachedtotheendofthelistswapptrTailandptrHeadpointers.Example7.34:Deleteasinglylinkedlist1.voiddeleteList(NodePtr*ptrHead,NodePtr*ptrTail)2.3.printf("\ndeleteList\n");4.NodePtrdeleteMe=*ptrHead;5.NodePtrnextNode;6.while(deleteMe!=NULL)7.8.nextNode=deleteMe->next;9.free(deleteMe);10.deleteMe=nextNode;11.
12.*ptrTail=ptrHead=NULL;13.Analysis:Traversethelistanddeleteeachnodeofthelistonebyone.
Deleteanodegivenitspointer
Example7.35:1.voiddeleteNodePtr(NodePtr*ptrHead,NodePtr*ptrTail,NodePtrptrDel)2.3.printf("\ndeleteNode\n");4.NodePtrcurrNode=*ptrHead;5.NodePtrnextNode;6.7.if(ptrDel==NULL||currNode==NULL)8.return;9.10.if(currNode==ptrDel)//firstnode11.12.deleteMe=currNode;
13.currNode=currNode->next;14.free(deleteMe);15.*ptrHead=currNode;16.if(currNode)17.currNode->prev=NULL;18.else19.*ptrTail=NULL;20.return;21.22.while(currNode!=NULL)23.24.nextNode=currNode->next;25.if(nextNode==ptrDel)//nodetobedeleated26.27.currNode->next=nextNode->next;28.nextNode->next->prev=currNode;29.if(nextNode==*ptrTail)30.*ptrTail=nextNode->prev;//lastnode31.free(nextNode);32.33.else34.35.currNode=nextNode;36.37.38.Analysis:Traversethelistfindthenodethatneedtobedeleted.Thenremoveitandadjustnextpointerofthenodebeforeitandprevpointerofthenodenexttoit.
RemoveDuplicateExample7.36:1./*RemoveDuplicate*/2.voidremoveDuplicate(Node*head)3.4.NodePtrdeleteMe;5.while(head)6.7.if((head->next)&&head->value==head->next->value)8.9.deleteMe=head->next;10.head->next=deleteMe->next;11.head->next->prev=head;12.if(deleteMe==tail)13.
14.tail=curr;15.16.free(deleteMe);17.18.else19.20.head=head->next;21.22.23.Analysis:Removeduplicate is same as single linked list case.Consider the list as sorted remove therepeatedvaluenodesofthelist.
CopyListReversedExample7.37:1.voidcopyListReversed(NodePtrhead,NodePtr*ptrHead2)2.3.printf("copylist");4.NodePtrtempNode=NULL;5.NodePtrtempNode2=NULL;6.while(head)7.8.tempNode2=(NodePtr)malloc(sizeof(Node));9.tempNode2->value=head->value;10.tempNode2->next=tempNode;11.tempNode2->prev=NULL;12.tempNode->prev=tempNode2;13.tempNode=tempNode2;14.head=head->next;15.16.*ptrHead2=tempNode;17.Analysis:Traversethroughthelistandcopythevalueofthenodesintoanotherlist.Alwaysaddthenextnodevaluetothestartofthenewlisttherebyformingareversedlist.
CopyListExample7.38:1.voidcopyList(NodePtrhead,NodePtr*ptrHead2)2.3.printf("copylist");4.NodePtrheadNode=NULL;
5.NodePtrtailNode=NULL;6.NodePtrtempNode=NULL;7.if(head==NULL)8.return;9.headNode=(NodePtr)malloc(sizeof(Node));10.tailNode=headNode;11.headNode->value=head->value;12.headNode->next=NULL;13.headNode->prev=NULL;14.head=head->next;15.while(head)16.17.tempNode=(NodePtr)malloc(sizeof(Node));18.tempNode->value=head->value;19.tempNode->next=NULL;20.tailNode->next=tempNode;21.tempNode->prev=tailNode;22.tailNode=tailNode->next;23.head=head->next;24.25.*ptrHead2=headNode;26.Analysis:Traversethelist,valueofthenodesareaddedtoanotherlistalwaysattheend.Usetailpointertokeeptrackoftheendofthenewlist.
CircularLinkedListThistypeissimilartothesinglylinkedlistexceptthatthelastelementpointstothefirstnodeofthelist.Thelinkportionofthelastnodecontainstheaddressofthefirstnode.
typedefstructNodeintvalue;structNode*next;
Node_t;typedefNode_t*NodePtr;
Insertelementinfront
Example7.39:1./*Insertelementinfront*/2.intinsertNodeAtFront(NodePtr*ptrHead,NodePtr*ptrTail,intvalue)3.4.printf("\nINSERTNODEATFRONT\n");5.NodePtrtemp=(NodePtr)malloc(sizeof(Node_t));6.if(!temp)7.return0;8.temp->value=value;9.NodePtrhead=*ptrHead;10.NodePtrtail=*ptrTail;11.if(!head)12.13.temp->next=temp;14.*ptrTail=temp;15.*ptrHead=temp;16.17.else18.19.temp->next=head;20.tail->next=temp;21.*ptrHead=temp;22.23.return1;24.Analysis:Line11-16:Whenthelistisempty,thenodeiscreated.Headandtailispointingtothisnode.Line17-22:Newnodeiscreateditsnextwillpoint toheadandheadwillpoint tothisnewnode.Tailnodesnextwillpointtothenewnode.
InsertelementattheendExample7.40:1./*Insertelementattheend*/2.intinsertNodeAtEnd(NodePtr*ptrHead,NodePtr*ptrTail,intvalue)
3.4.printf("\nINSERTNODEATEND\n");5.NodePtrtemp=(NodePtr)malloc(sizeof(Node_t));6.if(!temp)7.return0;8.temp->value=value;9.NodePtrhead=*ptrHead;10.NodePtrtail=*ptrTail;11.if(!head)12.13.temp->next=temp;14.*ptrTail=temp;15.*ptrHead=temp;16.17.else18.19.temp->next=head;20.tail->next=temp;21.*ptrTail=temp;22.23.return1;24.Analysis:Addingnodeattheendissameasaddingatthebeginning.Justneedtomodifytailpointerinplaceoftheheadpointer.
PrintListExample7.41:1./*PrintAsinglylinkedlist*/2.voidprintList(NodePtrhead)3.4.printf("LISTIS::");5.NodePtrcurrNode=head;6.if(currNode!=NULL)7.8.printf("%d",currNode->value);9.currNode=currNode->next;10.11.while(currNode!=head)12.13.printf("%d",currNode->value);14.currNode=currNode->next;15.16.
Analysis:Incircularlist,endoflistisnottheresowecannotcheckwithNULL.InplaceofNULLheadisusedtocheckendofthelist.
DeleteListExample7.42:1./*Deleteacircularlinkedlist*/2.voiddeleteList(NodePtr*ptrHead)3.4.printf("\nDELETELIST\n");5.NodePtrconsthead=*ptrHead;6.NodePtrcurrNode=*ptrHead;7.NodePtrnextNode;8.if(currNode!=NULL)9.10.nextNode=currNode->next;11.free(currNode);12.currNode=nextNode;13.14.while(currNode!=head)15.16.nextNode=currNode->next;17.free(currNode);18.currNode=nextNode;19.20.*ptrHead=NULL;21.Analysis:Traversethroughthelistanddeletenodesuntilyoureachthestartofthelist.
Deleteanodegivenitspointer
Example7.43:1./*Deleteanodegivenitspointer*/2.voiddeleteNodePtr(NodePtr*ptrHead,NodePtr*ptrTail,NodePtrptrDel)3.4.printf("\nDELETENODEGIVENITSPOINTER\n");5.if(ptrDel==NULL||ptrHead==NULL||ptrTail==NULL)6.return;7.NodePtrhead=*ptrHead;8.NodePtrtail=*ptrTail;9.NodePtrcurrNode=head;10.NodePtrprevNode;11.if(head==NULL||tail==NULL)12.return;13.if(currNode==ptrDel)/*oneelementandfirstelementcase*/14.15.if(currNode->next==currNode)16.17.*ptrHead=NULL;18.*ptrTail=NULL;19.free(currNode);20.return;21.22.else23.24.*ptrHead=currNode->next;25.tail->next=currNode->next;26.free(currNode);27.return;28.29.
30.prevNode=currNode;31.currNode=currNode->next;32.while(currNode!=head)33.34.if(currNode==ptrDel)35.36.if(currNode==tail)/*tailchangecase*/37.*ptrTail=prevNode;38.prevNode->next=currNode->next;39.free(currNode);40.return;41.42.prevNode=currNode;43.currNode=currNode->next;44.45.Analysis:Findthenodethatneedtofree.Onlydifference is thatwhile traversingthe listendof list istrackedbytheheadpointerinplaceofNULL.
DeleteanodegivenitsvalueExample7.44:1./*Deleteanodegivenitsvalue*/2.voiddeleteNodeValue(NodePtr*ptrHead,NodePtr*ptrTail,intvalue)3.4.printf("\nDELETENODEGIVENITSVALUE\n");5.if(ptrHead==NULL||ptrTail==NULL)6.return;7.NodePtrhead=*ptrHead;8.NodePtrtail=*ptrTail;9.NodePtrcurrNode=head;10.NodePtrprevNode;11.if(head==NULL||tail==NULL)12.return;13.if(currNode->value==value)/*oneelementandfirstelementcase*/14.15.if(currNode->next==currNode)16.17.*ptrHead=NULL;18.*ptrTail=NULL;19.free(currNode);20.return;21.22.else
23.24.*ptrHead=currNode->next;25.tail->next=currNode->next;26.free(currNode);27.return;28.29.30.prevNode=currNode;31.currNode=currNode->next;32.while(currNode!=head)33.34.if(currNode->value==value)35.36.if(currNode==tail)/*tailchangecase*/37.*ptrTail=prevNode;38.prevNode->next=currNode->next;39.free(currNode);40.return;41.42.prevNode=currNode;43.currNode=currNode->next;44.45.Analysis:Findthenodethatneedtofree.Onlydifference is thatwhile traversingthe listendof list istrackedbytheheadpointerinplaceofNULL.
RemoveDuplicateExample7.45:1./*RemoveDuplicate*/2.voidremoveDuplicate(NodePtr*ptrHead,NodePtr*ptrTail)3.4.printf("\nRemoveDuplicate\n");5.NodePtrhead=*ptrHead;6.NodePtrcurrent=head;7.NodePtrdeleteMe;8.if(!head)9.return;10.while(current->next!=head)11.12.if(current->value==current->next->value)13.14.deleteMe=current->next;15.current->next=deleteMe->next;
16.if(deleteMe==*ptrTail)17.18.*ptrTail=current;19.20.free(deleteMe);21.22.else23.24.current=current->next;25.26.27.Analysis:Consideringthelistissorted,repeatedvaluenodeswouldberemoved.OnlydifferenceisthatwhiletraversingthelistendofthetraversalisnotcheckedwithNULLitischeckedbyheadpointer.
CopyListReversedExample7.46:1./*CopyList*/2.voidcopyListReversed(NodePtrhead,NodePtr*newPtrHead,NodePtr*newPtrTail)3.4.printf("\nCOPYLISTREVERSED\n");5.NodePtrcurr=head;6.if(curr)7.8.insertNodeAtFront(newPtrHead,newPtrTail,curr->value);9.curr=curr->next;10.11.while(curr!=head)12.13.insertNodeAtFront(newPtrHead,newPtrTail,curr->value);14.curr=curr->next;15.16.Analysis:Thelististraversedandnodesareaddedtonewlistatthebeginning.Therebymakingthenewlistreverseofthegivenlist.
CopyListExample7.47:1.voidcopyList(NodePtrhead,NodePtr*newPtrHead,NodePtr*newPtrTail)2.3.printf("\nCOPYLIST\n");4.NodePtrcurr=head;
5.if(curr)6.7.insertNodeAtEnd(newPtrHead,newPtrTail,curr->value);8.curr=curr->next;9.10.while(curr!=head)11.12.insertNodeAtEnd(newPtrHead,newPtrTail,curr->value);13.curr=curr->next;14.15.Analysis:Lististraversedandnodesareaddedtothenewlistattheend.Therebymakingthelistwhosevaluearesameastheinputlist.
DoublyCircularlist1.Foranylinkedlistthereareonlythreecaseszeroelement,oneelement,generalcase2.Anyprogramwhichislikelytochangetheheadpointeristobepassedadoublepointer3.Todoublylinkedlistwehaveafewmorethings
a)NULLvaluesb)Onlyelement(itgenerallyintroducesanifstatementwithnull)c)Alwaysan“if”before“while”.Whichwillcheckfromthishead.d)Generalcase(checkwiththeinitialheadkept)e)Avoidusingrecursionsolutionsitmakeslifeharder
STLimplementlistusingadoublecircularlist.
InsertNodeExample7.48:Insertvalueatthefrontofthelist.1.#include<stdio.h>2.#include<stdlib.h>3.4.structNode5.intvalue;6.structNode*next;7.structNode*prev;8.;9.typedefNode*NodePtr;10.11.//Insertvalueinfront12.intinsertNode(NodePtr*ptrHead,intvalue)13.14.printf("\nINSERTNODE\n");15.NodePtrtemp=(NodePtr)malloc(sizeof(Node));16.if(!temp)17.return0;18.NodePtrhead=*ptrHead;19.if(!head)20.21.temp->value=value;22.temp->next=temp;23.temp->prev=temp;24.*ptrHead=temp;
25.26.else27.28.temp->value=value;29.temp->next=head;30.temp->prev=head->prev;31.temp->prev->next=temp32.head->prev=temp;33.*ptrHead=temp;34.35.return1;36.
DeleteNodePtrExample7.49:1./*2.Deleteanodegivenitspointer3.*/4.5.voiddeleteNodePtr(NodePtr*ptrHead,NodePtrptrDel)6.7.printf("\ndeleteNode\n");8.NodePtrcurrNode=*ptrHead;9.NodePtrhead=*ptrHead;10.NodePtrnextNode,prevNode;11.12.if(ptrDel==NULL||!(*ptrHead))13.return;14.15.if(currNode==ptrDel)//firstnode16.17.if(currNode->next==currNode)//onlynode18.19.*ptrHead=NULL;20.free(currNode);21.return;22.23.else24.25.prevNode=currNode->prev;26.*ptrHead=nextNode=currNode->next;27.prevNode->next=nextNode;28.nextNode->prev=prevNode;29.return;
30.31.32.currNode=currNode->next;33.while(currNode!=head)34.35.if(currNode==ptrDel)36.37.prevNode=currNode->prev;38.nextNode=currNode->next;39.prevNode->next=nextNode;40.nextNode->prev=prevNode;41.free(currNode);42.return;43.44.currNode=currNode->next;45.46.Analysis:Deletenodeinadoublycircularlinkedlistisjustsameasdeletenodeinacircularlinkedlist.Justfewextraprevpointerneedtobeadjusted.
RemoveDuplicateExample7.50:1.voidremoveDuplicate(NodePtrhead)2.3.NodePtrdeleteMe;4.NodePtrconsttagHead=head;5.6.if(!head)7.return;8.9.head=head->next;10.11.while(head!=tagHead)12.13.if(head==head->next)//thischeckistopreventonlyonenodetobedeleted14.break;15.if((head->next)&&head->value==head->next->value)16.17.deleteMe=head->next;18.head->next=deleteMe->next;19.head->next->prev=head;20.free(deleteMe);21.
22.else23.24.//checkfortail25.head=head->next;26.27.28.Analysis:Removeduplicate is sameas removeduplicate in a circular linked list. Just onemoreprevpointerneedtobeadjusted.
CopyListReversedExample7.51:1.voidcopyListReversed(NodePtrhead,NodePtr*ptrHead2)2.3.printf("copylist");4.NodePtrtempNode=NULL;5.NodePtrtempNode2=NULL;6.NodePtrconsthead2=head;7.8.if(head)9.10.tempNode2=(NodePtr)malloc(sizeof(Node));11.tempNode2->value=head->value;12.tempNode2->next=tempNode;13.tempNode2->prev=NULL;14.tempNode=tempNode2;15.head=head->next;16.17.18.while(head!=head2)19.20.tempNode2=(NodePtr)malloc(sizeof(Node));21.tempNode2->value=head->value;22.tempNode2->next=tempNode;23.tempNode2->prev=NULL;24.tempNode=tempNode2;25.head=head->next;26.27.*ptrHead2=tempNode;28.Analysis:Copylistissimilartocopylistinacircularlist.Justprevpointeralsoneedtobeadjusted.
CopyListExample7.52:1.voidcopyList(NodePtrhead,NodePtr*ptrHead2)2.3.printf("copylist");4.NodePtrheadNode=NULL5.NodePtrtempNode=NULL;6.NodePtrconsttagHead=head;7.NodePtrtailNode;8.9.if(head==NULL)10.return;11.12.tailNode=headNode=(NodePtr)malloc(sizeof(Node));13.headNode->value=head->value;14.headNode->next=headNode;15.headNode->prev=headNode;16.head=head->next;17.18.while(head!=tagHead)19.20.tempNode=(NodePtr)malloc(sizeof(Node));21.tempNode->value=head->value;22.tailNode->next=tempNode;23.tempNode->prev=tailNode;24.tempNode->next=headNode;25.headNode->prev=tempNode;26.head=head->next;27.28.*ptrHead2=headNode;29.Analysis:Copylistissimilartocopylistinacircularlist.Justprevpointeralsoneedtobeadjusted.
Exercise
1) Insertanelementkth position from the startof linked list.Return1 if successand if list isnot longenough,thenreturn-1.Hint:TakeapointeradvanceitKstepsforward,theninsertsthenode.
2) Insert an element kth position from the endof linked list.Return1 if success and if list is not long
enough,thenreturn-1.Hint:TakeapointeradvanceitKstepsforward,thentakeanotherpointerandadvancebothofthemsimultaneously,sothatwhenthefirstpointerreachtheendofalinkedlistthatisthepointwhereyouneedtoinsertthenode.
3)Considerthereisaloopinalinkedlist,Writeaprogramtoremoveloopifthereisaloopinthislinked
list.4)IntheaboveSearchListprogramreturn,thecountofhowmanyinstancesofsamevaluefoundelseif
valuenot found then return 0.For example, if the value passed is “4”.The elements in the list are1,2,4,3&4.Theprogramshouldreturn2.Hint:Inplaceofreturn1intheaboveprogramincrementacounterandthenreturncounterattheend.
5) Given two linked list head, pointer and they meet at some point and need to find the point of
intersection.However,inplaceoftheendofboththelinkedlisttobeanullpointerthereisaloop.
6)Iflinkedlisthavingaloopisgiven.Countthenumberofnodesinthelinkedlist7)WeweresupposedtowritethecompletecodefortheadditionofpolynomialsusingLinkedLists.8)Giventwolinkedlists.Wehavetofindthatwhetherthedatainoneisreversethatofdatainanother.
Noextraspaceshouldbeusedandtraversethelinkedlistsonlyonce.9)Findthemiddleelementinasinglylinkedlist.Tellthecomplexityofyoursolution.
Hint:-·Approach1:findthelengthoflinkedlist.Thenfindthemiddleelementandreturnit.·Approach2: use two pointer onewillmove fast and onewillmove slowmake sure you handle
bordercaseproperly.(Evenlengthandoddlengthlinkedlistcases.)10)Printlistinreverseorder.
Hint:Userecursion.
CHAPTER8:STACK
IntroductionA stack is a basic data structure that organized items in last-in-first-out (LIFO)manner. Last elementinsertedinastackwillbethefirsttoberemovedfromit.Thereal-lifeanalogyofthestackis"chapattisinhotpot","stackofplates".Imagineastackofplatesinadiningareaeverybodytakesaplateatthetopofthestack,therebyuncoveringthenextplateforthenextperson.Stackallowtoonlyaccessthetopelement.Theelementsthatareatthebottomofthestackaretheonethatisgoingtostayinthestackforthelongesttime.
Computersciencealsohas thecommonexampleofastack.Functioncallstack isagoodexampleofastack. Function main() calls function foo() and then foo() calls bar(). These function calls areimplementedusingstackfirstbar()exists,thengo()andthenfinallymain().Aswenavigatefromwebpagetowebpage,theURLofwebpagesarekeptinastack,withthecurrentpageURLatthetop.Ifweclickbackbutton,theneachURLentryispoppedonebyone.
TheStackAbstractDataTypeStackabstractdatatypeisdefinedasastructure,whichfollowsLIFOorlast-in-first-outfortheelements,addedtoit.Thestackshouldsupportthefollowingoperation:
1.Push():whichaddasingleelementatthetopofthestack2.Pop():whichremoveasingleelementfromthetopofastack.3.Top():Readsthevalueofthetopelementofthestack(doesnotremoveit)4.isEmpty():Returns1ifstackisempty5.Size():returnsthenumberofelementsinastack.
voidpush(intn);Addntothetopofastack.intpop();Removethetopelementofthestackandreturnittothecallerfunction.Thestackcanbeimplementedusinganarrayoralinkedlist.Inarraycase,therearetwotypesofimplementations
·Oneinwhicharraysizeisfixed,soitthecapacityofthestack.·Anotherapproachisvariablesizearrayinwhichmemoryofthearrayisallocatedusingmallocand
whenthearrayisfilledthesizeifdoubledusingrealloc(whenthestacksizedecreasesbelowhalfthecapacityisagainreducedusingrealloc).
Incaseofalinkedlist,thereisnolimitofnumberofelements.Whenastackisimplemented,usinganarraytopofthestackismanagedusinganindexvariablecalledtop.Whenastackisimplementedusingalinkedlist,push()andpop()isimplementedusinginsertattheheadofthelinkedlistandremovefromtheheadofthelinkedlist.
StackusingArray(Macro)
Implementastackusingafixedlengtharray.Example8.1:#defineMAX_WIDTH50typedefstructstack
inttop;intdata[MAX_WIDTH];
Stack;ThecapacityofthestackisdefinedbyMAX_WIDTHcompiletimeconstant.Stack is defined which will contain the array to store the data and index to indicate the number ofelementsinthestack.voidStackInitialize(Stack*stk)
stk->top=-1;Numberofelementsinthestackisgovernedbythe“top”indexandtopisinitializedto-1whenastackisinitialized.Topindexvalueof-1indicatesthatthestackisemptyinthebeginning.voidStackPush(Stack*stk,intvalue)
if(stk->top<MAX_WIDTH-1)stk->top++;stk->data[stk->top]=value;//printf("valuepush:%d\n",value);elseprintf("stackoverflow\n");
StackPush() function checks whether the stack has enough space to store one more element, then itincreasesthe"top"byone.Finallysortthedatainthestack"data"array.Incase,stackisfullthen"stackoverflow"messageisprintedandthatvaluewillnotbeaddedtothestackandwillbeignored.intStackPop(Stack*stk)
if(stk->top>=0)
intvalue=stk->data[stk->top];stk->top--;//printf("valuepop:%d\n",value);returnvalue;printf("stackempty\n");return0;
TheStackPop()functionis implemented,first itwillcheckthat therearesomeelementsinthestackbycheckingitstopindex.Ifsomeelementisthereinthestack,thenitwillstorethetopmostelementvalueinavariable"value".Thetopindexisreducedbyone.Finally,thatvalueisreturned.intStackTop(Stack*stk)
intvalue=stk->data[stk->top];returnvalue;
StackTop()functionreturnsthevalueofstoredinthetopelementofstack(doesnotremoveit)intStackIsEmpty(Stack*stk)
return(stk->top==-1);StackIsEmpty() function returns1 if stack isemptyor0 inallothercases.Bycomparing the top indexvaluewith-1.intStackSize(Stack*stk)
return(stk->top+1);StackSize() function returns the number of elements in the stack. It just returns "top+1".As the top isreferringthearrayindexofthestacktopvariablesoweneedtoaddonetoit.Analysis:·Theuserofthestackwillcreateastacklocalvariable.·Thenwillinitit.·Usepush()andpop()functionstoadd/removevariablestothestack.·Readthetopelementusingthetop()functioncall.·Queryregardingsizeofthestackusingsize()functioncall·QueryifstackisemptyusingisEmpty()functioncall
StackusingArray(Dynamicmemory)
Createastackusinganarray,butthememoryofthearrayneedtobetakenfromheap/orthememoryneedtobedynamicallyallocated.Example8.2:1.#include"stdio.h"2.#include"stdlib.h"3.4.typedefstructstack5.inttop;6.int*data;7.intmax;8.Stack;9.10.voidStackInitialize(Stack*stk,intsize)11.12.stk->data=(int*)malloc(size*sizeof(int));13.stk->top=-1;14.stk->max=size;15.16.17.intmain()18.19.Stacks;20.StackInitialize(&s,10);21.for(inti=0;i<20;i++)22.StackPush(&s,i);23.for(inti=0;i<20;i++)24.StackPop(&s);25.return0;26.Analysis:Line1:4-8:stackstructureisdefinedwhichcontainits“top”index,Apointerdatathatwillbeusedtopoint tomemoryallocatedusingmalloc inLine12.Wekeeptrackof thecapacityofanarray inamaxvariableofastackstructure.Line 10: in init() function, we are passing the stack pointer and the size of memory that need to beallocatedtothestack.Line 12: we are allocating the memory sufficient to store "size" number of integer s. Moreover, thismemorylocationisstoredinthedatafieldofastackstructure.Maxissettothesizeofthestackpassed.Line19-20:Stackvariableiscreatedandinitialized,astackisinitializedsothatitwillbeabletocontain
atmost10elements.Line22:pushisusedtoaddelementstothestack.Line24:popisusedtoreadandremovethetopelementofthestack.Note:All theother function like,StackPush(),StackPop(),StackIsEmpty(),StackSize() andStackTop()willworkfordynamicallyallocatedstackarraytoo.
StackusingArray(Growingcapacityimplementation)
Intheabovedynamicarrayimplementationofastack.Makethecapacityofstackvariablesothatwhenitisnearlyfilled,thendoublethecapacityofthestack.Example8.3:1.voidStackPush(Stack*stk,intvalue)2.3.4.if(stk->top<stk->max-1)5.6.stk->top++;7.stk->data[stk->top]=value;8.printf("valuepush:%d\n",value);9.10.else11.12.stk->max=stk->max*2;13.stk->data=(int*)realloc(stk->data,stk->max*sizeof(int));14.printf("stacksizedoubled");15.StackPush(stk,value);16.17.Analysis:Line12:Inthisline,wearedoublingthemaxcapacityvariableofthestack.Line13:Sizeofthememoryisreallocatedtoabiggervalueinthisline.Thereallocfunctionisusedtoincrease or decrease the size of memory allocated. All the data on the stack is copied into the newlocation.Line15:finallytheStackPush()functioniscalledrecursivelytopushthevalueintotheincreasedcapacitystack.
StackusingArray(Growing-Reducingcapacityimplementation)
Youcanalsowriteaprogram,whichcanreducethesizeofadynamicarraybytwowhenthenumberofelementsfallbelowmax/2.Youdonotwant to let the capacityof the stackbelow the initially allocated size.Youcandefineminlengthwheninitiscalled.Example8.4:1.typedefstructstack2.inttop;3.int*data;4.intmax;5.intmin;6.Stack;7.8.voidinitStack(Stack*stk,intsize)9.10.stk->data=(int*)malloc(size*sizeof(int));11.stk->top=-1;12.stk->max=size;13.stk->min=size;14.15.16.intStackPop(Stack*stk)17.18.if(stk->top>=0)19.20.intvalue=stk->data[stk->top];21.stk->top--;22.if(stk->top<(stk->max/2)&&stk->max>stk->min)23.24.stk->max=stk->max/2;25.stk->data=(int*)realloc(stk->data,stk->max*sizeof(int));26.printf("stacksizehalfed");27.28.printf("valuepop:%d\n",value);29.returnvalue;30.31.printf("stackempty\n");32.33.34.intmain()35.36.Stacks;37.initStack(&s,10);
38.for(inti=0;i<20;i++)39.StackPush(&s,i);40.for(inti=0;i<20;i++)41.StackPop(&s);42.return0;43.Analysis:Line 22: the size of the stack is checked if it goes below half of the capacity and is greater than theminimumsize.Line 23-25: the capacity of the stack is divided by 2 andmemory is reallocated to point to half sizememory.
Stackusinglinkedlist
Example8.5:1.#include<stdlib.h>2.#include<stdio.h>3.4.structstackNode_t5.intvalue;6.stackNode_t*next;7.;8.typedefstackNode_t*stackPtr;9.#defineERROR_VALUE-999991.voidStackPush(stackPtr*dPtrHead,intvalue)2.3.stackPtrtempNode=(stackPtr)malloc(sizeof(stackNode_t));4.if(!tempNode)5.6.Printf("memoryshortageunabletopush");7.return;8.9.tempNode->value=value;10.tempNode->next=*dPtrHead;11.*dPtrHead=tempNode;12.Analysis:Stackimplementedusingalinkedlistissimplyinsertionanddeletionattheheadofthelinkedlist.InStackPush()function,memoryiscreatedforonenode.Thenthevalueisstoredintothatnode.Finally,thenodeisinsertedatthebeginningofthelist.
PopNodeImplementpopNodefunctionthatwillreturnafirstnodepointerofthestackExample8.6:1.stackPtrpopNode(stackPtr*dPtrHead)//freethereturnednodeyourself2.3.stackPtrdeleteMe;4.if(*dPtrHead)5.6.deleteMe=*dPtrHead;7.*dPtrHead=deleteMe->next;8.returndeleteMe;9.
10.else11.12.printf("stackempty\n");13.returnNULL;14.15.Analysis:Line 1-15: popNode() functionwill take the first element of the list andwill return its pointer to thecaller.Thecallerofthisfunctionwillberesponsibleforthedeletionofthememory.Line13:Incasethestackisempty,NULLwillbereturned.
PopExample8.7:Implementthepopfunctionthatwillreturnthevalueatthetopofthestack.1.intpop(stackPtr*dPtrHead)//freethereturnednodeyourself2.3.stackPtrdeleteMe;4.intvalue;5.if(*dPtrHead)6.7.deleteMe=*dPtrHead;8.*dPtrHead=deleteMe->next;9.value=deleteMe->data;10.free(deleteMe);11.returnvalue;12.13.else14.15.printf("stackempty\n");16.returnERROR_VALUE;17.18.Analysis:Line 1-18: in pop() function first element of the list is pointed by a temporary pointer variable"deleteMe".Thevalueofthefirstnodeisstoredintoalocalvariable"value".Headofthelistwillpointtothenextelement.ThenthememorypointedbydeleteMewillbefreedusingfree()andfinallyvalueisreturned.Line16:Incasestackisempty,ERROR_VALUEwillbereturned.
ProblemsinStack
BalancedParenthesisExample8.8: Stacks can be used to check a program for balanced symbols (such as , (), []). Theclosingsymbolshouldbematchedwiththemostrecentlyseenopeningsymbol.Example:() is legal,()()islegal,but((and()arenotlegalintisBalancedParenthesis(char*expn)
Stackstk;StackInitialize(&stk);inti=0;charch;while((ch=expn[i++])!='\0')switch(ch)case'':case'[':case'(':StackPush(&stk,ch);break;
case'':if(StackPop(&stk)!='')return0;break;
case']':if(StackPop(&stk)!='[')return0;break;
case')':if(StackPop(&stk)!='(')return0;break;returnStackIsEmpty(&stk);
Analysis:Traversetheinputstringwhenwegetanopeningparenthesiswepushitintostack.Whenwegetaclosing
parenthesisthenwepopaparenthesisfromthestackandcompareifitisthecorrespondingtotheoneontheclosingparenthesis.Wereturn0/errorifthereisamismatchofparenthesis.Ifattheendofthewholestaringtraversal,wereachedtotheendofthestringandthestackisemptythenwehavebalancedparenthesis.intmain()
charexpn[50]="()";intvalue=isBalancedParenthesis(expn);printf("\nGivenExpn:%s\n",expn);printf("\nResultafterisParenthesisMatched:%d\n",value);return0;
Infix,PrefixandPostfixExpressionsWhenwe have an algebraic expression likeA +B thenwe know that the variable is being added tovariable B. This type of expression is called infix expression because the operator “+” is betweenoperandsAandoperandB.NowconsideranotherinfixexpressionA+B*C.Intheexpressionthereisaproblemthatinwhichorder+and*works.DoesAandBareaddedfirstandthentheresultismultiplied.OrBandCaremultipliedfirstandthentheresultisaddedtoA.Thismakestheexpressionambiguous.Todealwiththisambiguitywedefinetheprecedenceruleoruseparenthesestoremoveambiguity.So ifwewant tomultiplyBandCfirstand thenaddtheresult toA.ThenthesameexpressioncanbewrittenunambiguouslyusingparenthesesasA+(B*C).Ontheotherhand,ifwewanttoaddAandBfirstand then thesumwillbemultipliedbyCwewillwrite itas (A+B)*C.Therefore, in the infixexpressiontomaketheexpressionunambiguous,weneedparenthesis.Infixexpression:Inthisnotation,weplaceoperatorinthemiddleoftheoperands.<operand><operator><operand>Prefixexpressions:Inthisnotation,weplaceoperatoratthebeginningoftheoperands.<operator><operand><operand>Postfixexpression:Inthisnotation,weplaceoperatorattheendoftheoperands.<operand><operand><operator>InfixExpression PrefixExpression PostfixExpression
A+B +AB AB+
A+(B*C) +A*BC ABC*+
(A+B)*C *+ABC AB+C*
NowcomesthemostobviousquestionwhyweneedsounnaturalPrefixorPostfixexpressionswhenwealreadyhaveinfixexpressionswhichwordsjustfineforus.The answer to this is that infix expressions are ambiguous and they need parenthesis to make themunambiguous.Whilepostfixandprefixnotationsdonotneedanyparenthesis.
Infix-to-PostfixConversionExample8.9:voidinfixToPostfix(char*expn,char*output)
Stackstk;StackInitialize(&stk);charch,op;inti=0;intindex=0;intdigit=0;while((ch=expn[i++])!='\0')if(isdigit(ch))output[index++]=ch;digit=1;elseif(digit)output[index++]='';digit=0;switch(ch)case'+':case'-':case'*':case'/':case'%':case' ':while(!StackIsEmpty(&stk)&&precedence(ch)<=precedence(StackTop(&stk)))op=StackPop(&stk);output[index++]=op;output[index++]='';StackPush(&stk,ch);
break;case'(':StackPush(&stk,ch);break;case')':while(!StackIsEmpty(&stk)&&(op=StackPop(&stk))!='(')output[index++]=op;output[index++]='';break;while(!StackIsEmpty(&stk))op=StackPop(&stk);output[index++]=op;output[index++]='';output[index++]='\0';
intprecedence(charx)
if(x=='(')return(0);if(x=='+'||x=='-')return(1);if(x=='*'||x=='/'||x=='%')return(2);if(x==' ')return(3);return(4);
Analysis:1.Printoperandsinthesameorderastheyarrive.2.Ifthestackisemptyorcontainsaleftparenthesis“(”ontop,weshouldpushtheincomingoperatorinthestack.3.Iftheincomingsymbolisaleftparenthesis”(”,pushleftparenthesisinthestack.4.Iftheincomingsymbolisarightparenthesis“)”,popfromthestackandprinttheoperatorstillyouseealeftparenthesis“)”.Discardthepairofparentheses.5.Iftheprecedenceofincomingsymbolishigherthantheoperatoratthetopofthestack,thenpushittothestack.6.Iftheincomingsymbolhasanequalprecedencecomparedtothetopofthestackuseassociation.Ifthe
associationislefttoright,thenpopandprintthesymbolatthetopofthestackandthenpushtheincomingoperator.Iftheassociationisrighttoleft,thenpushtheincomingoperator.7.Iftheprecedenceofincomingsymbolislowerthantheoperatoronthetopofthestack,thenpopandprint the top operator. Then compare the incoming operator against the new operator at the top of thestack.8.Attheendoftheexpression,popandprintalloperatorsonthestack.
Infix-to-PrefixConversionExample8.10:voidinfixToPrefix(char*expn,char*output)
reverseString(expn);replaceParanthesis(expn);infixToPostfix(expn,output);reverseString(output);
voidreverseString(char*a)
intlower=0;intupper=strlen(a)-1;chartempChar;while(lower<upper)tempChar=a[lower];a[lower]=a[upper];a[upper]=tempChar;
lower++;upper--;
voidreplaceParanthesis(char*a)
intlower=0;intupper=strlen(a)-1;chartempChar;while(lower<=upper)if(a[lower]=='(')a[lower]=')';elseif(a[lower]==')')a[lower]='(';
lower++;
intmain()
charexpn[50]="10+((3))*5/(16-4)";printf("\nGivenExpn:%s\n",expn);infixToPostFix(expn);
charexpn2[50]="(5*3) (4-2)";printf("\nGivenExpn:%s\n",expn2);infixToPostFix(expn2);
return0;
Analysis:1.Reversethegiveninfixexpression.2.Replace'('with')'and')'with'('inthereversedexpression.3.Nowapplyinfixtopostfixsubroutinealreadydiscussed.4.Reversethegeneratedpostfixexpressionandthiswillgiverequiredprefixexpression.
PostfixEvaluateWriteapostfixEvaluate()functiontoevaluateapostfixexpression.Suchas:12+34+*Example8.11:intpostfixEvaluate(char*postfx)
Stacks;StackInitialize(&s);inti=0,op1,op2;charch;intdigit=0;intvalue=0;while((ch=postfx[i++])!='\0')if(isdigit(ch))digit=1;value=value*10+(ch-'0');elseif(ch=='')
if(digit==1)StackPush(&s,value);/*Pushtheoperand*/digit=0;value=0;elseop2=StackPop(&s);op1=StackPop(&s);switch(ch)case'+':StackPush(&s,op1+op2);break;case'-':StackPush(&s,op1-op2);break;case'*':StackPush(&s,op1*op2);break;case'/':StackPush(&s,op1/op2);break;returnStackTop(&s);
intmain()
charpostfx[50]="6523+8*+3+*";intvalue=postfixEvaluate(postfx);printf("\nGivenPostfixExpn:%s\n",postfx);printf("\nResultafterEvaluation:%d\n",value);return0;
Analysis:1)Createastacktostorevaluesoroperands.2)Scanthroughthegivenexpressionanddofollowingforeachelement:
a)Iftheelementisanumber,thenpushitintothestack.b) If the element is an operator, then pop values from the stack. Evaluate the operator over the
valuesandpushtheresultintothestack.3)Whentheexpressionisscannedcompletely,thenumberinthestackistheresult.
MinstackDesignastackinwhichgetminimumvalueinstackshouldalsoworkinO(1)TimeComplexity.Hint:Keeptwostackonewillbegeneralstack,whichwilljustkeeptheelements.Thesecondwillkeep
theminvalue.1.Push:Pushanelementtothetopofstack1.Comparethenewvaluewiththevalueatthetopofthe
stack2. If the new value is smaller, then push the new value into stack2.Alternatively, push thevalueatthetopofthestack2toitselfoncemore.
2.Pop:Popanelementfromtopofstack1andreturn.Popanelementfromtopofstack2too.3.Min:Readfromthetopofthestack2thisvaluewillbethemin.
PalindromestringFindifgivenstringisapalindromeornotusingastack.Definition of palindrome: A palindrome is a sequence of characters, which is same backward, orforward.Eg.“AAABBBCCCBBBAAA”,“ABA”&“ABBA”Hint:Pushcharacters to thestackuntil thehalf-lengthof thestring.Thenpop thesecharactersand thencompare.Makesureyoutakecareoftheoddlengthandevenlength.
ReverseStackGivenastackhowtoreversetheelementsofthestackwithoutusinganyotherdata-structure.Youcannotuseanotherstacktoo.TimeComplexityandSpaceComplexityiswrong,itisO(n)forbothcases.Hint: Use recursion (system stack.) When you go inside the stack pop elements from stack in eachsubsequent call until stack is empty. Then push these elements one by one when coming out of therecursion.Theelementswillbereversed.Example8.12:voidreverseStack(Stack*stk)
intdata;if(StackIsEmpty(stk))return;data=StackPop(stk);reverseStack(stk);insertAtBottom(stk,data);
InsertAtBottomExample8.13:voidinsertAtBottom(Stack*stk,intvalue)
if(StackIsEmpty(stk))StackPush(stk,value);
elseinttemp=StackPop(stk);insertAtBottom(stk,value);StackPush(stk,temp);
Depth-FirstSearchwithaStackInadepth-firstsearch,wetraversedownapathuntilwegetadeadend;thenwebacktrackbypoppingastacktogetanalternativepath.·Createastack·Createastartpoint·Pushthestartpointontothestack·While(valuesearchingnotfoundandthestackisnotempty)
oPopthestackoFindallpossiblepointsaftertheonewhichwejusttriedoPushthesepointsontothestack
StackusingaqueueHowtoimplementastackusingaqueue.Analysetherunningtimeofthestackoperations.Seequeuechapterforthis.
TwostacksusingsinglearrayExample8.14:Howtoimplementtwostacksusingonesinglearray.#defineMAX_SIZE50typedefstructstack
inttop1;inttop2;intdata[MAX_SIZE];
Stack;voidStackInitialize(Stack*stk)
stk->top1=-1;stk->top2=MAX_SIZE;
voidStackPush1(Stack*stk,intdata)
if(stk->top1<stk->top2-1)stk->data[++stk->top1]=data;elseprintf("StackisFull!\n");
intStackPop1(Stack*stk)
if(stk->top1>=0)intvalue=stk->data[stk->top1--];printf("%disbeingpoppedfromStack1\n",value);returnvalue;elseprintf("StackEmpty!CannotPop\n");returnINT_MIN;
voidStackPush2(Stack*stk,intdata)
if(stk->top1<stk->top2-1)stk->data[--stk->top2]=data;elseprintf("StackisFull!\n");
intStackPop2(Stack*stk)
if(stk->top2<MAX_SIZE)intvalue=stk->data[stk->top2++];printf("%disbeingpoppedfromStack2\n",value);returnvalue;
elseprintf("StackEmpty!CannotPop\n");returnINT_MIN;
Analysis:Samearrayisusedtoimplementtwostack.Firststackisfilledfromthebeginningofthearrayandsecondstackisfilledfromtheendofthearray.Overflowandunderflowconditionsneedtobetakencareofcarefully.
StockSpanProblem
Approach1:int*StockSpanRange(intarr[],intsize)
int*SR=(int*)malloc(size*sizeof(int));SR[0]=1;
for(inti=1;i<size;i++)SR[i]=1;for(intj=i-1;(j>=0)&&(arr[i]>=arr[j]);j--)SR[i]++;returnSR;
Approach2:int*StockSpanRange(intarr[],intsize)
Stackstk;int*SR=(int*)malloc(size*sizeof(int));StackInitialize(&stk);
StackPush(&stk,0);SR[0]=1;
for(inti=1;i<size;i++)while(!StackIsEmpty(&stk)&&arr[StackTop(&stk)]<=arr[i])StackPop(&stk);SR[i]=(StackIsEmpty(&stk))?(i+1):(i-StackTop(&stk));StackPush(&stk,i);returnSR;
GetMaxRectangularAreainaHistogram
Approach1intGetMaxArea(intarr[],intsize)
intmaxArea=-1;intcurrArea;intminHeight=0;for(inti=1;i<size;i++)minHeight=arr[i];for(intj=i-1;j>=0;j--)if(minHeight>arr[j])minHeight=arr[j];
currArea=minHeight*(i-j+1);
if(maxArea<currArea)maxArea=currArea;
returnmaxArea;
Approach2:DivideandconquerApproach3intGetMaxArea(intarr[],intsize)
Stackstk;StackInitialize(&stk);intmaxArea=0;inttop;inttopArea;inti=0;while(i<size)while((i<size)&&(StackIsEmpty(&stk)||arr[StackTop(&stk)]<=arr[i]))StackPush(&stk,i);i++;while(!StackIsEmpty(&stk)&&(i==size||arr[StackTop(&stk)]>arr[i]))top=StackTop(&stk);StackPop(&stk);topArea=arr[top]*(StackIsEmpty(&stk)?i:i-StackTop(&stk)-1);if(maxArea<topArea)maxArea=topArea;returnmaxArea;
Prosandconsofarrayandlinkedlistimplementationofstack.Linkedlists:Listimplementationuses1pointerextramemoryperitem.Thereisnosizerestriction.Arrays:Allocatedaconstantamountofspace,whenthestackisnearlyempty,thenlostofspaceiswasteasitisnotused.Maximumsizeisdeterminedwhenthestackiscreated.
UsesofStack·Recursioncanalsobedoneusingstack.(Inplaceofthesystemstack)·Thefunctioncallisimplementedusingstack.·Someproblemswhenwewanttoreverseasequence,wejustpusheverythinginstackandpopfromit.·Grammar checking, balance parenthesis, infix to postfix conversion, postfix evaluation of expression
etc.
ExerciseEx1:ConvertingDecimalNumberstoBinaryNumbersusingstackdatastructure.Hint:storeremindersintothestackandthenprintthestack.Ex2:Convertaninfixexpressiontoprefixexpression.Hint:Reversegivenexpression,Applyinfixtopostfix,andthenreversetheexpressionagain.
Step1.Reversetheinfixexpression.5^E+D*)C^B+A(
Step2.MakeEvery'('as')'andevery')'as'('5^E+D*(C^B+A)
Step3.Convertanexpressiontopostfixform.Step4.Reversetheexpression.
+*+A^BCD^E5Ex3:WriteanHTMLopeningtagandclosingtag-matchingprogram.Hint:parenthesismatching.Ex4:WriteafunctionthatwilldoPostfixtoInfixConversionEx5::WriteafunctionthatwilldoPrefixtoInfixConversionEx6 :Write a palindromematching function,which ignore characters other thanEnglish alphabet anddigits.String"Madam,I'mAdam."shouldreturntrue.
CHAPTER9:QUEUE
IntroductionAqueue is abasicdata structure thatorganized items in first-in-first-out (FIFO)manner.First elementinsertedintoaqueuewillbethefirsttoberemoved.Itisalsoknownas"first-come-first-served".Thereallifeanalogyofqueueistypicallinesinwhichweallparticipatetimetotime.
·Wewaitinalineofrailwayreservationcounter.·Wewaitinthecafeterialine(topopaplatefrom“stackofplates”).·Wewaitinaqueuewhenwecalltosomecustomercase.
Theelements,whichareatthefrontofthequeue,aretheone,whichstayedinthequeueforthelongesttime.
Computersciencealsohascommonexamplesofqueues.Weissueaprintcommandfromourofficetoasingleprinterperfloor,printtaskislinedupinaprinterqueue.Theprintcommandthatwasissuedfirstwillbeprintedbeforethenextcommandsinline.In addition to printing queues, operating system is also using different queues to control processscheduling.Processesareaddedtoprocessingqueue,whichisusedbyanoperatingsystemforvariousschedulingalgorithms.Soonwewillbereadingaboutgraphsandwillcometoknowaboutbreadth-firsttraversal,whichusesaqueue.
TheQueueAbstractDataTypeQueueabstractdatatypeisdefinedasastructurethatfollowsFIFOorfirst-in-first-outfortheelementsaddedtoit.Queueshouldsupportthefollowingoperation:1.enqueue():Whichaddasingleelementatthebackofaqueue2.dequeue():Whichremoveasingleelementfromthefrontofaqueue.3.isEmpty():Returns1ifthequeueisempty4.size():Returnsthenumberofelementsinaqueue.
QueueUsingArray
Example9.1:#defineSIZE100#defineERROR_VALUE-999typedefstructQueue_t
intfront;intback;intsize;intdata[SIZE];
Queue;voidQueueInitialize(Queue*que)
que->back=0;que->front=0;que->size=0;
intQueueIsEmpty(Queue*que)
returnque->size==0;intQueueSize(Queue*que)
returnque->size;voidEnqueue(Queue*que,intvalue)
if(que->size>=SIZE)
printf("\nQueueisfull.");return;elseque->size++;que->data[que->back]=value;que->back=(++(que->back))%(SIZE-1);//printf("\nenqueue:%d",value);
intDequeue(Queue*que)
intvalue;if(que->size<=0)printf("\nQueueisempty.");returnERROR_VALUE;elseque->size--;value=que->data[que->front];que->front=(++(que->front))%(SIZE-1);returnvalue;
intmain()
Queueque;QueueInitialize(&que);for(inti=0;i<20;i++)Enqueue(&que,i);for(inti=0;i<20;i++)Dequeue(&que);return0;
Analysis:1.Hearqueueiscreatedfromanarrayofsize100.2.QueueInitializefunctioninitializethenumberofelementinqueuetozero.Byassigningfront,backand
sizeofqueuetozero.3.Enqueueinsertoneelementatthebackofthequeue.4.Dequeuedeleteoneelementfromthefrontofthequeue.
QueueUsinglinkedliststructqueueNode_t
intvalue;queueNode_t*next;
;typedefqueueNode_t*queuePtr;#defineERROR_VALUE-99999Anodeofaqueueusinglinkedlistlookssameaslinkedlist.
EnqueueEnqueue into a queue using linked list.Nodes are added to the end of the linked list.Belowdiagramindicateshowanewnodeisaddedtothelist.Thetailismodifiedeverytimewhenanewvalueisaddedto thequeue.However, thehead isalsoupdated in thecasewhen there isnoelement in thequeueandwhenthatfirstelementisaddedtothequeuebothheadandtailwillbepointingtoit.
Example9.2:1.voidenqueue(queuePtr*dPtrHead,queuePtr*dPtrTail,intvalue)2.
3.queuePtrtempNode=(queuePtr)malloc(sizeof(queueNode_t));4.if(!tempNode)5.6.Printf("memoryshortageunabletoenqueue");7.return;8.9.tempNode->value=value;10.tempNode->next=NULL;11.if(*dPtrHead==NULL)12.13.*dPtrTail=tempNode;14.*dPtrHead=tempNode;15.16.else17.18.(*dPtrTail)->next=tempNode;19.*dPtrTail=tempNode;20.21.Analysis:EnqueueoperationaddoneelementattheendoftheQueue(linkedlist).
Dequeue
Inthisweneedthetailpointerasitmaybethecasetherewasonlyoneelementinthelistandthetailpointerwillalsobemodifiedincaseofthedequeue.Example9.3:1.intdequeue(queuePtr*dPtrHead,queuePtr*dPtrTail)
2.3.intvalue;4.queuePtrdeleteMe;5.if(*dPtrHead)6.7.deleteMe=*dPtrHead;8.*dPtrHead=deleteMe->next;9.value=deleteMe->value;10.free(deleteMe);11.if(*dPtrHead==NULL)12.*dPtrTail=NULL;13.returnvalue;14.15.else16.17.Printf("queueempty\n");18.returnERROR_VALUE;19.20.Analysis:Dequeueoperationremovesfirstnodefromthestartofthequeue(linkedlist).
ProblemsinQueue
QueueusingastackHowtoimplementaqueueusingastack.Youcanusemorethanonestack.Solution:Wecanusetwostacktoimplementqueue.Weneedtosimulatefirstinfirstourusingstack.a)EnqueueOperation:newelementsareaddedtothetopoffirststack.b)DequeueOperation:elementsarepoppedfromthesecondstack.Whensecondstackisemptythenall
theelementsoffirststackarepoppedandpushedintosecondstackonebyone.Example9.4:#include"Stack.h"typedefstructSQueue_t
Stackstk1;Stackstk2;
SQueue;voidQueueInitialize(SQueue*que)
StackInitialize(&que->stk1);StackInitialize(&que->stk2);
voidEnqueue(SQueue*que,intvalue)
StackPush(&que->stk1,value);intDequeue(SQueue*que)
intvalue;if(!StackIsEmpty(&que->stk2))returnStackPop(&que->stk2);while(!StackIsEmpty(&que->stk1))value=StackPop(&que->stk1);StackPush(&que->stk2,value);returnStackPop(&que->stk2);
intmain()
SQueueque;QueueInitialize(&que);Enqueue(&que,1);Enqueue(&que,11);Enqueue(&que,111);printf("%d",Dequeue(&que));printf("%d",Dequeue(&que));printf("%d",Dequeue(&que));
StackusingaQueueImplementstackusingaqueue.Solution1:usetwoqueuePush:enqueuenewelementstoqueue1.Pop:whilesizeofqueue1isbiggerthan1.Pushallitemsfromqueue1toqueue2exceptthelastitem.Switchthenameofqueue1andqueue2.Andreturnthelastitem.PushoperationisO(1)andPopoperationisO(n)Solution2:Thissamecanbedoneusingjustonequeue.Push:enqueuetheelementtoqueue.Pop: find the size of queue. If size is zero then return error. If size is positive then dequeue size- 1elementsfromthequeueandagainenqueuetothesamequeue.Atlast,dequeuethenextelementandreturnit.PushoperationisO(1)andPopoperationisO(n)Solution3:IntheabovesolutionsthepushisefficientandpopisunefficientcanwemakepopefficientO(1)andpushinefficientO(n)Push : enqueue new elements to queue2. Then enqueue all the elements of queue 1 to queue 2. Thenswitchnamesofqueue1andqueue2.Pop:dequeuefromqueue1PushoperationisO(n)andPopoperationisO(1)
ReverseastackReverseastackusingaqueueSolution:a)Popalltheelementsofstackandenqueuethemintoaqueue.b)Thendequeuealltheelementsofthequeueintostackc)Wehavetheelementsofthestackreversed.
Reverseaqueue
ReverseaqueueusingastackSolution:a)Dequeuealltheelementsofthequeueintostackb)Thenpopalltheelementsofstackandenqueuethemintoaqueue.c)Wehavetheelementsofthequeuereversed.
Breadth-FirstSearchwithaQueueIn breadth-first search, we explore all the nearest nodes first by finding all possible successors andenqueuethemtoaqueue.a)Createaqueueb)Createastartpointc)Enqueuethestartpointontothequeued)while(valuesearchingnotfoundandthequeueisnotempty)
oDequeuefromthequeueoFindallpossiblepointsafterthelastonetriedoEnqueuethesepointsontothequeue
JosephusproblemTherearenpeople standing in aqueuewaiting tobe executed.Thecountingbegins at the frontof thequeue. Ineachstep,knumberofpeoplearedequeuedandagainenqueuedonebyone from thequeue.Then the next person is executed. The execution proceeds around the circle until only the last personremains,whoisgivenfreedom.Findthatpositionwhereyouwanttostandandgainyourfreedom.Solution:1)Justinsertintegerfor1tokinaqueue.(correspondstokpeople)2)DefineaKpop()functionsuchthatitwilldequeueandenqueuethequeuek-1timesandthendequeue
onemoretime.(Thismanisdead.)3)Repeatsecondstepuntilsizeofqueueis1.4)Printthevalueinthelastelement.Thisisthesolution.
Exercise
1) Implement queue using dynamicmemory allocation. Such that the implementation should follow thefollowingconstraints.a)Theuser shouldusememoryallocation from theheapusingmalloc() function. In theabovecode
makethischangeintheinit()function.Inthis,youneedtotakecareofthemaxvalueinthequeuestructure.(Youdonotneed#define).
b)Onceyouaredonewiththeaboveexerciseandyouareabletotestyourqueue.Thenyoucanadd
some more complexity to your code. In enqueue() function when the queue is full in place ofprinting“Queueisfull”youshouldallocatemorespaceusingrealloc()functioncall.
c)Onceyouaredonewiththeaboveexercise.Nowindequeuefunctiononceyouarebelowhalfofthe
capacityofthequeue,youneedtodecreasethesizeofthequeuebyhalf.Youshouldaddonemorevariable"min"toqueuestructuresothatyoucantrackwhatistheoriginalvaluecapacitypassedatinit()function.Moreover,thecapacityofthequeuewillnotgobelowthevaluepassedintheinit()function.
(Ifyouarenotabletosolvetheaboveexercise,thenhavealookintostackchapter,wherewehavedonesimilarforstack)
2)Implementthebelowfunctionforthequeue:
a.IsEmpty:Thisisleftasanexercisefortheuser.Takeavariable,whichwilltakecareofthesizeofaqueueifthevalueofthatvariableiszero,isEmptyshouldreturn1(true).Ifthequeueisnotempty,thenitshouldreturn0(false).
b.Size:Usethesizevariabletobeusedundersizefunctioncall.Size()functionshouldreturnthe
numberofelementsinthequeue.3)Implementstackusingaqueue.Writeaprogramforthisproblem.Youcanusejustonequeue.4)WriteaprogramtoReverseastackusingqueue5)WriteaprogramtoReverseaqueueusingstack6)WriteaprogramtosolveJosephusproblem(algoalreadydiscussed.).Therearenpeoplestandingina
queuewaitingtobeexecuted.Thecountingbeginsatthefrontofthequeue.Ineachstep,knumberofpeoplearedequeuedandagainenqueuedonebyonefromthequeue.Thenthenextpersonisexecuted.Theeliminationproceedsaroundthecircleuntilonlythelastpersonremains,whoisgivenfreedom.Findthatpositionwhereyouwanttostandandgainyourfreedom.
7)WriteaCompStack()functionwhichtakesapointertotwostackstructureasanargumentandreturn
true or false depending uponwhether all the elements of the stack are equal or not.You are givenisEqual(int,int)whichwillcompareandreturn1ifbothvaluesareequaland0iftheyaredifferent.
CHAPTER10:TREE
IntroductionWehavealreadyreadaboutvariouslineardatastructureslikeanarray,linkedlist,stack,queueetc.Botharrayandlinkedlisthaveadrawbackoflineartimerequiredforsearchinganelement.A tree is anonlineardata structure,which is used to represent hierarchical relationships (parent-childrelationship).Eachnodeisconnectedbyanothernodebydirectededges.Example1:Treeinorganization
Example2:Treeinafilesystem
Terminologyintree
Root:Therootofthetreeistheonlynodewithoutanyincomingedges.Itisthetopnodeofatree.Node:Itisafundamentalelementofatree.Eachnodehasdataandtwopointers,whichpointtonulloritschild.Edge:Itisalsoafundamentalpartofatree,whichisusedtoconnecttwonodes.Path:Apathisanorderedlistofnodesthatareconnectedbyedges.Leaf:Aleafnodeisanodethathasnochildren.Heightofthetree:Theheightofatreeisthenumberofedgesonthelongestpathbetweentherootandaleaf.Thelevelofnode:Thelevelofanodeisthenumberofedgesonthepathfromtherootnodetothatnode.
Children:Nodesthathaveincomingedgesfromthesamenodetobesaidtobethechildrenofthatnode.Parent:Nodeisaparentofallthechildnodes,whicharelinkedbyoutgoingedges.Sibling:NodesinthetreethatarechildrenofthesameparentaresaidtobesiblingsAncestor:Anodereachablebyrepeatedmovingfromchildtoparent.
BinaryTreeAbinarytreeisatypetreeinwhicheachnodehasatmosttwochildren(0,1or2),whicharereferredtoastheleftchildandtherightchild.Belowisanodeofthebinarytreewith"a"storedasdataandwhoseleftchild(lChild)andwhoserightchild(rchild)bothpointingtowardsNULL.
Belowisastructureusedtodefinenode,wehavetypedefnodepointertotreeptrstructtreeNode_t
intvalue;treeNode_t*lChild;treeNode_t*rChild;
;typedeftreeNode_t*treePtr;Belowisabinarytreewhosenodescontainsdatafrom1to10
Intherestofthebook,binarytreewillberepresentedasbelow:
PropertiesofBinarytreeare:1.Themaximumnumberofnodesonleveliofabinarytreeis2i,wherei>=12.Themaximumnumberofnodesinabinarytreeofdepthkis2k+1,wherek>=13.Thereisexactlyonepathfromtheroottoanynodesinatree.4.AtreewithNnodeshaveexactlyN-1edgesconnectingthesenodes.5.TheheightofacompletebinarytreeofNnodesislog2N.
TypesofBinarytrees
CompletebinarytreeInacompletebinary tree,every levelexcept the lastone iscompletelyfilled.Allnodes in the leftarefilledfirst,thentherightone.Abinaryheapisanexampleofacompletebinarytree.
Full/StrictlybinarytreeThefullbinarytreeisabinarytreeinwhicheachnodehasexactlyzeroortwochildren.
PerfectbinarytreeTheperfectbinary tree is a typeof fullbinary tree inwhicheachnon-leafnodehasexactly twochildnodes.Allleafnodeshaveidenticalpathlengthandallpossiblenodeslotsareoccupied
RightskewedbinarytreeAbinarytreeinwhicheachnodeishavingeitherarightchildornochild(leaf)iscalledasrightskewedbinarytree
LeftskewedbinarytreeAbinarytreeinwhicheachnodeishavingeitheraleftchildornochild(leaf)iscalledasLeftskewedbinarytree
Height-balancedBinaryTreeAheight-balancedbinarytreeisabinarytreesuchthattheleft&rightsubtreesforanygivennodedifferinheightbymaxone.Note:Eachcompletebinarytreeisaheight-balancedbinarytree
AVLtreeandRBtreeareanexampleofheightbalancedtreewewilldiscussthesetreesinadvancetreetopic.
ProblemsinBinaryTree
CreateaCompletebinarytreeCreateabinarytreegivenalistofvaluesinanarray.Solution:Sincethereisnoorderdefinedinabinarytree,sonodescanbeinsertedinanyordersoitcanbeaskewedbinarytree.ButitisinefficienttodoanythinginaskewedbinarytreesowewillcreateaCompletebinarytree.Ateachnode,themiddlevaluestoredinthearrayisassignedtonodeandleftofarrayispassedtotheleftchildofthenodetocreateleftsub-tree.Andrightportionofarrayispassedtorightchildofthenodetocraterightsub-tree.Example10.1:treePtrCompleteBinaryTree(int*arr,intstart,intsize)
treePtrroot=NULL;
root=createNode(arr[start]);
intleft=2*start+1;intright=2*start+2;
if(left<size)root->lChild=CompleteBinaryTree(arr,left,size);if(right<size)root->rChild=CompleteBinaryTree(arr,right,size);
returnroot;
treePtrcreateNode(intval)
treePtrroot=(treePtr)malloc(sizeof(treeNode_t));root->value=val;root->lChild=root->rChild=NULL;returnroot;
intmain()
treePtrroot=NULL;intA[]=1,2,3,4,5,6,7,8,9,10;root=CompleteBinaryTree(A,0,10);return0;
ComplexityAnalysis:Thisisanefficientalgorithmforcreatingacompletebinarytree.TimeComplexity:O(n),SpaceComplexity:O(n)
Pre-OrderTraversalTraversalisaprocessofvisitingeachnodeofatree.InPre-OrderTraversalparentisvisited/traversedfirst,thenleftchildandrightchild.Pre-Ordertraversalisatypeofdepth-firsttraversal.
Solution:Preordertraversalisdoneusingrecursion.Ateachnode,firstthevaluestoredinitisprintedandthenfollowedbythevalueofleftchildandrightchild.AteachnodeitsvalueisprintedfollowedbycallingprintTree()functiontoitsleftandrightchildtoprintleftandrightsub-tree.Example10.2:1.voidprintPreorder(treePtrroot)/*preorder*/2.3.if(root)4.5.printf("%d",root->value);6.printPreorder(root->lChild);7.printPreorder(root->rChild);8.9.Output:64213587910ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)Note:Whenthereisanalgorithminwhichallnodesaretraversed,thencomplexitycannotbelessthanO(n).Whenthereisalargeportionofthetreethatisnottraversedthencomplexityreduces.
Post-OrderTraversalInPost-OrderTraversalleftchildisvisited/traversedfirst,thenrightchildandlastparentPost-Ordertraversalisatypeofdepth-firsttraversal.
Solution:Inpostordertraversal,first, theleftchildistraversedthenrightchildandintheend,currentnodevalueisprintedtothescreen.Example10.3:1.voidprintPostorder(treePtrroot)/*postorder*/2.3.if(root)4.5.printPostorder(root->lChild);6.printPostorder(root->rChild);7.printf("%d",root->value);8.9.Output:13254710986ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
In-OrderTraversalInIn-OrderTraversal,leftchildisvisited/traversedfirst,thentheparentandlastrightchildIn-Ordertraversalisatypeofdepth-firsttraversal.TheoutputofIn-OrdertraversalofBSTisasortedlist.Solution:InIn-Ordertraversal,firstthevalueofleftchildistraversed,thenthevalueofnodeisprintedtothescreenandthenthevalueofrightchildistraversed.
Example10.4:1.voidprintInorder(treePtrroot)/*inorder*/2.3.if(root)4.5.printInorder(root->lChild);6.printf(“%d“,root->value);7.printInorder(root->rChild);8.9.Output:12345678910ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)Note:Pre-Order,Post-Order,andIn-Ordertraversalareforallbinarytrees.Theycanbeusedtotraverseanykindofabinarytree.
Levelordertraversal/BreadthFirsttraversalWriteCcode to implement levelorder traversalofa tree.Such thatnodesatdepthk isprintedbeforenodesatdepthk+1.
Solution:LevelordertraversalorBreadthFirsttraversalofatreeisdoneusingaqueue.Atthestart,theroot node pointer is added to queue.The traversal of tree happens until its queue is empty.Whenwetraversethetree,wefirstdequeueanelementfromthequeue,printthevaluestoredinthatnodeandthenitsleftchildandrightchildwillbeaddedtothequeue.Example10.5:1.voidprintBredthFirst(treePtrroot)2.3.std::queue<treePtr>que;4.treePtrtemp;5.que.push(root);6.while(!que.empty())7.8.temp=que.front();9.que.pop();10.printf(“%d”,temp->value);11.if(temp->lChild)12.que.push(temp->lChild);13.if(temp->rChild)14.que.push(temp->rChild);15.16.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
PrintDepthFirstwithoutusingtherecursion/systemstack.Solution:Depthfirsttraversalofthetreeisdoneusingrecursionbyusingsystemstack.Thesamecanbedoneusingstack.Inthestart,rootnodepointerisaddedtothestack.Thewholetreeistraverseduntilthestackisempty.Ineachiteration,anelementispoppedfromthestackitsvalueisprintedtoscreen.Thenrightchildandthenleftchildofthenodeisaddedtostack.
Example10.6:1.voidprintDepthFirst(treePtrroot)2.3.std::stack<treePtr>st;4.treePtrtemp;5.st.push(root);6.while(!st.empty())7.8.temp=st.top();9.st.pop();10.printf(“%d”,temp->value);11.if(temp->rChild)12.st.push(temp->rChild);13.if(temp->lChild)14.st.push(temp->lChild);15.16.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
TreeDepthSolution:Depth is tree is calculated recursivelyby traversing left and right child of the root.At eachleveloftraversaldepthofleftchildiscalculatedanddepthofrightchildiscalculated.Thegreaterdepthamongtheleftandrightchildisaddedbyone(whichisthedepthofthecurrentnode)andthisvalueisreturned.Example10.7:1.inttreeDepth(treePtrroot)2.3.if(!root)4.return0;5.else6.7.intlDepth=treeDepth(root->lChild);8.intrDepth=treeDepth(root->rChild);9.if(lDepth>rDepth)10.returnlDepth+1;11.else12.returnrDepth+1;13.14.
ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
NthPre-OrderSolution:Wewant to print the node thatwill be at the nth indexwhenwe print the tree in PreOrdertraversal.Therefore,wekeepacounter tokeep trackof the index.When thecounter isequal to index,thenweprintthevalueandreturntheNthPreOrderindexnode.Example10.8:1.treePtrNthPreOrder(treePtrroot,intindex)/*preorder*/2.3.staticintcounter=0;4.treePtrtemp=NULL;5.if(root)6.7.counter++;8.if(counter==index)9.10.printf("%d",root->value);11.returnroot;12.13.temp=NthPreOrder(root->lChild,index);14.if(temp)15.returntemp;16.temp=NthPreOrder(root->rChild,index);17.if(temp)18.returntemp;19.20.returnNULL;21.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
NthPostOrderSolution:Wewant toprint thenode thatwill be at thenth indexwhenweprint the tree inpost ordertraversal.Therefore,wekeepacountertokeeptrackoftheindex,butatthistime,wewillincrementthecounterafterleftchildandrightchildtraversal.Whenthecounterisequaltoindex,weprintthevalueandreturnthenthpostordernode.Example10.91.treePtrNthPostOrder(treePtrroot,intindex)/*postorder*/2.3.staticintcount=0;4.treePtrtemp=NULL;
5.if(root)6.7.temp=NthPostOrder(root->lChild,index);8.if(temp)9.returntemp;10.temp=NthPostOrder(root->rChild,index);11.if(temp)12.returntemp;13.count++;14.if(count==index)15.16.printf("%d“,root->value);17.returnroot;18.19.20.returnNULL;21.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
NthInOrderSolution:Wewant toprint thenode,whichwill be at thenth indexwhenweprint the tree in in-ordertraversal.Therefore,wekeepacountertokeeptrackoftheindex,butatthistime,wewillincrementthecounterafterleftchildtraversalbutbeforetherightchildtraversal.Whenthecounterisequaltoindex,thenweprintthevalueandreturnthenthin-orderindexnode.Program10.10:1.treePtrNthInOrder(treePtrroot,intindex)/*inorder*/2.3.staticintcount=0;4.treePtrtemp=NULL;5.if(root)6.7.temp=NthInOrder(root->lChild,index);8.if(temp)9.returntemp;10.count++;11.if(count==index)12.13.printf("%d“,root->value);14.returnroot;15.16.
17.temp=NthInOrder(root->rChild,index);18.if(temp)19.returntemp;20.21.returnNULL;22.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(1)
CopyTreeSolution:Copytreeisdonebycopynodesoftheinputtreeateachlevelofthetraversalofthetree.Ateachlevelofthetraversalofnodesoftree,anewnodeiscreatedandthevalueoftheinputtreenodeiscopiedtoit.Theleftchildtreeiscopiedrecursivelyandthenpointertonewsubtreeisreturnedwhichwillbeassignedtotheleftchildofthecurrentnewnode.Similarlyfortherightchildtoo.Finally,thetreeiscopied.Example10.11:1.treePtrcopyTree(treePtrroot)2.3.treePtrtemp;4.if(root!=NULL)5.6.temp=(treePtr)malloc(sizeof(treeNode_t));7.if(!temp)8.returnNULL;9.temp->value=root->value;10.temp->lChild=copyTree(root->lChild);11.temp->rChild=copyTree(root->rChild);12.returntemp;13.14.else15.returnNULL;16.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
CopyMirrorTreeSolution:Copy,mirrorimageofthetreeisdonesameascopytree,butinplaceofleftchildpointingtothetreeformedbyleftchildtraversalofinputtree,thistimeleftchildpointstothetreeformedbyrightchildtraversal.Similarlyrightchildpointtothetraversaloftheleftchildoftheinputtree.Example10.12:1.treePtrcopyMirrorTree(treePtrroot)2.3.treePtrtemp;
4.if(root!=NULL)5.6.temp=(treePtr)malloc(sizeof(treeNode_t));7.if(!temp)8.returnNULL;9.temp->value=root->value;10.temp->lChild=copyMirrorTree(root->rChild);11.temp->rChild=copyMirrorTree(root->lChild);12.returntemp;13.14.else15.returnNULL;16.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
NumberofElementSolution:Numberofnodesattherightchildandthenumberofnodesattheleftchildisaddedbyoneandwegetthetotalnumberofnodesinanytree/sub-tree.Example10.13:1.intnumElement(treePtrroot)2.3.if(!root)4.return0;5.else6.return(1+numElement(root->rChild)+numElement(root->lChild));7ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
NumberofLeafnodesSolution:Ifweadd thenumberof leafnode in therightchildwith thenumberof leafnodes in the leftchild,wewillgetthetotalnumberofleafnodeinanytreeorsubtree.Example10.14:1.intnumLeafs(treePtrroot)2.3.if(!root)4.return0;5.if(!root->lChild&&!root->rChild)6.return1;7.else
8.return(numLeafs(root->rChild)+numLeafs(root->lChild));9.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
PrintMirrorSolution:PrintmirrortreeisverysimpleinplaceofprintingtheleftchildfirstprinttherightchildfirstsothatthetreeisprintedisPreOrdertraversalofmirroroftheinputtree.Example10.15:1.voidprintMirror(treePtrroot)/*preorder*/2.3.if(root!=NULL)4.5.printf("%d",root->value);6.printMirror(root->rChild);7.printMirror(root->lChild);8.return;9.10.else11.return;12.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
IdenticalSolution:Twotreeshaveidenticalvaluesifateachlevelthevalueisequal.Example10.16:1.intidentical(treePtrroot1,treePtrroot2)2.3.if(!root1&&!root2)4.return1;5.elseif(!root1||!root2)6.return0;7.else8. return ( identical(root1->lChild, root2->lChild) && identical(root1->rChild, root2->rChild) &&(root1->value==root2->value));9.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
FreeTree
Solution:Thetreeistraversedandnodesoftreearefreedinsuchamannersuchthatallchildnodesarefreedbeforeit.Example10.17:1.treePtrfreeTree(treePtrroot)2.3.if(root)4.5.root->lChild=freeTree(root->lChild);6.root->rChild=freeTree(root->rChild);7.if(root->lChild==NULL&&root->rChild==NULL)8.9.free(root);10.returnNULL;11.12.13.returnNULL;14.1.voidfreeTree(treePtr*rootPtr)2.3.*rootPtr=freeTree(*rootPtr);4.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
TreetoListRecSolution:Treetothelistisdonerecursively.Ateachnode,wewillsupposethatthetreetolistfunctionwilldoitsjobfortheleftchildandrightchild.Thenwewillcombinetheresultoftheleftchildandrightchild traversal.Weneedaheadand tail pointerof the left list and right list to combine themwith thecurrentnode.Intheprocessof integration, thecurrentnodewillbeaddedtothetailof theleft listandcurrentnodewillbeaddedtotheheadtotherightlist.Headoftheleftlistwillbecometheheadofthenewlyformedlistandtailoftherightlistwillbecomethetailofthenewlycreatedlist.Example10.18:treePtrtreeToListRec(treePtrcurr)
treePtrHead,Tail,tempHead;if(!curr)returnNULL;
if(curr->lChild==NULL&&curr->rChild==NULL)curr->lChild=curr;
curr->rChild=curr;returnroot;if(curr->lChild)Head=treeToListRec(curr->lChild);Tail=Head->lChild;curr->lChild=Tail;Tail->rChild=curr;elseHead=curr;
if(curr->rChild)tempHead=treeToListRec(curr->rChild);Tail=tempHead->lChild;curr->rChild=tempHead;tempHead->lChild=curr;elseTail=curr;
Head->lChild=Tail;Tail->rChild=Head;returnHead;
ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
PrintallthepathsPrintallthepathsfromtherootstotheleafSolution:Wheneverwetraverseanode,weaddthatnodetothelist.Whenwereachaleaf,weprintthewholelist.Whenwereturnfromafunction,thenweremovetheelementthatwasaddedtothelistwhenweenteredthisfunction.Example10.19:1.voidprintList(list<int>L1)2.3.list<int>::iteratoriterB=L1.begin();4.list<int>::iteratoriterE=L1.end();5.while(iterB!=iterE)6.7.cout<<(*iterB)<<"";
8.iterB++;9.10.cout<<endl;11.1.voidprintPath(treePtrhead,list<int>&L1)2.3.if(!head)4.return;5.If(head->rChild==null||head->lChild==null)6.7.printList(L1);8.L1.pop_back();9.return;10.11.L1.push_back(head->value);12.printPath(head->rChild,L1);13.printPath(head->lChild,L1);14.L1.pop_back();15.1.voidprintPathWrapper(treePtrhead)2.3.list<int>L1;4.printPath(head,L1);5.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
LeastCommonAncestorSolution:We recursively traverse the nodes of a binary tree. If we find any one of the nodewe aresearching for thenwe return that node.Andwhenwegetboth the left and right as somevalidpointerlocationotherthanNULL,wewillreturnthatnodeasthecommonancestor.Example10.20:1.treePtrLCA(treePtrroot,treePtrfirstPtr,treePtrsecondPtr)2.3.treePtrleft,right;4.5.if(root==NULL)6.returnNULL;7.8.if(root==firstPtr||root==secondPtr)9.returnroot;
10.11.left=LCA(root->lChild,firstPtr,secondPtr);12.right=LCA(root->rChild,firstPtr,secondPtr);13.14.if(left&&right)15.returnroot;16.elseif(left)17.returnleft;18.else19.returnright;20.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
FindMaxinBinaryTreeSolution:Werecursivelytraversethenodesofabinarytree.Wewillfindthemaximumvalueintheleftandrightsubtreeofanynodethenwillcomparethevaluewiththevalueofthecurrentnodeandfinallyreturnthelargestofthethreevalues.Example10.21:1.intfindMaxBT(treePtrroot)2.3.intmax;4.intleft,right;5.6.if(root==NULL)7.returnINT_MIN;8.9.max=root->value;10.11.left=findMaxBT(root->lChild);12.right=findMaxBT(root->rChild);13.14.if(left>max)15.max=left;16.if(right>max)17.max=right;18.19.returnmax;20.
SearchvalueinaBinaryTreeSolution:To find if somevalue is there in a binary treeor not is doneusing exhaustive searchof the
binarytree.First,thevalueofcurrentnodeiscomparedwiththevaluethatwearelookingfor.Thenitiscomparedrecursivelyinsidetheleftchildandrightchild.Example10.22:1.intsearchBT(treePtrroot,intvalue)2.3.intmax;4.intleft,right;5.6.if(root==NULL)7.return0;8.9.if(root->value==value)10.return1;11.12.left=searchBT(root->lChild,value);13.if(left)14.return1;15.16.right=searchBT(root->rChild,value);17.if(right)18.return1;19.20.return0;21.
MaximumDepthinaBinaryTreeSolution:Tofindthemaximumdepthofabinarytreeweneedtofindthedepthofthelefttreeanddepthofrighttreethenweneedtostorethevalueandincrementitbyonesothatwegetdepthofthegivennode.Example10.23:1.intmaxDepthBT(treePtrroot)2.3.intmax;4.intleft,right;5.6.if(root==NULL)7.return0;8.9.left=findMaxBT(root->lChild);10.right=findMaxBT(root->rChild);11.12.if(left>max)13.max=left;
14.if(right>max)15.max=right;16.17.returnmax+1;18.
NumberofFullNodesinaBTSolution:Afullnodeisanodethathavebothleftandrightchild.Wewillrecursivelytraversthewholetreeandwillincreasethecountoffullnodeaswefindthem.Example10.24:1.intnumFullNodesBT(treePtrroot)2.3.intcount=0;4.intleft,right;5.6.if(root==NULL)7.return0;8.9.left=findMaxBT(root->lChild);10.right=findMaxBT(root->rChild);11.12.count=left+right;13.14.if(root->lChild&&root->rChild)//thislinecanbechangedtosolvemanyproblems.15.count++;16.17.returncount;18.
MaximumLengthPathinaBT/DiameterofBTSolution:TofindthediameterofBTweneedtofindthedepthofleftchildandrightchildthenwilladdthese two values and increment it by one so that we will get the maximum length path (diametercandidate)whichcontainsthecurrentnode.Thenwewillfindmaxlengthpathintheleftchildsub-tree.Andwillfindthemaxlengthpathintherightchildsub-tree.Finally,wewillcomparethethreevaluesandreturnthemaximumvalueoutofthesethiswillbethediameteroftheBinarytree.Example10.25:1.intmaxLengthPathBT(treePtrroot)//diameter2.3.intmax;4.intleftPath,rightPath;5.intleftMax,rightMax;
6.7.if(root==NULL)8.return0;9.10.leftPath=maxDepthBT(root->lChild);11.rightPath=maxDepthBT(root->rChild);12.13.max=leftPath+rightPath+1;14.15.leftMax=maxLengthPathBT(root->lChild);16.rightMax=maxLengthPathBT(root->rChild);17.18.if(leftMax>max)19.max=leftMax;20.21.if(rightMax>max)22.max=rightMax;23.24.returnmax;25.
SumofAllnodesinaBTSolution:Wewill find the sumofall thenodes recursively. sumAllBT()will return the sumofall thenodeofleftandrightsubtreethenwilladdthevalueofcurrentnodeandwillreturnthefinalsum.Example10.26:1.intsumAllBT(treePtrroot)2.3.intsum;4.intleft,right;5.6.if(root==NULL)7.return0;8.9.left=sumAllBT(root->lChild);10.right=sumAllBT(root->rChild);11.12.sum=left+right+root->value;13.14.returnsum;15.
IterativePre-order
Solution:Inplaceofusingsystemstackinrecursion,wecantraversethetreeusingstackdatastructure.Example10.27:1.#definemaxStackSize1002.voiditerativePreorder(treePtrroot)3.4.structstructStack5.treePtrptr;6.intvisited;7.;8.treePtrtempRoot=NULL;9.inttempVisit=0;10.structStackstack[maxStackSize];11.inttop=1;12.stack[top].ptr=root;13.stack[top].visited=0;14.while(top)15.16.tempRoot=stack[top].ptr;17.tempVisit=stack[top].visited;18.top--;19.if(tempVisit)20.21.printf("[%d]",tempRoot->value);22.23.else24.25.if(tempRoot->rChild!=NULL)/*rightchildalwaysgoesfirstthenleftchild*/26.27.top++;28.stack[top].ptr=tempRoot->rChild;29.stack[top].visited=0;30.31.if(tempRoot->lChild!=NULL)32.33.top++;34.stack[top].ptr=tempRoot->lChild;35.stack[top].visited=0;36.37.top++;38.stack[top].ptr=tempRoot;39.stack[top].visited=1;40.41.42.
ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
IterativePost-orderSolution:Inplaceofusingsystemstackinrecursion,wecantraversethetreeusingstackdatastructure.Example10.28:1.voiditerativePostorder(treePtrroot)2.3.structstructStack4.treePtrptr;5.intvisited;6.;7.treePtrtempRoot=NULL;8.inttempVisit=0;9.structStackstack[maxStackSize];10.inttop=1;11.stack[top].ptr=root;12.stack[top].visited=0;13.while(top)14.15.tempRoot=stack[top].ptr;16.tempVisit=stack[top].visited;17.top--;18.if(tempVisit)19.20.printf("[%d]",tempRoot->value);21.22.else23.24.top++;25.stack[top].ptr=tempRoot;26.stack[top].visited=1;27.if(tempRoot->rChild!=NULL)/*rightchldalwaysgoesfirstthenrightchild*/28.29.top++;30.stack[top].ptr=tempRoot->rChild;31.stack[top].visited=0;32.33.if(tempRoot->lChild!=NULL)34.35.top++;36.stack[top].ptr=tempRoot->lChild;37.stack[top].visited=0;38.39.
40.41.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
IterativeIn-orderSolution:Inplaceofusingsystemstackinrecursion,wecantraversethetreeusingstackdatastructure.Example10.29:1.voiditerativeInorder(treePtrroot)2.3.structstructStack4.treePtrptr;5.intvisited;6.;7.treePtrtempRoot=NULL;8.inttempVisit=0;9.structStackstack[maxStackSize];10.inttop=1;11.stack[top].ptr=root;12.stack[top].visited=0;13.while(top)14.15.tempRoot=stack[top].ptr;16.tempVisit=stack[top].visited;17.top--;18.if(tempVisit)19.20.printf("[%d]",tempRoot->value);21.22.else23.24.if(tempRoot->rChild!=NULL)/*rightchldalwaysgoesfirstthenrightchild*/25.26.top++;27.stack[top].ptr=tempRoot->rChild;28.stack[top].visited=0;29.30.top++;31.stack[top].ptr=tempRoot;32.stack[top].visited=1;33.if(tempRoot->lChild!=NULL)34.35.top++;36.stack[top].ptr=tempRoot->lChild;37.stack[top].visited=0;
38.39.40.41.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
BinarySearchTree(BST)Abinarysearchtree(BST)isabinarytreeonwhichnodesareorderedinthefollowingway:
·Thekeyintheleftsubtreeislessthanthekeyinitsparentnode.·Thekeyintherightsubtreeisgreaterthekeyinitsparentnode.·Noduplicatekeyallowed.
Note: there can be two separate key and value fields in the tree node. But for simplicity, we areconsideringvalueasthekey.Allproblemsinthebinarysearchtreearesolvedusingthissuppositionthatthevalueinthenodeiskeyforthetree.Note:Sincebinarysearchtreeisabinarytreetoalltheabovealgorithmofabinarytreeareapplicabletoabinarysearchtree.
ProblemsinBinarySearchTree(BST)Allbinarytreealgorithmsarevalidforbinarysearchtreetoo.
CreateabinarysearchtreeCreateabinarytreegivenlistofvaluesinanarrayinsortedorderSolution:Sincetheelementsinthearrayareinsortedorderandwewanttocreateabinarysearchtreeinwhichleftsubtreenodesarehavingvalueslessthanthecurrentnodeandrightsubtreenodeshavevaluegreaterthanthevalueofthecurrentnode.Wehavetofindthemiddlenodetocreateacurrentnodeandsendtherestofthearraytoconstructleftandrightsubtree.Example10.30:treePtrCreateBinaryTree(int*arr,intstart,intend)
treePtrroot=NULL;if(start>end)returnNULL;intmid=(start+end)/2;root=createNode(arr[mid]);root->left=CreateBinaryTree(arr,start,mid-1);root->right=CreateBinaryTree(arr,mid+1,end);returnroot;
treePtrcreateNode(intval)
treePtrroot=(treePtr)malloc(sizeof(treeNode_t));root->value=val;root->lChild=root->rChild=NULL;returnroot;
intmain()
treePtrroot=NULL;intA[]=1,2,3,4,5,6,7,8,9,10;root=CreateBinarySearchTree(A,0,10);return0;
InsertionNodeswithkey6,4,2,5,1,3,8,7,9,10areinsertedinatree.Belowisstepbysteptreeafterinsertingnodesintheorder.
Solution: Smaller valueswill be added to the left child sub-tree of a node and greater valuewill beaddedtotherightchildsub-treeofthecurrentnode.Example10.31:1.treePtrinsertNode(intvalue,treePtrroot)2.3.if(root==NULL)4.5.root=(treePtr)malloc(sizeof(treeNode_t));6.if(root==NULL)7.8.printf("filledmemoryshortage");
9.returnroot;10.11.root->value=value;12.root->lChild=root->rChild=NULL;13.14.else15.16.if(root->value>value)17.18.root->lChild=insertNode(value,root->lChild);19.20.else21.22.root->rChild=insertNode(value,root->rChild);23.24.25.returnroot;26.1.voidinsertNode(intvalue,treePtr*ptrRoot)2.3.*ptrRoot=insertNode(value,*ptrRoot);4.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
FindNodeFindthenodewiththevaluegiven.Solution:Thevaluegreaterthanthecurrentnodevaluewillbeintherightchildsub-treeandthevaluesmallerthanthecurrentnodeisintheleftchildsub-tree.Example10.32:1.treePtrfindNode(treePtrroot,intvalue)2.3.if(!root)4.returnNULL;5.if(root->value==value)6.returnroot;7.else8.9.if(root->value>value)10.returnfindNode(root->lChild,value);11.else
12.returnfindNode(root->rChild,value);13.14.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(1)
FindNodeIterativeSolution:Thevaluegreaterthanthecurrentnodevaluewillbeintherightchildsub-treeandthevaluesmallerthanthecurrentnodeisintheleftchildsub-tree.Wecanfindavaluebytraversingtheleftorrightsubtreeiteratively.Example10.33:1.treePtrfindNodeIterative(treePtrroot,intvalue)/*iterative*/2.3.while(root)4.5.if(root->value==value)6.returnroot;7.elseif(root->value>value)8.root=root->lChild;9.else10.root=root->rChild;11.12.returnNULL;13.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(1)Example10.34:Operatorsaregenerallyreadfromlefttoright1.treePtrfindNodeIterative_optimized(treePtrroot,intvalue)2.3.while(root&&root->value!=value)4.(root->value>value)?root=root->lChild:root=root->rChild;5.returnroot;6.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
FindMinFindthenodewiththeminimumvalue.
Solution:leftmostchildofthetreewillbethenodewiththeminimumvalue.Example10.35:1.treePtrfindMin(treePtrroot)2.3.if(root)4.5.while(root->lChild)6.7.root=root->lChild;8.9.10.returnroot;11.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(1)
FindMinRecursiveFindthenodewithaminimumvalue,providearecursivesolution.Example10.36:1.treePtrfindMinRec(treePtrroot)2.3.if(!root)4.returnNULL;5.if(root->lChild==NULL)6.returnroot;7.else8.returnfindMinRec(root->lChild);9.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
FindMax
Findthenodeinthetreewiththemaximumvalue.
Solution:Rightmostnodeofthetreewillbethenodewiththemaximumvalue.Example10.37:1.treePtrfindMax(treePtrroot)2.3.if(root)4.5.while(root->rChild)6.7.root=root->rChild;8.9.10.returnroot;11.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(1)
FindMaxRecursiveFindthenodeinthetreewiththemaximumvalue,providearecursivesolution.Example10.38:1.treePtrfindMaxRec(treePtrroot)2.3.if(!root)4.returnNULL;5.if(root->rChild==NULL)6.returnroot;7.else8.returnfindMaxRec(root->rChild);9.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)
MaxValue
Example10.39:1.intmaxValue(treePtrroot)2.3.if(root)4.5.while(root->rChild)6.7.root=root->rChild;8.9.returnroot->value;10.11.return-999;12.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(1)
MinValueExample10.40:1.intminValue(treePtrroot)2.3.if(root)4.5.while(root->lChild)6.7.root=root->lChild;8.9.returnroot->value;10.11.return-999;12.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(1)
IstreeaBSTSolution:Ateachnodewecheck,maxvalueofleftsubtreeissmallerthanthevalueofcurrentnodeandminvalueofrightsubtreeisgreaterthanthecurrentnode.Example10.41:1.intisBST(treePtrroot)2.3.if(!root)4.return1;5.if(root->lChild&&maxValue(root->lChild)>root->value)6.return0;
7.if(root->rChild&&minValue(root->rChild)<=root->value)8.return0;9.return(isBST(root->lChild)&&isBST(root->rChild));10.ComplexityAnalysis:TimeComplexity:O(n2),SpaceComplexity:O(n)Theabovesolutioniscorrectbutitisnotefficientassametreenodesaretraversedmanytimes.Solution:Anotherbettersolutionwillbetheoneinwhichwewilllookintoeachnodeonlyonce.Thisisdonebynarrowingtherange.WewillbeusinganisBSTUtil()functionwhichtakethemaxandminrangeofthevaluesofthenodes.TheinitialvalueofminandmaxwillbeINT_MINandINT_MAX.Example10.42:1.intisBSTUtil(treePtrroot,intmin,intmax)2.3.if(!root)4.return1;5.6.if(root->value<min||root->value>max)7.return0;8.9.returnisBSTUtil(root->lChild,min,root->value)&&isBSTUtil(root->rChild,root->value+1,max);10.11.12.intisBST(treePtrroot)13.14.returnisBSTUtil(root,INT_MIN,INT_MAX);15.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)forstackSolution:Abovemethodiscorrectandefficientbutthereisanothereasymethodtodothesame.Wecandoin-ordertraversalofnodesandseeifwearegettingastrictlyincreasingsequenceExample10.43:1.intisBST(treePtrroot,int*value)/*inordertraversal*/2.3.intret;4.if(root)5.6.ret=isBST(root->lChild,value);7.if(!ret)8.return0;9.
10.if(*value>root->value)11.return0;12.13.*value=root->value;14.15.ret=isBST(root->rChild,value);16.if(!ret)17.return0;18.19.return1;20.ComplexityAnalysis:TimeComplexity:O(n),SpaceComplexity:O(n)forstack
DeleteNodeDescription:Removethenodexfromthebinarysearchtree,makingthenecessary,reorganizenodesofbinarysearchtreetomaintainitsproperties.Therearethreecasesindeletenode,letuscallthenodethatneedtobedeletedasx.Case1:nodexhasnochildren.Justdeleteit(i.e.Changeparentnodesothatitdoesnotpointtox)Case2:nodexhasonechild.Spliceoutxbylinkingx’sparenttox’schildCase3:nodexhastwochildren.Spliceoutthex’ssuccessorandreplacexwithx’ssuccessorWhenthenodetobedeletedhavenochildrenThisisatrivialcasewedirectlydeletethenodeandreturnnull.Whenthenodetobedeletedhaveonlyonechild.Inthiscase,wesavethechildinatempvariable,thendeletecurrentnode,andfinallyreturnthechild.
Whenthenodetobedeletedhastwochildren.
Example10.44:1.voiddeleteNode(treePtr*rootPtr,intvalue)2.3.*rootPtr=deleteNode(*rootPtr,value);4.1.treePtrdeleteNode(treePtrroot,intvalue)2.3.treePtrtemp=NULL;4.if(root)5.6.if(root->value==value)
7.8.if(root->lChild==NULL&&root->rChild==NULL)9.10.free(root);11.returnNULL;12.13.else14.15.if(root->lChild==NULL)16.17.temp=root->rChild;18.free(root);19.returntemp;20.21.if(root->rChild==NULL)22.23.temp=root->lChild;24.free(root);25.returntemp;26.27.temp=findMin(root->rChild);28.root->value=temp->value;29.root->rChild=deleteNode(root->rChild,temp->value);30.31.32.else33.34.if(root->value>value)35.36.root->lChild=deleteNode(root->lChild,value);37.38.else39.40.root->rChild=deleteNode(root->rChild,value);41.42.43.44.returnroot;45.Analysis:TimeComplexity:O(n),SpaceComplexity:O(n)
LeastCommonAncestorInatreeT.Theleastcommonancestorbetweentwonodesn1andn2isdefinedasthelowestnodeinT
thathasbothn1andn2asdescendants.Example10.45:1.treePtrLcaBST(treePtrroot,treePtrfirstPtr,treePtrsecondPtr)2.3.if(!firstPtr||!secondPtr||!root)4.5.returnroot;6.7.8.if(root->value>firstPtr->value&&9.root->value>secondPtr->value)10.11.returnLcaBST(root->lChild,firstPtr,secondPtr);12.13.if(root->value<firstPtr->value&&14.root->value<secondPtr->value)15.16.returnLcaBST(root->rChild,firstPtr,secondPtr);17.18.returnroot;19.
PathLengthGiventwonodes,findthepathlengthbetweenthem.Solution:Thisproblemsolutionissimplyfindingtheleastcommonancestorofthetwonodeandthenfindthelengthofeachofnodefromtheirparent.Example10.46:1.intpathLength(treePtrroot,treePtrfirstPtr,treePtrsecondPtr)2.3.treePtrparent=LcaBST(root,firstPtr,secondPtr);4.intfirst=findNodeDepth(parent,firstPtr);5.intsecond=findNodeDepth(parent,secondPtr);6.returnfirst+second;7.1.intfindNodeDepth(treePtrroot,treePtrdstPtr)2.3.intvalue;4.if(!root||!dstPtr)5.return-1;6.
7.if(root->value==dstPtr->value)8.return0;9.else10.11.if(root->value>dstPtr->value)12.value=findNodeDepth(root->lChild,dstPtr);13.else14.value=findNodeDepth(root->rChild,dstPtr);15.16.if(value!=-1)17.return(value+1);18.19.
TrimtheTreenodeswhichareOutsideRangeGivenarangeasmin,max.Weneedtodeleteallthenodesofthetreethatareoutofthisrange.Solution:Traversethetreeandeachnodethatishavingvalueoutsidetherangewilldeleteitself.Allthedeletionwillhappenfrominsideoutsowedonothavetocareaboutthechildrenofanodeasiftheyareoutofrangethentheyalreadyhaddeletedthemselves.Example10.47:1.treePtrtrimOutsideRange(treePtrroot,intmin,intmax)2.3.treePtrtempNode;4.if(root==NULL)5.returnNULL;6.7.root->lChild=trimOutsideRange(root->lChild,min,max);8.root->rChild=trimOutsideRange(root->rChild,min,max);9.10.if(root->value<min)11.12.tempNode=root->rChild;13.deleteroot;14.returntempNode;15.16.17.if(root->value>max)18.19.tempNode=root->lChild;20.deleteroot;21.returntempNode;22.
23.24.returnroot;25.
PrintTreenodeswhichareinRangePrintonlythosenodesofthetreewhosevalueisintherangegiven.Solution:Justnormalinordertraversalandatthetimeofprintingwewillcheckifthevalueisinsidetherangeprovided.Example10.48:1.voidprintInRange(treePtrroot,intmin,intmax)2.3.if(!root)4.return;5.6.printInRange(root->lChild,min,max);7.8.if(root->value>=min&&root->value<=max)9.printf("%d",root->value);10.11.printInRange(root->rChild,min,max);12.
FindCeilandFloorvalueinsideBSTgivenkeyGivenatreeandavalueweneedtofindtheceilvalueofnodeintreewhichissmallerthanthegivenvalueandneedtofindthefloorvalueofnodeintreewhichisbigger.Ouraimistofindceilandfloorvalueascloseaspossiblethenthegivenvalue.Example10.49:1.voidCeilFloorBST(treePtrroot,intvalue,int*ceil,int*floor)2.3.while(root)4.5.if(root->value==value)6.7.*ceil=root->value;8.*floor=root->value;9.break;10.11.elseif(root->value>value)12.13.*ceil=root->value;
14.root=root->lChild;15.16.else17.18.*floor=root->value;19.root=root->rChild;20.21.22.
TreetoDoublyLinkedlistWeneedtoconvertabinarytreetodoublelinkedlist.Suchthattheinordertraversalofthetreeissavedasasequenceofnodeindoublelinkedlist.Solution:WewilluseTreeToList()function.AteachnodewewillcallTreeToList()functionforbothleftchildandrightchild.Thesetwofunctioncallwillcreatedoublelinkedlistofleftchildandrightchildrespectively. Thenwe just have toworry about the current node.Wewill append the two linked listformedbytherecursivecallandthejointhemandthefinallistisprepared.Example10.50:1.voidconnect(treePtra,treePtrb)2.3.a->rChild=b;4.b->lChild=a;5.1.treePtrappend(treePtra,treePtrb)2.3.treePtraLast,bLast;4.if(a==NULL)5.return(b);6.if(b==NULL)7.return(a);8.aLast=a->lChild;9.bLast=b->lChild;10.connect(aLast,b);11.connect(bLast,a);12.return(a);13.1.treePtrtreeToList(treePtrroot)2.3.treePtraList,bList;4.if(root==NULL)
5.return(NULL);6.aList=treeToList(root->lChild);7.bList=treeToList(root->rChild);8.root->lChild=root;9.root->rChild=root;10.aList=append(aList,root);11.aList=append(aList,bList);12.return(aList);13.Analysis:TimeComplexity:O(n),SpaceComplexity:O(1)
Exercise1.Constructatreegivenitsin-orderandpre-ordertraversalstrings.
oinorder:12345678910opre-order:64213587910
2.Constructatreegivenitsin-orderandpost-ordertraversalstrings.
oinorder:12345678910opost-order:13254710986
3.WriteadeletenodefunctioninBinarytree.4.Writeafunctionprintdepthfirst inabinarytreewithoutusingsystemstack(useSTLqueueorstack
etc.)Hint:youmaywanttokeepanotherelementtotreenodelikevisitedflag.
5.CheckwhetheragivenBinaryTreeisCompleteornot
oInacompletebinarytree,everylevelexceptthelastoneiscompletelyfilled.Allnodesintheleftarefilledfirst,thentherightone.
6.CheckwhetheragivenBinaryTreeisFull/Strictlybinarytreeornot
oThefullbinarytreeisabinarytreeinwhicheachnodehaszeroortwochildren.
7.CheckwhetheragivenBinaryTreeisaPerfectbinarytreeornotoTheperfectbinarytree-isatypeoffullbinarytreesinwhicheachnon-leafnodehasexactlytwo
childnodes.
8.CheckwhetheragivenBinaryTreeisHeight-balancedBinaryTreeornotoAheight-balancedbinarytreeisabinarytreesuchthattheleft&rightsubtreesforanygivennode
differinheightbynomorethanone
9.Isomorphic:twotreesareisomorphiciftheyhavethesameshape,itdoesnotmatterwhatthevalueis.Writeaprogramtofindiftwogiventreeareisomorphicornot.
10.Theworst-caseruntimeComplexityofbuildingaBSTwithnnodes
oO(n2)oO(n*logn)oO(n)oO(logn)
11.Theworst-caseruntimeComplexityofinsertionintoaBSTwithnnodesis
oO(n2)oO(n*logn)oO(n)oO(logn)
12.Theworst-caseruntimeComplexityofasearchofavalueinaBSTwithnnodes.
oO(n2)oO(n*logn)oO(n)oO(logn)
13.WhichofthefollowingtraversalsalwaysgivesthesortedsequenceoftheelementsinaBST?
oPreorderoIgnoredoPostorderoUndefined
14.TheheightofaBinarySearchTreewithnnodesintheworstcase?
oO(n*logn)oO(n)oO(logn)oO(1)
15.TrytooptimizetheabovesolutiontogiveaDFStraversalwithoutusingrecursionusesomestackor
queue.16.Thisisanopenexerciseforthereaders.Everyalgorithmthatissolvedusingrecursion(systemstack)
can also be solved using user defined or library defined (STL) stack. So try to figure outwhat allalgorithms,whichareusingrecursionandtrytofigureouthowyouwilldothissameissueusinguserlayerstack.
17.Inabinarytree,printthenodesinzigzagorder.Inthefirstlevel,nodesareprintedinthelefttoright
order.Inthesecondlevel,nodesareprintedinrighttoleftandinthethirdlevelagainintheorderlefttoright.Hint:Usetwostacks.Popfromfirststackandpushintoanotherstack.Swapthestacksalternatively.
18.Findnthsmallestelementinabinarysearchtree.
Hint:Nthinorderinabinarytree.19.Findthefloorvalueofkey,whichisinsideaBST.20.FindtheCeilvalueofkey,whichisinsideaBST.21.WhatisTimeComplexityofthebelowcode:voidDFS(treePtrhead)
treePtrcurr=head,*prev;intcount=0;while(curr&&!curr->visited)count++;if(curr->lChild&&!curr->lChild->visited)curr=curr->lChild;elseif(curr->rChild&&!curr->rChild->visited)curr=curr->rChild;else
printf("%d",curr->value);curr->visited=1;curr=head;printf("\ncountis:%d",count);
CHAPTER11:PRIORITYQUEUE
IntroductionAPriority-QueuealsoknowsasBinary-Heap,isavariantofqueue.Itemsareremovedfromthestartofthe queue but in a Priority-Queue, the logical ordering of objects is determined by their priority. ThehighestpriorityitemareatthefrontofthePriority-Queue.WhenyouenqueueanitemtoPriority-Queuethe new item canmore to the front of the queue.A Priority-Queue is a very important data structure.Priority-Queue is used in various Graph algorithms like Prim’s Algorithm and Dijkstra’s algorithm.Priority-Queueisalsousedinthetimerimplementationetc.A Priority-Queue is implemented using a Heap (Binary Heap). A Heap data structure is an array ofelementsthatcanbeobservedasacompletebinarytree.Thetreeiscompletelyfilledonalllevelsexceptpossiblythelowest.Heapsatisfiestheheaporderingproperty.Aheapisacompletebinarytreeso theheightoftreewithNnodesisalwaysO(logn).
Aheapisnotasortedstructureandcanberegardedaspartiallyordered.Asyouseefromthepicture,thereisnorelationshipamongnodesatanygivenlevel,evenamongthesiblings.Heapisimplementedusinganarray.Inaddition,becauseheapisacompletebinarytree,theleftchildofaparent(atpositionx)isthenodethatisfoundinposition2xinthearray.Similarly,therightchildoftheparentisatposition2x+1inthearray.Tofindtheparentofanynodeintheheap,wecansimplydivision.Giventheindexyofanode,theparentindexwillbyy/2.
TypesofHeapTherearetwotypesofheapandthetypedependsontheorderingoftheelements.Theorderingcanbedoneintwoways:Min-HeapandMax-Heap
MaxHeapMax-Heap:thevalueofeachnodeislessthanorequaltothevalueofitsparent,withthelargest-valueelementattheroot.
MaxHeapOperationsInsert O(logn)DeleteMax O(logn)Remove O(logn)FindMax O(1)
MinHeapMin-Heap:thevalueofeachnodeisgreaterthanorequaltothevalueofitsparent,withtheminimum-valueelementattheroot.
Useitwheneveryouneedquickaccesstothesmallestitem,becausethatitemwillalwaysbeattherootofthetreeorthefirstelementinthearray.However,theremainderofthearrayiskeptpartiallysorted.Thus,instantaccessisonlypossibleforthesmallestitem.MinHeapOperations
Insert O(logn)DeleteMin O(logn)Remove O(logn)FindMin O(1)Throughout thischapter, theword"heap"willalwaysrefer toamax-heap.The implementationofmin-heapisleftfortheusertodoitasanexercise.
HeapADTOperationsThebasicoperationsofbinaryheapareasfollows:BinaryHeap Createanewemptybinaryheap O(1)Insert Addinganewelementtotheheap O(logn)DeleteMax Deletethemaximumelementformtheheap. O(logn)FindMax Findthemaximumelementintheheap. O(1)isEmpty returntrueiftheheapisemptyelsereturnfalse O(1)Size Returnthenumberofelementsintheheap. O(1)BuildHeap Buildanewheapfromthearrayofelements O(logn)
OperationonHeap
CreateHeapfromanarray1.Startsbyputtingtheelementstoanarray.2.Starting from themiddle of the arraymove downward towards the start of the array.At each step,
compareparentvaluewithitsleftchildandrightchild.Thenrestoretheheappropertybyshiftingtheparent valuewith its largest-value child. Such that the parent valuewill always be greater than orequaltoleftchildandrightchild.
3.Forallelementsfrommiddleofthearraytothestartofthearray.Wewillcompareandshiftuntilwereachtotheleafnodes.TheTimeComplexityofbuildheapisO(N).
Given an array as input to create heapfunction. Value of index i is comparedwithvalueofitschildrennodesthatisatindex(i*2+1)and(i*2+2).MiddleofarrayN/2thatisindex3iscomapredwithindex7. If the childrennode valueis greater than parent node then thevaluewillbeswapped.
Similarly, value of index 2 is comparedwith index 5 and 6. The largest of thevalue is 7 which will be swapped withthevalueattheindex2.
Similarly, value of index 1 is comparedwith index 3 and 4 The largest of thevalue is 8 which will be swapped withthevalueattheindex1.
Proclate down function is used tosubsequentlyadjestthevaluereplasedinthe previous step by comparing it withitschildrennodes.
Now value at index 0 is comared withindex1and2.8isthelargestvaluesoitswappedwiththevalueatindex0.
Proclate down function is used tofurther compare the value at index 1withitschildrennodesatindex3and4.
Intheendmaxheapiscreated.
Example11.1:PtrHeapBuildHeap(int*arr,intsize,intcapacity)
PtrHeappHeap=(PtrHeap)malloc(sizeof(Heap));pHeap->Size=size;pHeap->Capacity=capacity;pHeap->Array=arr;
for(inti=(pHeap->Size)/2;i>0;i--)ProclateDown(pHeap->Array,i,pHeap->Size);
returnpHeap;
voidProclateDown(int*arr,intposition,intsize)
intlChild=2*position+1;
intrChild=lChild+1;intsmall=-1;
if(lChild<size)small=lChild;
if(rChild<size&&arr[rChild]<arr[lChild])small=rChild;
if(small!=-1&&arr[small]<arr[position])inttemp=arr[position];arr[position]=arr[small];arr[small]=temp;ProclateDown(arr,small,size);
InitializinganemptyHeapExample11.2:voidHeapInitialize(PtrHeappHeap,intcapacity)
pHeap->Size=0;pHeap->Capacity=capacity;pHeap->Array=(int*)malloc((capacity)*sizeof(int));
Enqueue/Insert1.Addthenewelementattheendofthearray.Thiskeepsthestructureasacompletebinarytree,butit
mightnolongerbeaheapsincethenewelementmighthaveavaluegreaterthanitsparentvalue.2.Swapthenewelementwithitsparentuntilithasvaluegreaterthanitsparentvalue.3.Step2willterminatewhenthenewelementreachestherootorwhenthenewelement'sparenthavea
valuegreaterthanorequaltothenewelement'svalue.LetustakeanexampleoftheMaxheapcreatedintheaboveexample.
Letustakeanexamplebyinsertingelementwithvalue9totheheap.Theelementisaddedtotheendoftheheaparray.Nowthevaluewillbeproclateupbycomparingitwiththeparent.Thevalueisaddedtoindex8anditsparentwillbe(N-1)/2=index3.
Sincethevalue9isgreaterthan4itwillbeswappedwithit.
Proclate up is used and the value is moved up tillheappropertyissatisfied.
Now the value at index 1 is comparedwith index 0andtosatisfyheappropertyitisfurtherswapped.
Now finally max heap is created by inserting newnode.
Example11.3:voidHeapInsert(PtrHeappHeap,intvalue)
pHeap->Array[pHeap->Size]=value;ProclateUp(pHeap->Array,pHeap->Size);pHeap->Size++;
voidProclateUp(int*arr,intposition)
intparent=(position–1)/2;
if(position!=0)if(arr[parent]>arr[position])inttemp=arr[position];arr[position]=arr[parent];arr[parent]=temp;
ProclateUp(arr,parent);
Dequeue/Delete1.Copythevalueattherootoftheheaptothevariableusedtoreturnavalue.2.Copythe lastelementof theheapto theroot,and thenreduce thesizeofheapby1.Thiselement is
calledthe"out-of-place"element.3.Restoreheappropertybyswappingtheout-of-placeelementwithitsgreatest-valuechild.Repeatthis
processuntiltheout-of-placeelementreachesaleaforithasavaluethatisgreaterorequaltoallitschildren.
4.ReturntheanswerthatwassavedinStep1.Todequeueanelementfromheapitstopvalueisswappedtotheendoftheheaparrayandsizeofheapisreducedby1.
Sinceendoftheheapvalueiscopiedtohead of heap. Heap property isdisturbed sowe need to again proclatedown by comparing node with itschildrennodestorestoreheapproperty.
Proclate downcontinuedby comparingwithitschildrennodes.
Proclatedown
ProclatedownComplete
Example11.4:intHeapDelete(PtrHeappHeap)
intvalue=(pHeap->Array[0);pHeap->Array[0]=pHeap->Array[pHeap->Size-1];pHeap->Size--;ProclateDown(pHeap->Array,0,pHeap->Size);returnvalue;
Heap-Sort1.Usecreateheapfunctiontobuildamaxheapfromthegivenarrayofelements.Thisoperationwilltake
O(N)time.2.Dequeuethemaxvaluefromtheheapandstorethisvaluetotheendofthearrayatlocationarr[size-1]
a)Copythevalueattherootoftheheaptoendofthearray.b)Copy the last element of the heap to the root, and then reduce the size of heap by 1.This
elementiscalledthe"out-of-place"element.c) Restore heap property by swapping the out-of-place elementwith its greatest-value child.
Repeat this process until the out-of-place element reaches a leaf or it has a value that isgreaterorequaltoallitschildren
3.Repeatthisoperationuntilthereisjustoneelementintheheap.Letustakeexampleoftheheapthatwehadcreatedatthestartofthechapter.Heapsortisalgorithmstartsbycreatingaheapof thegivenarray that isdone in linear time.Thenateachstepheadof theheap isswappedwiththeendoftheheapandtheheapsizeisreducedby1.Thenproclatedownisusedtorestoretheheapproperty.Inaddition,thissameisdonemultipletimesuntiltheheapcontainjustoneelement.
Example11.5:voidHeapSort(int*arr,intsize)
inttemp;for(inti=size/2;i>0;i--)ProclateDown(arr,i,size);while(size)temp=arr[0];arr[0]=arr[size-1];arr[size-1]=temp;size--;ProclateDown(arr,0,size);
voidProclateDown(int*arr,intposition,intsize)
intlChild=2*position+1;intrChild=lChild+1;intsmall=-1;
if(lChild<size)small=lChild;if(rChild<size&&compare(arr[rChild],arr[lChild]))//comparefunctiondecideminsmall=rChild;
if(small!=-1&&compare(arr[small],arr[position]))inttemp=arr[position];arr[position]=arr[small];arr[small]=temp;ProclateDown(arr,small,size);
//Swap,whentheparentislessthanchild,willgivemaxheap.//Swapwhenaparentisgreaterthenachild.itwillgiveminheap.intcompare(intparentVal,intchildVal)
return(parentVal<childVal);intmain()
inta[10]=4,5,3,2,6,7,11,8,9,10;heapSort(a,sizeof(a)/sizeof(int));
Datastructure ArrayWorstCaseTimeComplexity O(nlogn)BestCaseTimeComplexity O(nlogn)AverageTimeComplexity O(nlogn)SpaceComplexity O(1)Note:Heap-SortisnotaStablesortanddonotrequireanyextraspaceforsortingalist.
UsesofHeap1.Heapsort:Oneofthebestsortingmethodsbeingin-placeandlog(N)timecomplexityinallscenarios.2.Selectionalgorithms: Finding themin,max, both themin andmax,median, or even the kth largest
elementcanbedoneinlineartime(oftenconstanttime)usingheaps.3. Priority Queues: Heap Implemented priority queues are used in Graph algorithms like Prim’s
AlgorithmandDijkstra’salgorithm.Aheap is a useful data structurewhen you need to remove theobjectwiththehighest(orlowest)priority.Schedulers,timers
4.Graphalgorithms:Byusingheapsas internal traversaldatastructures, run timewillbereducedby
polynomialorder.ExamplesofsuchproblemsarePrim'sminimal5. Because of the lack of pointers, the operations are faster than a binary tree. Also, some more
complicatedheaps(suchasbinomial)canbemergedefficiently,whichisnoteasytodoforabinarytree.
ProblemsinHeap
KthSmallestinaMinHeapJustcallDeleteMin()operationK-1timesandthenagaincallDeleteMin() this lastoperationwillgiveKthsmallestvalue.TimeComplexityO(KlogN)
KthLargestinaMaxHeapJustcallDeleteMax()operationK-1timesandthenagaincallDeleteMax()thislastoperationwillgiveKthsmallestvalue.TimeComplexityO(KlogN)
100LargestinaStreamTherearebillionsofintegerscomingoutofastreamsomegetInt()functionisprovidingintegersonebyone.Howwouldyoudeterminethelargest100numbers?Solution:Largehundred(orsmallesthundredetc.)suchproblemsaresolvedveryeasilyusingaHeap.Inourcase,wewillcreateaminheap.1.Firstfrom100firstintegersbuildsaminheap.2.Thenforeachcomingintegercompareifitisgreaterthanthetopoftheminheap.3.Ifnot,thenlookfornextinteger.Ifyes,thenremovethetopminvaluefromtheminheaptheninsertthe
newvalueatthetopoftheheapanduseprocolateDownandmoveittoitsproperpositiondowntheheap.
4.Everytimeyouhavelargest100valuesstoredinyourhead
MergetwoHeapHowcanwemergetwoheaps?Solution:Thereisnosinglesolutionforthis.LetussupposethesizeofthebiggerheapisNandthesizeofthesmallerheapisM.1.Ifbothheapsarecomparablesize,thenputbothheaparraysinsamebiggerarrays.Alternatively,inone
of thearrays if theyarebig enough.ThenapplyCreateHeap() functionwhichwill take theta(N+M)time.
2.IfMismuchsmallerthanNthenenqueue()eachelementofMarrayonebyonetoNheap.ThiswilltakeO(MlogN)theworstcaseorO(M)bestcase.
Min/MaxHeapusingfunctionpointerExample11.6: Same heap data structure can be used to create bothmin heap andmax heap only thecomparefunctionwillchange.typedefstructHeap_t
intCapacity;intSize;int*Array;
bool(*compare)(int,int);Heap;voidHeapInitialize(PtrHeappHeap,intcapacity,bool(*comp)(int,int))
pHeap->Size=0;pHeap->Capacity=capacity;pHeap->Array=(int*)malloc((capacity)*sizeof(int));pHeap->compare=comp;
voidHeapInsert(PtrHeappHeap,intvalue)
pHeap->Array[pHeap->Size]=value;ProclateUp(pHeap->Array,pHeap->Size,pHeap->compare);pHeap->Size++;
voidProclateUp(int*arr,intposition,bool(*comp)(int,int))
intparent=(position-1)/2;
if(position!=0)if(comp(arr[parent],arr[position]))inttemp=arr[position];arr[position]=arr[parent];arr[parent]=temp;ProclateUp(arr,parent);
voidProclateDown(int*arr,intposition,intsize,bool(*comp)(int,int))
intlChild=2*position+1;intrChild=lChild+1;intsmall=-1;
if(lChild<size)small=lChild;
if(rChild<size&&comp(arr[rChild],arr[lChild]))small=rChild;
if(small!=-1&&comp(arr[small],arr[position]))inttemp=arr[position];arr[position]=arr[small];arr[small]=temp;ProclateDown(arr,small,size);
intHeapDelete(PtrHeappHeap)
intvalue=(pHeap->Array[0]);pHeap->Array[0]=pHeap->Array[pHeap->Size-1];pHeap->Size--;ProclateDown(pHeap->Array,0,pHeap->Size,pHeap->compare);returnvalue;
//Greaterusedascomparator,swapwhenparentisgreaterthenchild,minheapboolGreater(inta,intb)
returna>b;//Smallerusedascomparator,swapwhenparentissmallerthenchild,maxheapboolSmaller(inta,intb)
returna<b;intHeapSize(PtrHeappHeap)
returnpHeap->Size;
GetMedianfunctionExample11.7:Giveadatastructure,whichwillprovidemedianofgivenvaluesinconstanttime.Solution:Wewillbeusingtwoheaponeminheapandothermaxheap.First,therewillbeamaxheap,whichwillcontainthefirsthalfofdata,andtherewillbeanotherminheap,whichwillcontainthesecondhalfofthedata.Maxheapwillcontainthesmallerhalfofthedataanditsmaxvalue,whichisatthetopoftheheap,willbethemediancontender.Similarly,theMinheapwillcontainthelargervaluesofthedataanditsminvalue,whichisatitstop,willcontainthemediancontender.Wewillkeeptrackofthesizeofheaps.Wheneverweinsertavaluetoheap,wewillmakesurethatthesizeoftwoheapsdiffersbymax
oneelement,otherwisewewillpoponeelementfromoneandinsertintoanothertokeepthembalanced.voidMedianHeapInit(MedianHeap*heap)
HeapInitialize(&heap->minHeap,100,Greater);HeapInitialize(&heap->maxHeap,100,Smaller);
voidMedianHeapInsert(MedianHeap*heap,intvalue)
if(HeapSize(&heap->maxHeap)==0||HeapTop(&heap->maxHeap)>=value)HeapInsert(&heap->maxHeap,value);elseHeapInsert(&heap->minHeap,value);//sizebalancingif(HeapSize(&heap->maxHeap)>HeapSize(&heap->minHeap)+1)value=HeapDelete(&heap->maxHeap);HeapInsert(&heap->minHeap,value);if(HeapSize(&heap->minHeap)>HeapSize(&heap->maxHeap)+1)value=HeapDelete(&heap->minHeap);HeapInsert(&heap->maxHeap,value);
intgetMedian(MedianHeap*heap)
if(HeapSize(&heap->maxHeap)==0&&HeapSize(&heap->minHeap)==0)returnINT_MIN;
if(HeapSize(&heap->maxHeap)==HeapSize(&heap->minHeap))return(HeapTop(&heap->maxHeap)+HeapTop(&heap->minHeap))/2;elseif(HeapSize(&heap->maxHeap)>HeapSize(&heap->minHeap))returnHeapTop(&heap->maxHeap);elsereturnHeapTop(&heap->minHeap);
intmain()
intarr[]=1,9,2,8,3,7,4,6,5,1,9,2,8,3,7,4,6,5,10,10;
MedianHeapheap;MedianHeapInit(&heap);for(inti=0;i<20;i++)MedianHeapInsert(&heap,arr[i]);printf("Medianafterinsertionof%dis%d\n",arr[i],getMedian(&heap));return0;
IsMinHeapGivenanarrayfindifitisabinaryHeapisMinHeapExample11.8:intIsMinHeap(intarr[],intsize)
for(inti=0;i<=(size-2)/2;i++)if(2*i+1<size)if(arr[i]>arr[2*i+1])return0;if(2*i+2<size)if(arr[i]>arr[2*i+2])return0;return1;
IsMaxHeapGivenanarrayfindifitisabinaryHeapMaxheapExample11.9:intIsMaxHeap(intarr[],intsize)
for(inti=0;i<=(size-2)/2;i++)
if(2*i+1<size)if(arr[i]<arr[2*i+1])return0;if(2*i+2<size)if(arr[i]<arr[2*i+2])return0;return1;
TraversalinHeapHeapsarenotdesignedto traverse tofindsomeelement theyaremadetogetminormaxelementfast.Stillifyouwanttotraverseaheapjusttraversethearraysequentially.Thistraversalwillbelevelordertraversal.ThistraversalwillhavelinearTimeComplexity.
DeletingArbitraryelementfromMinHeapHeap isnotdesigned todeleteanarbitraryelement,butstill ifyouwant todoso.Find theelementbylinearsearchintheheaparray.ReplaceitwiththevaluestoredattheendoftheHeapvalue.Reducethesizeoftheheapbyone.Comparethenewinsertedvaluewithitsparent.If itsvalueissmallerthantheparentvalue,thenpercolateup.Elseifitsvalueisgreaterthanitsleftandrightchildthenpercolatedown.TimeComplexityisO(logn)
DeletingKthelementfromMinHeapHeapisnotdesignedtodeleteanarbitraryelement,butstillifyouwanttodoso.ReplacethekthvaluewiththevaluestoredattheendoftheHeapvalue.Reducethesizeoftheheapbyone.Comparethenewinsertedvaluewithitsparent.Ifitsvalueissmallerthantheparentvalue,thenpercolateup.Elseif itsvalueisgreaterthanitsleftandrightchildthenpercolatedown.TimeComplexityisO(logn)
PrintvalueinRangeinMinHeapLinearlytraversethroughtheheapandprintthevaluethatareinthegivenrange.
Exercise1.Whatistheworst-caseruntimeComplexityoffindingthesmallestiteminamin-heap?2.Findmaxinaminheap.
Hint:normalsearchinthecompletearray.ThereisonemoreoptimizationyoucansearchfromthemidofthearrayatindexN/2
3.Whatistheworst-casetimeComplexityoffindingthelargestiteminamin-heap?4.Whatistheworst-casetimeComplexityofdeleteMininamin-heap?5.Whatistheworst-casetimeComplexityofbuildingaheapbyinsertion?6.Isaheapfullorcompletebinarytree?7.WhatistheworsttimeruntimeComplexityofsortinganarrayofNelementsusingheapsort?8.Givenasequenceofnumbers:1,2,3,4,5,6,7,8,9
a.DrawabinaryMin-heapbyinsertingtheabovenumbersonebyoneb.AlsodrawthetreethatwillbeformedaftercallingDequeue()onthisheap
9.Givenasequenceofnumbers:1,2,3,4,5,6,7,8,9
a.DrawabinaryMax-heapbyinsertingtheabovenumbersonebyoneb.AlsodrawthetreethatwillbeformedaftercallingDequeue()onthisheap
10.Givenasequenceofnumbers:3,9,5,4,8,1,5,2,7,6.ConstructaMin-heapbycallingCreateHeap
function.11.ShowanarraythatwouldbetheresultafterthecalltodeleteMin()onthisheap12.Givenanarray:[3,9,5,4,8,1,5,2,7,6].Applyheapifyoverthistomakeaminheapandsortthe
elementsindecreasingorder?13.InHeap-Sortoncearootelementhasbeenputinitsfinalposition,howmuchtimedoesittaketore-
heapify the structure so that the next removal can take place? In other words, what is the TimeComplexityofasingleelementremovalfromtheheapofsizeN?
14.WhatdoyouthinktheoverallTimeComplexityforheapsortis?Whydoyoufeelthisway?
CHAPTER12:HASH-TABLE
IntroductionIn the previous chapter, we have looked into various searching techniques. Consider a problem ofsearchingavalueinanarray.Ifthearrayisnotsortedthenwehavenootheroptionbuttolookintoeveryelementonebyoneso thesearchingTimeComplexitywillbeO(n). If thearray is sorted thenwecansearchthevaluewearelookingforinO(logn)logarithmictimeusingbinarysearch.Whatifwehaveafunctionthatcantellusthelocation/indexofthevaluewearelookingforinthearray?WecandirectlygointothatlocationandtellwhetherourobjectwearesearchingforispresentornotinjustO(1)constanttime.SuchafunctioniscalledaHashfunction.In real lifewhena letter ishandedover toapostman,by lookingat theaddresson the letter,postmanpreciselyknowstowhichhousethisletterneedstobedelivered.Heisnotgoingtoaskforapersondoortodoor.
Theprocessofstoringobjectsusingahashfunctionisasfollows:1.CreateanarrayofsizeMtostoreobjects,thisarrayiscalledHash-Table.2.Findahashcodeofanobjectbypassingitthroughthehashfunction.3.TakemoduleofhashcodebythesizeofHashtabletogettheindexofthetablewhereobjectswillbestored.4.Finallystoretheseobjectsinthedesignatedindex.TheprocessofsearchingobjectsinHash-Tableusingahashfunctionisasfollows:1.Findahashcodeoftheobjectwearesearchingforbypassingitthroughthehashfunction.2. Takemodule of hash code by the size ofHashtable to get the index of the tablewhere objects arestored.3.Finally,retrievetheobjectfromthedesignatedindex.
Hash-TableAHash-Tableisadatastructure,whichmapskeystovalues.EachpositionoftheHash-Tableiscalledaslot.TheHash-Tableusesahash function tocalculatean indexofanarrayof slots.Weuse theHash-Tablewhenthenumberofkeysactuallystoredissmallrelativelytothenumberofpossiblekeys.
Hash-TableAbstractDataType(ADT)ADTofHash-Tablecontainsthefollowingfunctions:1.Insert(x),addobjectxtothedataset.2.Delete(x),deleteobjectxfromthedataset.3.Search(x),searchobjectxindataset.
HashFunctionAhashfunctionisafunction,whichgeneratesanindexinatableforagivenobject.Anidealhashfunctionshouldgenerateauniqueindexforeveryobjectiscalledtheperfecthashfunction.Example12.1:MostsimplehashfunctionunsignedintHash(intkey,inttableSize)//divisionmethod
unsignedinthashValue=0;hashValue=key;returnhashValue%tableSize;
Therearemanyhashfunctions,butthisistheminimumthatitshoulddo.Varioushashgenerationlogicswillbeaddedtothisfunctiontogenerateabetterhash.
Propertiesofgoodhashfunction:1. It should provide a uniform distribution of hash values. A non-uniform distribution increased the
numberofcollisionsandthecostofresolvingthem.2.Choose a hash function,which can be computed quickly and returns valueswithin the range of the
Hash-Table.3. Chose a hash function with a good collision resolution algorithm which can be used to compute
alternativeindexifthecollisionoccurs.4.Chooseahashfunction,whichusesthenecessaryinformationprovidedinthekey.5.Itshouldhavehighloadfactorforagivensetofkeys.
LoadFactorLoadfactor=NumberofelementsinHash-Table/Hash-TablesizeBased on the above definition, Load factor tells whether the hash function is distributing the keysuniformlyornot.Therefore,ithelpsindeterminingtheefficiencyofthehashingfunction.Italsoworksas
decisionparameterwhenwewanttoexpandorrehashtheexistingHash-Tableentries.
CollisionsWhenahashfunctiongeneratesthesameindexforthetwoormoredifferentobjects,theproblemknownasthecollision.Ideally,hashfunctionshouldreturnauniqueaddressforeachkey,butpracticallyitisnotpossible.
CollisionResolutionTechniquesHashcollisionsarepracticallyunavoidablewhenhashinglargenumberofobjects.Techniques,whichareusedtofindthealternatelocationintheHash-Table,iscalledcollisionresolution.Thereareanumberofcollisionresolutiontechniquestohandlethecollisioninhashing.Mostcommonandwidelyusedtechniquesare:·Openaddressing·Separatechaining
HashingwithOpenAddressing
When using linear open addressing, the Hash-Table is represented by a one-dimensional array withindicesthatrangefrom0tothedesiredtablesize-1.Onemethodof resolvingcollision is the look intoaHash-Tableandfindanother freeslot thehold theobjectthathavecausedthecollision.Asimplewayistomovefromoneslottoanotherinsomesequentialorderuntilwefindafreespace.ThiscollisionresolutionprocessiscalledOpenAddressing.
LinearProbingInLinearProbing,wetrytoresolvethecollisionofanindexofaHash-TablebysequentiallysearchingtheHash-Tablefreelocation.Letussuppose,ifkistheindexretrievedfromthehashfunction.Ifthekthindexisalreadyfilledthenwewilllookfor(k+1)%M,then(k+2)%Mandsoon.Whenwegetafreeslot,wewillinserttheobjectintothatfreeslot.Example12.2:TheresolverfunctionoflinearprobingintresolverFun(inti)
returni;
QuadraticProbingIn Quadratic Probing, we try to resolve the collision of the index of a Hash-Table by quadratic allyincreasingthesearchindexfreelocation.Letussuppose,ifkistheindexretrievedfromthehashfunction.Ifthekthindexisalreadyfilledthenwewilllookfor(k+1^2)%M,then(k+2^2)%Mandsoon.Whenwegetafreeslot,wewillinserttheobjectintothatfreeslot.Example12.3:TheresolverfunctionofquadraticprobingintresolverFun(inti)
returni*i;Tablesizeshouldbeaprimenumbertopreventearlyloopingshouldnotbetoocloseto2powN
LinearProbingimplementationExample12.4:BelowisalinearprobingcollisionresolutionHash-Tableimplementation.#defineTABLE_SIZE50#defineEMPTY_NODE-1#defineLAZY_DELETED-2Table array size will be 50 and we have defined two constant values EMPTY_NODE andLAZY_DELETED.
unsignedintComputeHash(intkey,inttableSize)//divisionmethod
unsignedinthashValue=0;hashValue=key;returnhashValue%tableSize;
Thisisthemostsimplehashgenerationfunction,whichjusttakethemodulusofthekey.intResolverFun(inti)
returni;Whenthehashindexisalreadyoccupiedbysomeelementthevaluewillbeplacedinsomeotherlocationtofindthatnewlocationresolverfunctionisused.structhashash_t
inttableSize;int*intArray;
;typedefhashash_t*hashPtr;Hash-Tablehastwocomponentoneistablesizeandotherispointertoarray.hashPtrHashInitialize()
hashPtrhash;hash=(hashPtr)malloc(sizeof(hashash_t));
hash->tableSize=TABLE_SIZE;hash->intArray=(int*)malloc(hash->tableSize*sizeof(int));
for(inti=0;i<hash->tableSize;i++)hash->intArray[i]=EMPTY_NODE;
returnhash;
HashInitializeis thefirstfunction,whichwill initialize theHash-Table.This is thefirstfunctioncalledafteraHash-Tablebeforeanyinsert,searchordeleteoperation.Example12.5:intHashInsert(hashPtrhash,intvalue)
inthashValue=ComputeHash(value,hash->tableSize);inti=0;for(i=0;i<hash->tableSize;i++)
if (hash->intArray[hashValue] == EMPTY_NODE || hash->intArray[hashValue] ==LAZY_DELETED)
hash->intArray[hashValue]=value;return1;hashValue=hashValue+ResolverFun(i);hashValue=hashValue%hash->tableSize;return-1;
Aninsertnodefunctionisusedtoaddvaluestothearray.Firsthashiscalculated.ThenwetrytoplacethatvalueintheHash-Table.Welookforemptynodeorlazydeletednodetoinsertvalue.Incaseinsertdidnotsuccess,wetrynewlocationusingaresolverfunction.Example12.6:intHashFind(hashPtrhash,intvalue)
inthashValue=ComputeHash(value,hash->tableSize);for(inti=0;i<hash->tableSize;i++)if(hash->intArray[hashValue]==value||hash->intArray[hashValue]==EMPTY_NODE)break;hashValue=hashValue+ResolverFun(i);hashValue=hashValue%hash->tableSize;
if(hash->intArray[hashValue]==value)returnhashValue;elsereturn-1;//valuenotfound
Findnodefunctionisusedtosearchvaluesinthearray.Firsthashiscalculated.ThenwetrytofindthatvalueintheHash-Table.Welookforoverdesiredvalueoremptynode.Incasewefindthevaluethatwearelooking,thenwereturnthatvalueorwereturn-1.Weusearesolverfunctiontofindthenextprobableindextosearch.Example12.7:intHashDelete(hashPtrhash,intvalue)
inthashValue=ComputeHash(value,hash->tableSize);
for(inti=0;i<hash->tableSize;i++)if(hash->intArray[hashValue]==EMPTY_NODE)return-1;
if(hash->intArray[hashValue]==value)hash->intArray[hashValue]=LAZY_DELETED;return1;//deletedproperly
hashValue=hashValue+ResolverFun(i);hashValue=hashValue%hash->tableSize;return-1;//valuenotfound
DeletenodefunctionisusedtodeletevaluesfromaHashtable.WedonotactuallydeletethevaluewejustmarkthatvalueasLAZY_DELETED.SameastheinsertandsearchweuseresolverFuntofindthenextprobablelocationofthekey.Example12.8:voidPrintHash(hashPtrhash,inttableSize)//printkeywisevalues
for(inti=0;i<hash->tableSize;i++)printf("index%dvalue::%d\n",i,hash->intArray[i]);
intmain()
hashPtrmyHash=HashInitialize();for(inti=100;i<110;i++)HashInsert(myHash,i);printf("search100::%d\n",HashFind(myHash,100));printf("remove100::%d\n",HashDelete(myHash,100));printf("search100::%d\n",HashFind(myHash,100));printf("remove100::%d\n",HashDelete(myHash,100));PrintHash(myHash,TABLE_SIZE);
QuadraticProbingimplementation.
Everythingwillbesameaslinearprobingimplementationonlyresolverfunctionwillbechanged.intresolverFun(inti)
returni*i;
HashingwithSeparate-ChainingAnothermethodforcollisionresolutionisbasedonanideaofputtingthekeysthatcollideinalinkedlist.Thismethodiscalledseparatechaining.TospeedupsearchweuseInsertion-Sortorkeepingthelinkedlistsorted.
SeparateChainingimplementation‘Example12.9:Belowisseparatechainingimplementationofhashtables.#defineTABLE_SIZE517#defineTABLE_BITS9structlistNode_t
intvalue;listNode_t*next;
;typedeflistNode_t*ptrList;structhashTable_t
inttableSize;ptrList*listArray;//doublepointer
;typedefhashTable_t*hashPtr;unsignedintComputeHash(intkey,inttableSize)//divisionmethod
unsignedinthashValue=0;hashValue=key;returnhashValue%tableSize;
hashPtrHashInitialize(intsize)
hashPtrhTable=(hashPtr)malloc(sizeof(hashTable_t));hTable->tableSize=size;hTable->listArray=(ptrList*)malloc(hTable->tableSize*sizeof(ptrList));
for(inti=0;i<hTable->tableSize;i++)hTable->listArray[i]=NULL;returnhTable;
voidPrintHash(hashPtrhTable,inttableSize)//printkeywisevalues
for(inti=0;i<hTable->tableSize;i++)printf("\nPrintingforindexvalue::(%d)Listofvalueprinting::",i);ptrListhead=hTable->listArray[i];while(head)printf("%d",head->value);head=head->next;
intHashFind(hashPtrhTable,intvalue)
ptrListpossition;intindex=ComputeHash(value,hTable->tableSize);possition=hTable->listArray[index];while(possition&&possition->value!=value)possition=possition->next;return!(possition==NULL);
voidHashInsert(hashPtrhTable,intvalue)
intindex=ComputeHash(value,hTable->tableSize);ptrListtempPtr=(ptrList)malloc(sizeof(listNode_t));tempPtr->value=value;tempPtr->next=hTable->listArray[index];hTable->listArray[index]=tempPtr;
intHashDelete(hashPtrhTable,intvalue)
ptrListcurrNode,nextNode;intindex=ComputeHash(value,hTable->tableSize);currNode=hTable->listArray[index];
if(currNode&&currNode->value==value)hTable->listArray[index]=currNode->next;free(currNode);return1;while(currNode)nextNode=currNode->next;if(nextNode&&nextNode->value==value)currNode->next=nextNode->next;free(nextNode);return1;elsecurrNode=nextNode;return0;
intmain()
hashPtrmyTable=HashInitialize(TABLE_SIZE);for(inti=100;i<110;i++)HashInsert(myTable,i);printf("search100::%d\n",HashFind(myTable,100));printf("remove100::%d\n",HashDelete(myTable,100));printf("search100::%d\n",HashFind(myTable,100));printf("remove100::%d\n",HashDelete(myTable,100));PrintHash(myTable,TABLE_SIZE);
Note: It is important tonote that the sizeof the“skip”mustbe such that all the slots in the tablewilleventuallybeoccupied.Otherwise,partofthetablewillbeunused.Toensurethis,itisoftensuggestedthatthetablesizebeingaprimenumber.Thisisthereasonwehavebeenusing11inourexamples.
ProblemsinHashing
AnagramsolverAnanagramisawordorphraseformedbyreorderingthelettersofanotherwordorphrase.Example12.10:Twowordsareanagramiftheyareofsamesizeandtheircharactersaresame.intisAnagram(charstr1[],charstr2[])
intcurr=0;intsize1=strlen(str1);intsize2=strlen(str2);if(size1!=size2)return0;hashPtrpHash=HashInitialize();
while(str1[curr])HashInsert(pHash,str1[curr]);curr++;curr=0;
while(str2[curr])if(!HashDelete(pHash,str2[curr]))return0;curr++;return1;
RemoveDuplicateRemoveduplicatesinanarrayofnumbers.Solution:Wecanuseasecondarrayorthesamearray,astheoutputarray.InthebelowexampleHash-Tableisusedtosolvethisproblem.Example12.11:voidremoveDuplicate(charstr[])
intcurr=0,end=0;intsize=strlen(str);hashPtrpHash=HashInitialize();
while(str[curr])if(!HashFind(pHash,str[curr]))str[end++]=str[curr];HashInsert(pHash,str[curr]);curr++;str[end]='\0';
FindMissingExample12.12:Findthemissingnumberinthelistofintegers.intfindMissing(intarr[],intsize)
intcurr=0;intn=size+1;hashPtrpHash=HashInitialize();
while(arr[curr])HashInsert(pHash,arr[curr]);curr++;
for(curr=1;curr<=n;curr++)if(!HashFind(pHash,curr))returncurr;return1;
PrintRepeatingExample12.13:Printtherepeatingintegerinalistofintegers.voidprintRepeating(intarr[],intsize)
inti;hashPtrpHash=HashInitialize();printf("Repeatingelementsare");
for(i=0;i<size;i++)if(HashFind(pHash,arr[i]))printf("%d",arr[i]);elseHashInsert(pHash,arr[i]);
PrintFirstRepeatingExample12.14:Sameastheaboveprobleminthisweneedtoprintthefirstrepeatingnumber.Cautionshouldbetakentofindthefirstrepeatingnumber.Itshouldbethenumberwhichisrepeating.Forexample1,2,3,2,1.Theanswershouldbe1asitisthefirstnumber,whichisrepeating.voidprintFirstRepeating(intarr[],intsize)
inti;hashPtrpHash=HashInitialize();
for(i=0;i<size;i++)HashInsert(pHash,arr[i]);for(i=0;i<size;i++)HashDelete(pHash,arr[i]);
if(HashFind(pHash,arr[i]))printf("FirstRepeatingnumberis:%d",arr[i]);return;
Exercise1.Designanumber(ID)generatorsystemthatgeneratenumbersbetween0-99999999(8-digits).
Thesystemshouldsupporttwofunctions:a)intgetNumber();b)intrequestNumber();
getNumber()functionshouldfindoutanumberthatisnotassigned,thenmarksitasassignedandreturnthatnumber.requestNumber()functionchecksthenumberisassignedornot.Ifitisassignedreturns0,elsemarksitasassignedandreturn1.Hint:Youcankeepacounter forassigningnumbers.Whenever there isagetNumber()callyouwillcheckif thatnumber isalreadyassignedinaHash-Table.If it isalreadyassigned, thenincrease thecounter and check again. If you find a number not in the Hash-Table then add it to Hashtable andincreasethecounter.requestNumber()willlookintheHash-Tableifthenumberisalreadytaken,thenitwillreturn0elseitwillreturn1andmarkthatnumberastakeninsidetheHash-Table.
2.Givenalargestring,findthemostoccurringwordsinthestring.WhatistheTimeComplexityofthe
abovesolution?Hint:-
a.CreateaHashtablewhichwillkeeptrackof<word,frequency>b.IteratethroughthestringandkeeptrackofwordfrequencybyinsertingintoHash-Table.c. When we have a new word, we will insert it into the Hashtable with frequency 1. For all
repetitionoftheword,wewillincreasethefrequency.d.Wecankeeptrackofthemostoccurringwordswheneverweareincreasingthefrequencywecan
seeifthisisthemostoccurringwordornot.e.TheTimeComplexityisO(n)wherenisthenumberofwordsinthestringandSpaceComplexity
istheO(m)wheremistheuniquewordsinthestring.3.Intheabovequestion,WhatifyouaregivenwholeworkofOSCARWILDE,mostpopularplaywrights
intheearly1890s.Hint:-
a.Whoknowshowmanybooksarethere,letusassumethereisalotandwecannotputeverythinginmemory.First,weneedaStreamingLibraryso thatwecanreadsectionbysection ineachdocument.Thenweneed a tokenizer,whichwill givewords to our program. In addition,weneedsomesortofdictionaryletussaywewilluseHashTable.
b.Whatyouneedis-1.Astreaminglibrarytokenizer,2.Atokenizer3.AhashmapMethod:1.Usestreamerstofindastreamofthegivenwords2.Tokenizetheinputtext3.Ifthestemmedwordisinhashmap,incrementitsfrequencycountelseaddsawordtohashmapwithfrequency1
c.Wecanimprovetheperformancebylookingintoparallelcomputing.Wecanusethemap-reduce
tosolvethisproblem.Multiplenodeswillreadandprocessmultipledocuments.Oncetheyaredonewiththeirprocessing,thenwecanusereducetomergethem.
4.Intheabovequestion,WhatifwewantedtofindthemostcommonPHRASEinhiswritings.
Hint:-We can keep <phrase, frequency> Hash-Table and do the same process of the 2nd and 3rdproblems.
5.Writeahashingalgorithmforstrings.
Hint:UseHorner'smethodinthornerHash(char*key,inttableSize)
intsize=strlen(key);inth=0;inti;for(i=0;i<size;i++)h=(32*h+key[i])%tableSize;returnh;
6.PicktwodatastructurestouseinimplementingaMap.Describelookup,insert,&deleteoperations.Givetime&SpaceComplexityforeach.Givepros&consforeach.Hint:-a)LinkedList
I.InsertisO(1)II.DeleteisO(1)III.LookupisO(1)auxiliaryandO(N)worstcase.IV.Pros:Fastinsertsanddeletes,canuseforanydatatype.V.Cons:Slowlookups.
b)BalancedSearchTree(RBTree)I.InsertisO(logn)II.DeleteisO(logn)III.LookupisO(logn)IV.Pros:Reasonablyfastinserts/deletesandlookups.V.Cons:Dataneedstohaveorderdefinedonit.
CHAPTER13:GRAPHS
IntroductionInthischapter,wewillstudyaboutGraphs.Graphscanbeusedtorepresentmanyinterestingthingsintherealworld.Flightsfromcities tocities, rodsconnectingvarious townandcities.Even thesequenceofstepsthatwetaketobecomereadyforjobsdaily,orevenasequenceofclassesthatwetaketobecomeagraduate incomputerscience.Oncewehaveagoodrepresentationof themap, thenweuseastandardgraphalgorithmstosolvemanyinterestingproblemsofreallife.The flight connectionbetweenmajor citiesof India canalsobe representedby thebelowgraph.Eachnodeisacityandeachedgeisastraightflightpathfromonecitytoanother.YoumaywanttogofromDelhi toChennai, ifgiven thisdata ingood representation to a computer, throughgraphalgorithms thecomputermayproposeshortest,quickestorcheapestpathfromsouredtodestination.
Googlemapthatweuseisalsoabiggraphoflotsofnodesandedges.Thensuggestshortestandquickestpathtotheuser.GraphDefinitionsAGraphisrepresentedbyGwhereG=(V,E),whereVisafinitesetofpointscalledVerticesandEisafinitesetofEdges.Eachedgeisatuple(u,v)whereu,v∈V.Therecanbeathirdcomponentweighttothetuple.Weightiscosttogofromonevertextoanother.Edge inagraphcanbedirectedorundirected. If theedgesofgraphareoneway it iscalledDirectedgraphorDigraph.ThegraphwhoseedgesaretwowaysarecalledUndirectedgraphorjustgraph.APath isasequenceofedgesbetween twovertices.The lengthofapath isdefinedas thesumof theweightofalltheedgesinthepath.Twoverticesuandvareadjacentifthereisanedgewhoseendpointsareuandv.Inthebelowgraph:V=V1,V2,V3,V4,V5,V6,V7,V8,V9,
E=
Thein-degreeofavertexv,Donetedbyindeg(v)isthenumberofincomingedgestothevertexv.Theout-degreeofavertexv,Donetedbyoutdeg(v)isthenumberofoutgoingedgesofavertexv.Thedegreeofavertexv,Donetedbydeg(v)isthetotalnumberofedgeswhoseoneendpointisv.deg(v)=Indeg(v)+outdeg(v)Intheabovegraphdeg(V4)=3,indeg(V4)=2andoutdeg(V4)=1ACycleisapaththatstartsandendsatthesamevertexandincludeatleastonevertex.AnedgeisaSelf-Loopiftwoifitstwoendpointscoincide.Thisisaformofacycle.AvertexvisReachable fromvertexuor“ureachesv”if thereisapathfromutov.Inanundirectedgraphifvisreachablefromuthenuisreachablefromv.However,inadirectedgraphitispossiblethatureachesvbutthereisnopathfromvtou.AgraphisConnectedifforanytwoverticesthereisapathbetweenthem.AForestisagraphwithoutcycles.ASub-GraphofagraphGisagraphwhoseverticesandedgesareasubsetoftheverticesandedgesofG.ASpanningSub-GraphofGisagraphthatconnectsalltheverticesofG.ATreeisaacyclicconnectedgraph.ASpanningtreeofagraphisaspanningsub-graph,whichisalsoatreethatmeans,aconnectedgraph,whichconnectsalltheverticesofgraphandthat,doesnothaveacycle.
GraphRepresentationInthissection,weintroducethedatastructureforrepresentingagraph.Inthebelowrepresentationswemaintainacollectiontostoreedgesandverticesofthegraph.
AdjacencyMatrixOneof theways to represent a graph is to use two-dimensionalmatrix.Each combinationof rowandcolumnrepresentavertexinthegraph.Thevaluestoredatthelocationrowvandcolumnwistheedgefromvertexvtovertexw.Thenodesthatareconnectedbyanedgearecalledadjacentnodes.ThismatrixisusedtostoreadjacentrelationsoitiscalledtheAdjacencyMatrix.Inthebelowdiagram,wehaveagraphanditsAdjacencymatrix.
Intheabovegraph,eachnodehasweight1sotheadjacencymatrixhasjust1sor0s.Iftheedgesareofdifferent,weightsthatthatweightwillbefilledinthematrix.Pros:Adjacencymatriximplementationissimple.Adding/RemovinganedgebetweentwoverticesisjustO(1).QueryifthereisanedgebetweentwoverticesisalsoO(1)Cons:ItalwaysconsumesO(V2)space,whichisaninefficientwaytostorewhenagraphisasparse.SparseMatrix: In a huge graph, each node is connected with fewer nodes. So most of the places inadjacencymatrix are empty. Suchmatrix is called sparsematrix. Inmost of the real world problemsadjacencymatrixisnotagoodchoiceforsoregraphdata.
AdjacencyListAmorespaceefficientwayofstoringgraphisadjacencylist.Inadjacencylistofpointerstoalinkedlistnode.Eachpointercorrespondstoverticesinagraph.Eachpointerwillthenpointtothevertices,whichareconnectedtoitandstorethisasalist.Inthebelowdiagramnode2isconnectedto1,3and4.Therefore,thepointeratlocation2ispointingtoalistthatcontain1,3and4.
Theadjacencylisthelpsustorepresentasparsegraph.Anadjacencylistrepresentationalsoallowsustofind all the vertices that are directly connected to any vertices by just one link list scan. In all ourprograms,wearegoingtousetheadjacencylisttostorethegraph.BelowisCcodeforadjacencylistrepresentationofanundirectedgraph:Example13.1:structListNode
intweight;intindex;ListNode*next;
;typedefListNode*ListPtr;typedefstructGraph_t
ListPtr*dpHead;intcount;
Graph;typedefGraph*GraphPtr;voidGraphInit(GraphPtrpGraph,intnodeCount)//oneextrasentinelforease
ListPtr*dpHead=(ListPtr*)malloc((nodeCount)*sizeof(ListPtr));for(inti=0;i<=nodeCount;i++)dpHead[i]=NULL;pGraph->dpHead=dpHead;pGraph->count=nodeCount;
intGraphInsert(GraphPtrG,intsrc,intdst,intweight)
ListPtrtemp=(ListPtr)malloc(sizeof(ListNode));temp->weight=weight;temp->index=dst;temp->next=G->dpHead[src];G->dpHead[src]=temp;return1;
voidUndirectedGraphInsert(GraphPtrG,intsrc,intdst,intweight)
GraphInsert(G,src,dst,weight);GraphInsert(G,dst,src,weight);
voidGraphPrint(GraphPtrG)
intcount=G->count,index;for(index=0;index<count;index++)ListPtrhead=G->dpHead[index];printf("\nAdjacencylistofindex%dare[",index);while(head)printf("%d",head->index);head=head->next;printf("]\n");
intmain()
intnodeCount=8;Graphgph;GraphInit(&gph,nodeCount);UndirectedGraphInsert(&gph,0,1,1);UndirectedGraphInsert(&gph,0,2,1);UndirectedGraphInsert(&gph,0,3,1);UndirectedGraphInsert(&gph,1,4,1);UndirectedGraphInsert(&gph,2,5,1);UndirectedGraphInsert(&gph,3,6,1);
UndirectedGraphInsert(&gph,4,7,1);UndirectedGraphInsert(&gph,5,7,1);UndirectedGraphInsert(&gph,6,7,1);GraphPrint(&gph);return0;
GraphtraversalsTheDepthfirstsearch(DFS)andBreadthfirstsearch(BFS)arethetwoalgorithmsusedtotraverseagraph.Thesesamealgorithmscanalsobeusedtofindsomenodeinthegraph,findifanodeisreachableetc.Traversalistheprocessofexploringagraphbyexaminingallitsedgesandvertices.Alistofsomeoftheproblemsthataresolvedusinggraphtraversalare:1.Determiningapathfromvertexutovertexv,orreportanerrorifthereisnosuchpath.2.Givenastartingvertexs,findingtheminimumnumberofedgesfromvertexstoalltheotherverticesof
thegraph.3.TestingofagraphGisconnected.4.FindingaspanningtreeofaGraph.5.Findingifthereissomecycleinthegraph.
DepthFirstTraversalTheDFSalgorithmwestartfromstartingpointandgointodepthofgraphuntilwereachadeadendandthenmoveup toparentnode(Backtrack). InDFS,weusestack toget thenextvertex tostartasearch.Alternatively,wecanuserecursion(systemstack)todothesame.
AlgorithmstepsforDFS1.Pushthestartingnodeinthestack.2.Loopuntilthestackisempty.3.Popthenodefromthestackinsideloopcallthisnodecurrent.4.Processthecurrentnode.//Print,etc.5.Traverseallthechildnodesofthecurrentnodeandpushthemintostack.6.Repeatsteps3to5untilthestackisempty.
StackbasedimplementationofDFSExample13.2:voidDFSStack(GraphPtrG)
intcount=G->count;int*visited=(int*)malloc((count)*sizeof(int));intcurr;Stackstk;for(inti=0;i<count;i++)visited[i]=0;StackInitialize(&stk);visited[0]=1;StackPush(&stk,0);
while(!StackIsEmpty(&stk))curr=StackPop(&stk);printf("%d",curr);
ListPtrhead=G->dpHead[curr];while(head)if(!visited[head->index])visited[head->index]=1;StackPush(&stk,head->index);head=head->next;
RecursionbasedimplementationofDFSExample13.3:voidDFSRec(GraphPtrG,intindex,int*visited)
printf("%d",index);ListPtrhead=G->dpHead[index];while(head)if(!visited[head->index])visited[head->index]=1;DFSRec(G,head->index,visited);head=head->next;
voidDFS(GraphPtrG)
intcount=G->count;int*visited=(int*)malloc((count)*sizeof(int));for(inti=0;i<count;i++)visited[i]=0;
for(inti=0;i<count;i++)if(visited[i]==0)visited[i]=1;DFSRec(G,i,visited);
BreadthFirstTraversalInBFS algorithm, a graph is traversed in layer-by-layer fashion. The graph is traversed closer to thestartingpoint.ThequeueisusedtoimplementBFS.
AlgorithmstepsforBFS1.PushthestartingnodeintotheQueue.2.LoopuntiltheQueueisempty.3.RemoveanodefromtheQueueinsideloop,callthisnodecurrent.4.Processthecurrentnode.//printetc.5.TraverseallthechildnodesofthecurrentnodeandpushthemintoQueue.6.Repeatsteps3to5untilQueueisempty.Example13.4:voidBFSQueue(GraphPtrG,intindex,int*visited)
intcurr;Queueque;QueueInitialize(&que);
visited[index]=1;Enqueue(&que,index);
while(!QueueIsEmpty(&que))curr=Dequeue(&que);printf("%d",curr);ListPtrhead=G->dpHead[curr];while(head)if(!visited[head->index])
visited[head->index]=1;Enqueue(&que,head->index);head=head->next;
voidBFS(GraphPtrG)
intcount=G->count;int*visited=(int*)malloc((count)*sizeof(int));for(inti=0;i<count;i++)visited[i]=0;
for(inti=0;i<count;i++)if(visited[i]==0)BFSQueue(G,i,visited);
AruntimeanalysisofDFSandBFStraversalisO(n+m)time,wherenisthenumberofedgesreachablefromsourcenodeandmisthenumberofedgesincidentons.ThefollowingproblemshaveO(m+n)timeperformance:1.Determiningapathfromvertexutovertexv,orreportanerrorifthereisnosuchpath.2.Givenastartingvertexs,findingtheminimumnumberofedgesfromvertexstoalltheotherverticesof
thegraph.3.TestingofagraphGisconnected.4.FindingaspanningtreeofaGraph.5.Findingifthereissomecycleinthegraph.
ProblemsinGraph
DeterminingapathfromvertexutovertexvIFthereisapathfromutovandwearedoingDFSfromuthenvmustbevisited.Ifthereisnopaththenreportanerror.Example13.5:intPathExist(GraphPtrG,intsrc,intdst)
intcount=G->count;int*visited=(int*)malloc((count)*sizeof(int));for(inti=0;i<count;i++)visited[i]=0;visited[src]=1;DFSRec(G,src,visited);returnvisited[dst];
Givenastartingvertexs,findingtheminimumnumberofedgesfromvertexstoalltheotherverticesofthegraphLookforsinglesourceshortestpathalgorithmforeachedgecostas1unit.
TestingofagraphGisconnected.DoDFS search start from any arbitrary vertex. Find if there is a vertex that is not visited. If all theverticesarevisitedthenitisaconnectedgraph.Example13.6:intisConnected(GraphPtrG)
intcount=G->count;int*visited=(int*)malloc((count)*sizeof(int));for(inti=0;i<count;i++)visited[i]=0;visited[0]=1;DFSRec(G,0,visited);for(inti=0;i<count;i++)if(visited[i]==0)return0;return1;
Findingifthereissomecycleinthegraph.
ModifyDFSproblemandgetthisdone.
DirectedAcyclicGraphADirected Acyclic Graph (DAG) is a directed graph with no cycle. A DAG represent relationship,whichismoregeneralthanatree.BelowisanexampleofDAG,thisishowsomeonebecomesreadyforwork.ThereareNother real lifeexamplesofDAGsuchascoercesselection tobeinggraduatedfromcollege
TopologicalSortA topological sort is a method of ordering the nodes of a directed graph in which nodes representactivities and the edges represent dependency among those tasks.For topological sorting towork it isrequiredthatthegraphshouldbeaDAGwhichmeansitshouldnothaveanycycle.JustuseDFStogettopologicalsorting.Example13.7:voidTopologicalSort(GraphPtrG)
Stackstk;StackInitialize(&stk);intcount=G->count;int*visited=(int*)malloc((count)*sizeof(int));for(inti=0;i<count;i++)visited[i]=0;for(inti=0;i<count;i++)if(visited[i]==0)visited[i]=1;TopologicalSortDFS(G,i,visited,&stk);while(!StackIsEmpty(&stk))printf("%d",StackPop(&stk));
voidTopologicalSortDFS(GraphPtrG,intindex,int*visited,Stack*stk)
ListPtrhead=G->dpHead[index];while(head)if(!visited[head->index])visited[head->index]=1;TopologicalSortDFS(G,head->index,visited,stk);head=head->next;StackPush(stk,index);
MinimumSpanningTrees(MST)ASpanningTreeofagraphGisatreethatcontainsalltheedgesoftheGraphG.AMinimumSpanningTreeisatreewhosesumoflength/weightofedgesisminimumaspossible.Forexample, ifyouwanttosetupcommunicationbetweenasetofcities, thenyoumaywant tousetheleastamountofwireaspossible.MSTcanbeusedtofindthenetworkpathandwirecostestimate.
Prim’sAlgorithmforMSTPrim’salgorithmgrowsasingletreeT,oneedgeatatime,untilitbecomesaspanningtree.WeinitializeTwithzeroedges.Uwithsinglenode.WhereTisspanningtreeedgessetandUisspanningtreevertexset.Ateachstep,Prim’salgorithmaddsthesmallestvalueedgewithoneendpointinUandothernotinus.SinceeachedgeaddsonenewvertextoU,aftern−1additions,UcontainalltheverticesofthespanningtreeandTbecomesaspanningtree.Example13.8://ReturnstheMSTbyPrim’sAlgorithm//Input:AweightedconnectedgraphG=(V,E)//Output:SetofedgescomprisingaMSTAlgorithmPrim(G)
T=LetrbeanyvertexinGU=rfori=1to|V|-1doe=minimum-weightedge(u,v)WithuinUandvinV-UU=U+v
T=T+ereturnT
Prim’sAlgorithmusingapriorityqueue(minheap)togettheclosestfringevertexTimeComplexitywillbeO(mlogn)wherenverticesandmedgesoftheMST.Example13.9:voidPrimMST(GraphPtrG)
inti,src=1;intcount=G->count;//Getthenumberofverticesingraphint*distance=(int*)malloc((count)*sizeof(int));int*path=(int*)malloc((count)*sizeof(int));
Heapque;HeapInitialize(&que,100);
for(i=0;i<count;i++)distance[i]=-1;
ListPtrhead;Edgee;e.src=src;e.dst=src;e.weight=0;HeapInsert(&que,e);distance[src]=0;path[src]=-1;Edgecurr;while(que.Size)curr=HeapDeleteMin(&que);head=G->dpHead[curr.dst];
while(head)intdisUpdate=head->weight;if(distance[head->index]==-1||distance[head->index]>disUpdate)distance[head->index]=disUpdate;path[head->index]=curr.dst;e.src=curr.dst;e.dst=head->index;e.weight=distance[head->index];HeapInsert(&que,e);
head=head->next;for(inti=0;i<count;i++)printf("%dto%dweight%d\n",path[i],i,distance[i]);
Kruskal’sAlgorithmKruskal’sAlgorithmrepeatedlychoosesthesmallest-weightedgethatdoesnotformacycle.Sorttheedgesinnon-decreasingorderofcost:c(e1)≤c(e2)≤···≤c(em).SetTtobetheemptytree.Addedgestotreeonebyoneifitdoesnotcreateacycle.Example13.10://ReturnstheMSTbyKruskal’sAlgorithm//Input:AweightedconnectedgraphG=(V,E)//Output:SetofedgescomprisingaMSTAlgorithmKruskal(G)
SorttheedgesEbytheirweightsT=while|T|+1<|V|doe=nextedgeinEifT+edoesnothaveacyclethenT=T+ereturnT
Kruskal’sAlgorithmisO(ElogV)usingefficientcycledetection.
ShortestPathAlgorithmsinGraph
SingleSourceShortestPathForagraphG=(V,E), thesinglesourceshortestpathproblemis tofindtheshortestpathfromagivensourcevertexstoalltheverticesofV.
SingleSourceShortestPathforunweightedGraph.Findsinglesourceshortestpathforunweightedgraphoragraphwhosealltheverticeshavesameweight.Example13.11:voidShortestPath(GraphPtrG,intsrc)
intcurr;intcount=G->count;int*distance=(int*)malloc((count)*sizeof(int));int*path=(int*)malloc((count)*sizeof(int));
Queueque;QueueInitialize(&que);
for(inti=0;i<count;i++)distance[i]=-1;
Enqueue(&que,src);distance[src]=0;
while(!QueueIsEmpty(&que))curr=Dequeue(&que);ListPtrhead=G->dpHead[curr];while(head)if(distance[head->index]==-1)distance[head->index]=distance[curr]+1;path[head->index]=curr;Enqueue(&que,head->index);head=head->next;for(inti=0;i<count;i++)printf("%dto%dweight%d\n",path[i],i,distance[i]);
Dijkstra’salgorithmDijkstra’salgorithmforsingle-sourceshortestpathproblemforweightededgeswithnonegativeweight.GivenaweightedconnectedgraphG, find shortest paths from the sourcevertex s to eachof theothervertices.Dijkstra’salgorithmissimilartoprimsalgorithm.Itmaintainsasetofnodesforwhichshortestpathisknown.
Thealgorithmstartsbykeepingtrackofthedistanceofeachnodeanditsparents.Allthedistanceissettoinfiniteinthebeginningaswedonotknowtheactualpathtothenodesandparentsofalltheverticesaresettonull.Alltheverticesareaddedtoapriorityqueue(minheapimplementation)Ateachstepalgorithmtakesonevertexfromthepriorityqueue(whichwillbethesourcevertexinthebeginning).Thenupdatethedistancearraycorrespondingtoalltheadjacentvertices.Whenthequeueisempty,thenwewillhavethedistanceandparentarrayfullypopulated.Example13.12://SolvesSSSPbyDijkstra’sAlgorithm//Input:AweightedconnectedgraphG=(V,E)//withnonegativeweights,andsourcevertexv//Output:ThelengthandpathfromstoeveryvAlgorithmDijkstra(G,s)
foreachvinVdoD[v]=infinite//UnknowndistanceP[v]=null//unknownpreviousnodeaddvtoPQ//addingallnodestopriorityqueue
D[source]=0//Distancefromsourcetosource
while(PQisnotempty)u=vertexfromPQwithsmallestD[u]removeufromPQforeachvadjacentfromudoalt=D[u]+length(u,v)
ifalt<D[v]thenD[v]=altP[v]=uReturnD[],P[]
TimeComplexitywillbeO(|E|log|V|)Note:Dijkstra’salgorithmdoesnotworkforgraphswithnegativeedgesweight.Note:Dijkstra’salgorithmisapplicabletobothundirectedanddirectedgraphs.Example13.13:voidDijkstraShortestPath(GraphPtrG,intsrc)
Edgecurr;inti;intcount=G->count;
int*distance=(int*)malloc((count)*sizeof(int));int*path=(int*)malloc((count)*sizeof(int));
Heapque;HeapInitialize(&que,100);
for(i=0;i<count;i++)distance[i]=-1;
ListPtrhead;Edgee;e.src=src;e.dst=src;e.weight=0;
HeapInsert(&que,e);distance[src]=0;path[src]=-1;
while(que.Size)curr=HeapDeleteMin(&que);head=G->dpHead[curr.dst];while(head)intdisUpdate=distance[curr.dst]+head->weight;if(distance[head->index]==-1||distance[head->index]>disUpdate)
distance[head->index]=disUpdate;path[head->index]=curr.dst;e.src=curr.dst;e.dst=head->index;e.weight=distance[head->index];HeapInsert(&que,e);head=head->next;for(inti=0;i<count;i++)printf("%dto%dweight%d\n",path[i],i,distance[i]);
BellmanFordShortestPathThebellmanfordalgorithmworksevenwhen therearenegativeweightedges in thegraph. Itdoesnotworkifthereissomecycleinthegraphwhosetotalweightisnegative.Example13.14:#defineINF123456789voidBellmanFordShortestPath(GraphPtrG,intsrc)
intcurr;intcount=G->count;
int*distance=(int*)malloc((count)*sizeof(int));int*path=(int*)malloc((count)*sizeof(int));
for(inti=0;i<count;i++)distance[i]=INF;
distance[src]=0;
for(inti=0;i<count-1;i++)for(intj=0;j<count;j++)ListPtrhead=G->dpHead[j];while(head)intnewDistance=distance[j]+head->weight;if(distance[head->index]>newDistance)distance[head->index]=newDistance;
path[head->index]=j;head=head->next;for(inti=0;i<count;i++)printf("%dto%dweight%d\n",path[i],i,distance[i]);
AllPairsShortestPathsGivenaweightedgraphG(V,E),theallpairshortestpathproblemistofindtheshortestpathbetweenallpairsofverticesu,vєV.Executeninstancesofsinglesourceshortestpathalgorithmforeachvertexofthegraph.ThecomplexityofthisalgorithmwillbeO(n3)
Exercise1.Intheentirepath-findingalgorithm,wehavecreatedapatharraythatjuststoreimmediateparentofa
node,printthecompletepathforit.2.Allthefunctionsareimplementedconsideringasifthegraphisrepresentedbyadjacencylist.Towrite
allthosefunctionsforgraphrepresentationasadjacencymatrix.3.Givenastartstring,endstringandasetofstrings,findifthereexistsapathbetweenthestartstringand
endstringviathesetofstrings.
Apathexistsifwecangetfromstartstringtoendthestringbychanging(noaddition/removal)onlyonecharacteratatime.Therestrictionisthatthenewstringgeneratedafterchangingonecharacterhastobeintheset.Start:"cog"End:"bad"Set:["bag","cag","cat","fag","con","rat","sat","fog"]Oneofthepaths:"cog"->"fog"->"fag"->"bag"->"bad"
CHAPTER14:STRINGALGORITHMS
IntroductionStringinClanguageisanarrayofcharacter.Weusestringalgorithminsomanytasks,whenweareusingsomecopy-paste,somestringreplacement,andsomestringsearch.Whenweareusingsomedictionaryprogram,we are using string algorithms.Whenwe are searching something in googlewe are passingsomeinformationthatisalsoastringandthatwillfurtherconvertandprocessedbygoogle.Note:Thischapterisveryimportantfortheinterviewpointofviewasmanyinterviewproblemsarefromthischapter.
StringMatchingEverywordprocessingprogramhas a search function inwhichyou can search all occurrencesof anyparticularwordinalongtextfile.Forthis,weneedstring-matchingalgorithms.
BruteForceSearchWehaveapatternthatwewanttosearchinthetext.Thepatternisoflengthmandthetextisoflengthn.Wherem<n.Thebruteforcesearchalgorithmwillcheckthepatternatallpossiblevalueof“i”inthetextwherethevalueof“i”rangefrom0ton-m.Thepatterniscomparedwiththetext,characterbycharacterfromlefttoright.When amismatch is detected, then pattern is compared by shifting the comparewindowby onecharacter.Example14.1:intBruteForceSearch(char*text,char*pattern)
inti=0,j=0,count=0;constintn=strlen(text);constintm=strlen(pattern);
while(i<=n-m)j=0;while(j<m&&pattern[j]==text[i+j])j++;if(j==m)return(i);i++;return-1;
WorstcaseTimeComplexityofthealgorithmisO(m*n),wegotthepatternattheendofthetextorwedidnotgetthepatternatall.BestcaseTimeComplexityofthisalgorithmisO(m),theaverageTimeComplexityofthisalgorithmisO(n)
Robin-KarpalgorithmRobin-Karpalgorithmissomewhatsimilartothebruteforcealgorithm.Becausethepatterniscomparedto each textbox. Instead of pattern at each position a hash code is compared, only one comparison isperformed.Thehashcodeof thepattern iscomparedwith thehashcodeof the textwindow.Wetry tokeepthehashcodeasuniqueaspossible.
Thetwofeaturesofgoodhashcodeare:·Thecollisionshouldbeexcludedasmuchaspossible.·Thehashcodeoftextmustbecalculatedinconstanttime.
Acollisionoccurswhenhashcodematches,butthepatterndoesnot.Calculationinconstanttime,onememberleavesthewindowandanewnumberentersawindow.Multiplicationby2issameasleftshiftoperation.Multiplicationby2m-1issameasleftshiftm-1times.Wewantthismultipletimessojuststoreitinvariablepow(m)=2m-1Wedonotwanttodobigmultiplicationoperationssomodularoperationwithaprimenumberisused.Example14.2:intRobinKarp(char*text,char*pattern)
intn=strlen(text);intm=strlen(pattern);inti,j;intprime=101;intpowm=1;intTextHash=0,PatternHash=0;
if(m==0||m>n)return-1;
for(i=0;i<m-1;i++)powm=(powm<<1)%prime;
for(i=0;i<m;i++)PatternHash=((PatternHash<<1)+pattern[i])%prime;TextHash=((TextHash<<1)+text[i])%prime;
for(i=0;i<=(n-m);i++)if(TextHash==PatternHash)for(j=0;j<m;j++)if(text[i+j]!=pattern[j])break;
if(j==m)
returni;TextHash=(((TextHash-text[i]*powm)<<1)+text[i+m])%prime;
if(TextHash<0)TextHash=(TextHash+prime);return-1;
Knuth-Morris-PrattalgorithmAfter a shift of the pattern, the brute force algorithm forgotten all the information about the previousmatchedsymbols.ThisisbecauseofwhichitsworstcaseTimeComplexityisO(mn).The Knuth-Morris-Pratt algorithm make use of this information that is computed in the previouscomparison.Itneverrecomparesthewholetext.Itusespre-processingofthepattern.Thepre-processingtakesO(m)timeandwholealgorithmisO(n)Pre-processingstep:wetrytofindtheborderofthepatternatadifferentprefixofthepattern.Aprefixisastringthatcomesatthestartofastring.Aproperprefixisaprefixthatisnotthecompletestring.Itslengthislessthanthelengthofthestring.Asuffixisastringthatcomesattheendofastring.Apropersuffixisasuffixthatisnotthecompletestring.Itslengthislessthanthelengthofthestring.Aborderisastringthatisbothproperprefixandapropersuffix.
Example14.3:voidKMPPreprocess(char*pattern,int*ShiftArr)
constintm=strlen(pattern);inti=0,j=-1;ShiftArr[i]=-1;while(i<m)while(j>=0&&pattern[i]!=pattern[j])j=ShiftArr[j];i++;
j++;ShiftArr[i]=j;
Wehavetoloopouterloopforthetextandinnerloopforthepatternwhenwehavematchedthetextandpatternmismatch,weshiftthetextsuchthatthewidestborderisconsideredandthentherestofthepatternmatchingisresumedafterthisshift.Ifagainamismatchhappensthenthenextmismatchistaken.
Example14.4:intKMP(char*text,char*pattern)
inti=0,j=0,count=0;constintn=strlen(text);constintm=strlen(pattern);
int*ShiftArr=(int*)calloc(m+1,sizeof(int));
KMPPreprocess(pattern,ShiftArr);
while(i<n)while(j>=0&&text[i]!=pattern[j])j=ShiftArr[j];i++;j++;if(j==m)return(i-m);return-1;
Example14.5:UsethesameKMPalgorithmtofindthenumberofoccurrencesofthepatterninatext.intKMPFindCount(char*text,char*pattern)
inti=0,j=0,count=0;constintn=strlen(text);constintm=strlen(pattern);int*ShiftArr=(int*)calloc(m+1,sizeof(int));
KMPPreprocess(pattern,ShiftArr);while(i<n)while(j>=0&&text[i]!=pattern[j])j=ShiftArr[j];
i++;j++;
if(j==m)count++;j=ShiftArr[j];returncount;
Dictionary/SymbolTableAsymboltableisamappingbetweenastring(key)andavaluethatcanbeofanytype.Avaluecanbeanintegersuchasoccurrencecount,dictionarymeaningofawordandsoon.
BinarySearchTree(BST)forStringsBinarySearchTree(BST)isthesimplestwaytoimplementsymboltable.Simplestrcmp()functioncanbeusedtocomparetwostrings.Ifallthekeysarerandom,andthetreeisbalanced.ThenonanaveragekeylookupcanbedoneinO(logn)time.
Below is an implementation of binary search tree to store string as key. This will keep track of theoccurrencecountofwordsinatext.Example14.6:1.#include<stdio.h>2.#include<stdlib.h>3.#include<string.h>4.#include<iostream>5.usingnamespacestd;6.7.structtreeNode_t8.char*value;9.intcount;10.treeNode_t*lChild;11.treeNode_t*rChild;12.;13.14.typedeftreeNode_t*treePtr;1.voidprintTree(treePtrroot)/*preorder*/2.3.if(root)4.5.cout<<"valueis::"<<root->value;
6.cout<<"countis::"<<root->count<<endl;7.printTree(root->lChild);8.printTree(root->rChild);9.10.1.treePtrinsertNode(char*value,treePtrroot)2.3.intcompare;4.if(root==NULL)5.6.root=(treePtr)malloc(sizeof(treeNode_t));7.if(root==NULL)8.9.printf("fallelmemoryshortage...");10.returnroot;11.12.root->value=(char*)malloc((1+strlen(value))*sizeof(char));13.strcpy(root->value,value);14.root->lChild=root->rChild=NULL;15.root->count=1;16.17.else18.19.compare=strcmp(root->value,value);20.if(compare==0)21.22.root->count++;23.24.elseif(compare==1)25.26.root->lChild=insertNode(value,root->lChild);27.28.else29.30.root->rChild=insertNode(value,root->rChild);31.32.33.returnroot;34.1.voidinsertNode(char*value,treePtr*ptrRoot)2.3.*ptrRoot=insertNode(value,*ptrRoot);4.
1.voidfreeTree(treePtr*rootPtr)2.3.*rootPtr=freeTree(*rootPtr);4.1.treePtrfreeTree(treePtrroot)2.3.if(root)4.5.freeTree(root->lChild);6.freeTree(root->rChild);7.free(root->value);8.free(root);9.10.returnNULL;11.1.treePtrfindNode(treePtrroot,char*value)2.3.intcompare;4.if(!root)5.returnNULL;6.compare=strcmp(root->value,value);7.if(compare==0)8.returnroot;9.else10.11.if(compare==1)12.returnfindNode(root->lChild,value);13.else14.returnfindNode(root->rChild,value);15.16.1.intfrequency(treePtrroot,char*value)2.3.intcompare;4.if(!root)5.return0;6.compare=strcmp(root->value,value);7.if(compare==0)8.returnroot->count;9.else10.
11.if(compare==1)12.returnfrequency(root->lChild,value);13.else14.returnfrequency(root->rChild,value);15.16.1.intgetword(char*a,FILE*fp)2.3.inti=0;4.while(1)5.6.a[i]=getc(fp);7.if(a[i]==EOF)8.9.a[i]='\0';10.return0;11.12.elseif(a[i]==''||a[i]=='\t'||a[i]=='\n')13.14.a[i]='\0';15.return1;16.17.i++;18.19.1.intmain()2.3.4.treePtrroot=NULL;5.treePtrtemp=NULL;6.FILE*fp=fopen("binaryString.cpp","r");7.chara[100];8.while(getword(a,fp))9.10.root=insertNode(a,root);11.12.printTree(root);13.printf("\n");14.cout<<"quencyreturned::"<<frequency(root,"&&");15.printf("\n");16.
Hash-TableTheHash-Tableisanotherdatastructurethatcanbeusedforsymboltableimplementation.BelowHash-Tablediagram,wecanseethenameofthatpersonistakenasthekey,andtheirmeaningisthevalueofthesearch.Thefirstkeyisconvertedintoahashcodebypassingittoappropriatehashfunction.InsidehashfunctionthesizeofHash-Tableisalsopassed,whichisusedtofindtheactualindexwherevalueswillbestored.Finally,thevaluethatismeaningofnameisstoredintheHash-Table,oryoucanstoreareferencetothestringwhichstoremeaningcanbestoredintotheHash-Table.
Hash-TablehasanexcellentlookupofO(1).Let us supposewewant to implement autocomplete the box feature ofGoogle search.When you typesomestringtosearchingoogle,itproposesomecompletestringevenbeforeyouhavedonetyping.BSTcannotsolvethisproblemasrelatedstringscanbeinbothrightandleftsubtree.TheHash-Tableisalsonotsuitedfor this job.OnecannotperformapartialmatchorrangequeryonaHash-Table.Hash function transforms string to a number.Agoodhash functionwill give a distributedhashcodeevenforpartialstringandthereisnowaytorelatetwostringsinaHash-Table.TrieandTernarySearchtreeareaspecialkindoftreethatsolvespartialmatchandrangequeryproblemwell.
TrieTrieisatree,inwhichwestoreonlyonecharacterateachnode.Thisfinalkeyvaluepairisstoredintheleaves.EachnodehasRchildren,oneforeachpossiblecharacter.Forsimplicitypurpose,letusconsiderthatthecharactersetis26,correspondstodifferentcharactersofEnglishalphabets.Trie is an efficient data structure. Using Trie, we can search the key in O(M) time.WhereM is themaximumstringlength.Trieisalsosuitableforsolvingpartialmatchandrangequeryproblems.
Example14.7:1.#include<iostream>2.usingnamespacestd;3.structtrieNode_t4.intflag;5.charch;6.trieNode_t*child[26];7.;8.typedeftrieNode_t*triePtr;1.triePtrcreateNode()2.3.triePtrtemp=(triePtr)malloc(sizeof(trieNode_t));4.for(inti=0;i<26;i++)5.temp->child[i]==NULL;6.returntemp;7.1.voidmyToLower(char*str)2.3.intlength=strlen(str);4.for(inti=0;i<length;i++)5.str[i]=tolower(str[i]);6.7.voidtrie(triePtrroot,char*str)8.
9.if(*(str+1)=='\0')10.11.if(root->child[*str-'a']==NULL)12.13.root->child[*str-'a']=createNode();14.15.root->child[*str-'a']->flag=1;16.root->child[*str-'a']->ch=*str;17.return;18.19.else20.21.if(root->child[*str-'a']==NULL)22.23.root->child[*str-'a']=createNode();24.root->child[*str-'a']->flag=0;25.26.root->child[*str-'a']->ch=*str;27.trie(root->child[*str-'a'],(str+1));28.29.1.triePtrtrieInsert(triePtrroot,char*str)2.3.myToLower(str);4.if(str==NULL||*str=='\0')5.returnroot;6.if(root==NULL)7.8.root=createNode();9.trie(root,str);10.11.else12.13.trie(root,str);14.15.returnroot;16.1.intfindNode(triePtrroot,char*str)2.3.myToLower(str);4.if(str==NULL)5.6.cout<<"nodefound"<<endl;
7.return0;8.9.if(root==NULL)10.11.cout<<"nodefound"<<endl;12.return0;13.14.while(root->child[*str-'a']&&*(str+1)!='\0'&&root->child[*str-'a']->ch==*str)15.16.root=root->child[*str-'a'];17.str++;18.19.//chardoesnotmatchorcharindexchilddoesnotexist20.if(!root->child[*str-'a']||root->child[*str-'a']->ch!=*str)21.22.cout<<"nodenotfound"<<endl;23.return0;24.25.if(*(str+1)=='\0')26.27.if(root->child[*str-'a']->ch==*str&&root->child[*str-'a']->flag==1)28.29.cout<<"nodefound"<<endl;30.return1;31.32.33.cout<<"nodenotfound"<<endl;34.return0;35.1.intmain()2.3.triePtrroot=NULL;4.chara[]="hemant";5.charb[]="heman";6.charc[]="hemantjain";7.chard[]="jain";8.root=trieInsert(root,a);9.root=trieInsert(root,d);10.printf(“%s”,findNode(root,a));11.printf(“%s”,findNode(root,b));12.printf(“%s”,findNode(root,c));13.printf(“%s”,findNode(root,d));14.
TernarySearchTrie/TernarySearchTreeTrieshaveaverygoodsearchperformanceofO(M)whereMisthemaximumsizeofthesearchstring.However, tries have a very high space requirement. Every node Trie contains references to multiplenodes,eachreferencecorrespondstopossiblecharactersofthekey.ToavoidthishighspacerequirementTernarySearchTrie(TST)isused.A TST avoid the heavy space requirement of the traditional Trie while still keeping many of itsadvantages. In aTST, eachnode contains a character, an endofkey indicator, and threepointers.Thethree pointers are corresponding to current char hold by the node (equal), characters less than andcharactergreaterthan.TheTimeComplexityofternarysearchtreeoperationisproportionaltotheheightoftheternarysearchtree.Intheworstcase,weneedtotraverseupto3timesthatmanylinks.However,thiscaseisrare.Therefore,TSTisaverygoodsolutionforimplementingSymbolTable,Partialmatchandrangequery.
Example14.8:structNode
chardata;unsignedisLastChar:1;structNode*left,*equal,*right;
;typedefNode*NodePtr;NodePtrnewNode(chardata)
NodePtrtemp=(NodePtr)malloc(sizeof(structNode));
temp->data=data;temp->isLastChar=0;temp->left=temp->equal=temp->right=NULL;returntemp;
voidinsert(NodePtr*root,char*word)
if(!(*root))*root=newNode(*word);
if((*word)<(*root)->data)insert(&((*root)->left),word);elseif((*word)>(*root)->data)insert(&((*root)->right),word);elseif(*(word+1))insert(&((*root)->equal),word+1);else(*root)->isLastChar=1;
intsearchTST(NodePtrroot,char*word)
if(!root)return0;
if(*word<(root)->data)returnsearchTST(root->left,word);elseif(*word>(root)->data)returnsearchTST(root->right,word);elseif(*(word+1)=='\0')returnroot->isLastChar;
returnsearchTST(root->equal,word+1);
intsearchTSTWrapper(NodePtrroot,char*word)
intret=searchTST(root,word);
printf("%s:",word);ret?printf("Found\n"):printf("NotFound\n");returnret;
intmain()
NodePtrroot=NULL;insert(&root,"banana");insert(&root,"apple");insert(&root,"mango");printf("\nSearchresultsforapple,banana,grapesandmango:\n");searchTSTWrapper(root,"apple");searchTSTWrapper(root,"banana");searchTSTWrapper(root,"grapes");searchTSTWrapper(root,"mango");return0;
ProblemsinString
RegularExpressionMatchingImplementregularexpressionmatchingwiththesupportof‘?’and‘*’specialcharacter.‘?’Matchesanysinglecharacter.‘*’Matcheszeroormoreoftheprecedingelement.Example14.9:intmatchExpUtil(char*exp,char*str,inti,intj)
if(i==strlen(exp)&&j==strlen(str))return1;
if((i==strlen(exp)&&j!=strlen(str))||(i!=strlen(exp)&&j==strlen(str)))return0;
if(exp[i]=='?'||exp[i]==str[j])returnmatchExpUtil(exp,str,i+1,j+1);
if(exp[i]=='*')returnmatchExpUtil(exp,str,i+1,j)||matchExpUtil(exp,str,i,j+1)||matchExpUtil(exp,str,i+1,j+1);return0;
intmatchExp(char*exp,char*str)
returnmatchExpUtil(exp,str,0,0);
OrderMatchingGivenalongtextstringandapatternstring.Findifthecharactersofpatternstringareinthesameorderintextstring.Eg.TextString:ABCDEFGHIJKLMNOPQRSTUVWXYZPatternstring:JOSTExample14.10:intmatch(char*source,char*pattern)
intiSource=0;intiPattern=0;intsourceLen=strlen(source);intpatternLen=strlen(pattern);
for(iSource=0;iSource<sourceLen;iSource++)if(source[iSource]==pattern[iPattern])iPattern++;if(iPattern==patternLen)return1;return0;
ASCIItoIntegerConversionWriteafunctionthattakeintegerasachararrayandconvertitintoanint.Example14.11:1.intmyAtoi(constchar*str)2.3.intvalue=0;4.while(*str)5.6.value=(value<<3)+(value<<1)+(*str-'0');7.str++;8.9.returnvalue;10.
UniqueCharactersWriteafunctionthatwilltakeastringasinputandreturn1ifitcontainuniquecharacterselsereturn0.Example14.12:intisUniqueChar(char*s)
intbitarr=0;intsize=strlen(s);for(inti=0;i<size;i++)charc=s[i];if('A'<=c&&'Z'>=c)c=c-'A';elseif('a'<=c&&'z'>=c)
c=c-'a';elseprintf("UnknownChar!\n");return0;
if(bitarr&(1<<c))printf("Duplicatedetected!\n");return0;bitarr|=(1<<c);printf("Noduplicatedetected!\n");return1;
ToUpperCaseWriteafunctionthatwillconvertalllowercaselettersinastringtouppercase.Example14.13:ToUpper1.charToUpper(chars)2.3.if(s>=97&&s<=(97+25))4.s=s-32;5.returns;6.
ToLowerCaseWriteafunctionthatwillconvertuppercaseletterinastringtolowercaseExample14.14:ToLower1.charToLower(chars)2.3.if(s>=65&&s<=(65+25))4.s=s+32;5.returns;6.
PermutationCheckWriteafunctiontocheckiftwostringsarepermutationofeachother.
Example14.15:boolisPermutation(char*s1,char*s2)
intcount[256];intlength=strlen(s1);if(strlen(s2)!=length)printf("ispermutationreturnfalse\n");returnfalse;for(inti=0;i<256;i++)count[i]=0;
for(inti=0;i<length;i++)charch=s1[i];count[ch]++;ch=s2[i];count[ch]--;
for(inti=0;i<length;i++)if(count[i])printf("ispermutationreturnfalse\n");returnfalse;printf("ispermutationreturntrue\n");returntrue;
PalindromeCheckGivenastringasanarrayofcharactersfindifthestringisapalindromeornot?Example14.16:intisPalindrome(char*str)
inti=0,j=strlen(str)-1;while(i<j&&str[i]==str[j])
i++;j--;if(i<j)printf("StringisnotaPalindrome");return0;elseprintf("StringisaPalindrome");return1;
TimeComplexityisO(n)andSpaceComplexityisO(1)
IntegertoASCIIConversionWriteafunctionthatconvertandintegerintoachararray.Example14.17:1.voidmyItoa(char*buffer,intvalue,)2.3.staticintindex=-1;4.intremender=value%10;5.value/=10;6.if(value)7.myItoa(buffer,value);8.buffer[++index]='0'+remender;9.buffer[index+1]='\0';10.
ASCIItofloatConversionWriteafunctionthattakefloatasachararrayandconvertitintoafloat.Example14.18:floatMyAtof(char*str)
floatnum=0.0F;floatfraction=0.1F;intdecimalStart=0;intsize=strlen(str);
for(inti=0;i<size;i++)if(str[i]=='.')decimalStart=1;continue;if(!decimalStart)num=(num*10)+(str[i]-'0');elsenum+=(str[i]-'0')*fraction;fraction*=0.1F;returnnum;
ReverseCasefunctionWrite a function thatwill convertLower case letter in a string to upper case and upper case letter tolowercase.Example14.19:/*lowertoupper*/charLowerUpper(chars)
if(s>=97&&s<=(97+25))s=s-32;elseif(s>=65&&s<=(65+25))s=s+32;returns;
StringCopyfunctionWriteafunctiontocopystringprovidedasasource in thearrayprovidedasdestination, including theterminatingnullcharacter.Example14.20:1.char*myStrcpy(char*dst,char*src)2.3.char*ptr=dst;4.while(*dst++=*src++);
5.returnptr;6.
PowerfunctionWriteafunctionwhichwillcalculatexn,Takingxandnasargument.Example14.21:Powerfunction1.intpow(intx,intn)2.3.intvalue;4.if(n==0)5.return(1);6.elseif(n%2==0)7.8.value=pow(x,n/2);9.return(value*value);10.11.else12.13.value=pow(x,n/2);14.return(x*value*value);15.16.
StringComparefunctionWriteafunctionstrcmp()tocomparetwostrings.Thefunctionreturnvaluesshouldbe:Thereturnvalueis0indicatesthatbothfirstandsecondstringsareequal.Thereturnvalueisnegativeindicatesthefirststringislessthanthesecondstring.Thereturnvalueispositiveindicatesthatthefirststringisgreaterthanthesecondstring.Example14.22:1.intmyStrcmp(char*a,char*b)2.3.while((*a)==(*b))4.5.if(*a=='\0')6.return0;7.a++;8.b++;9.10.if(*a=='\0')11.return-1;12.if(*b=='\0')
13.return1;14.intvalue=(*a-*b);15.returnvalue;16.
SubstringfunctionWriteaCfunctionthatwillcopyfirstncharacterfromasourcestringtothedestinationstring.Example14.23:1.char*mySubstr(char*src,char*dst,intn)2.3.intcount=n;4.char*ptr=dst;5.do6.7.n--;8.if(n==0)9.10.ptr[count-1]='\0';11.break;12.13.else14.15.dst++=src++;16.17.while(src);18.returnptr;19.
StringduplicatefunctionWriteaCfunctionthatwillreturnapointertoanewstringthatisaduplicateoftheinputstring.Memoryforthenewstringisobtainedwithmallocandfreedwithfree.Example14.24:1.char*myStrdup(char*src)2.3.char*dst=(char*)malloc((strlen(src)+1)*sizeof(char));4.char*ptr=dst;5.while(*dst++=*src++);6.returnptr;7.
MemcpyFunction
ThefunctionMemcpyfunctioncopiesncharactersfromthesourcearraytodestinationarray.Example14.25:1.voidmymemcpy(void*destPtr,constvoid*srcPtr,intsize)2.3.char*destTemp=(char*)destPtr;4.constchar*srcTemp=(char*)srcPtr;5.while(size--)6.7.*destTemp++=*srcTemp++;8.9.
ReverseNcharactersExample14.26:ReversefirstNcharactersofastring1.voidreverse(char*s,intN)2.3.char*p,*q;4.chart;5.for(p=s,q=s+N-1;p<q;++p,--q)6.t=*q;7.*q=*p;8.*p=t;9.10.
MemmoveFunctionThe functionMemmove function copies n characters from the source array todestination array. It alsotakescaseifthesourceanddestinationmemoryarraysoverlaps.Example14.27:1.voidmymemmove(void*from,void*to,size_tsize)2.3.char*destTemp=(char*)to;4.char*srcTemp=(char*)from;5.size_ti;6.if(from==to)7.8.//Nothingtocopy!9.10.elseif(from>to)11.
12.for(i=0;i<size;i++)13.destTemp[i]=srcTemp[i];14.15.else16.17.for(i=size-1;i>=0;i--)18.destTemp[i]=srcTemp[i];19.20.#ifdefBIG_ENDIAN21.reverse(destTemp,size);22.#endif23.
StringLengthfunctionWriteaCfunctiontofindthelengthofthechararray,stringpassedasargument.Example14.28:1.intmyStrlen(char*src)2.3.intlength=0;4.while(*src!='\0')5.6.length++;7.8.returnlength;16.
StringConcatenatefunctionWriteacfunctiontoconcatenatetwostrings.Assumingthatfirststringhasenoughspaceforthesecondstring.Example14.29:1.char*strcat(char*s1,char*s2)2.3.char*ptr=s1;4.while(*s1++);5.while(*s2!='\0')6.*s1++=*s2++;7.returnptr;8.
ReverseStringExample14.30:Reverseallthecharactersofastring.voidreverseString(char*a,intlower,intupper)
chartempChar;while(lower<upper)tempChar=a[lower];a[lower]=a[upper];a[upper]=tempChar;lower++;upper--;
ReverseWordsExample14.31:Reverseorderofwordsinastringsentence.voidreverseWords(char*a)
intlength=strlen(a);intlower,upper=-1;lower=0;
for(inti=0;i<=length;i++)if(a[i]==''||a[i]=='\0')reverseString(a,lower,upper);lower=i+1;upper=i;elseupper++;reverseString(a,0,length-1);//-1becausewedonotwanttoreverse‘\0’
PrintAnagramExample14.32:Givenastringascharacterarray,printalltheanagramofthestring.voidprintAnagram(char*a)
intn=strlen(a);printAnagram(a,n,n);
voidprintAnagram(char*a,intmax,intn)
if(max==1)printString(a,n);
for(inti=-1;i<max-1;i++)if(i!=-1)a[i]^=a[max-1]^=a[i]^=a[max-1];printAnagram(a,max-1,n);if(i!=-1)a[i]^=a[max-1]^=a[i]^=a[max-1];
ShuffleStringExample14.33:WriteaprogramtoconvertarrayABCDE12345toA1B2C3D4E5voidshuffle(charar[],intn)
intcount=0;intk=1;chartemp='\0';for(inti=1;i<n;i=i+2)temp=ar[i];k=i;dok=(2*k)%(2*n-1);temp^=ar[k]^=temp^=ar[k];count++;while(i!=k);if(count==(2*n-2))break;
BinaryAdditionExample14.34:Giventwobinarystring,findthesumofthesetwobinarystrings.char*addBinary(char*first,char*second)
intsize1=strlen(first);intsize2=strlen(second);inttotalIndex;char*total;if(size1>size2)total=(char*)malloc((size1+2)*sizeof(char));totalIndex=size1;elsetotal=(char*)malloc((size2+2)*sizeof(char));totalIndex=size2;total[totalIndex+1]='\0';intcarry=0;intcurr=0;size1--;size2--;while(size1>=0||size2>=0)intfirstValue=(size1<0)?0:first[size1]-'0';intsecondValue=(size2<0)?0:second[size2]-'0';intsum=firstValue+secondValue+carry;
carry=sum>>1;sum=sum&1;
total[totalIndex]=(sum==0)?'0':'1';
totalIndex--;size1--;size2--;total[totalIndex]=(carry==0)?'0':'1';returntotal;
RemoveAllSpacesExample14.35:WriteaCfunctiontoremoveallspacesfromastring.voidremoveSpaces(char*str)
char*to=str;char*from=str;
if(str==NULL)return;
while(*from!='\0')if(*from=='')from++;continue;
*to=*from;from++;to++;*to='\0';
Exercise1.Givenastring,findthelongestsubstringwithoutreputedcharacters.2.Thefunctionmemset()copieschintothefirst'n'charactersofthestring3.Serializeacollectionof string intoa single stringanddeserializes thestring into thatcollectionof
strings.4.Writeasmartinputfunctionthattake20charactersasinputfromtheuser.Withoutcuttingsomeword.
Userinput:“HarryPottermustnotgo”First20chars:“HarryPottermustno”Smartinput:“HarryPottermust”
5.Writeacodethatreturnsifastringispalindromeanditshouldreturntrueforbelowinputstoo.
Stellawonnowallets.No,itisopenononeposition.Risetovote,Sir.Won'tloversrevoltnow?
6.Write anASCII to integer function that ignore the non-integral character and give the integer . For
example,iftheinputis“12AS5”itshouldreturn125.7.WritecodethatwouldparseaBashbraceexpansion.
Example:theexpression"(a,b,c)d,e"andwouldoutputallthepossiblestrings:ad,bd,cd,e8.Givenastringwriteafunctiontoreturnthelengthofthelongestsubstringwithonlyuniquecharacters9.Replacealloccurrencesof"a"with"the"10.Replacealloccurrencesof%20with''.
E.g.Input:www.Hello%20World.comOutput:www.HelloWorld.com
11.Write anexpansion function thatwill takean input string like "1..5,8,11..14,18,20,26..30"andwill
print"1,2,3,4,5,8,11,12,13,14,18,20,26,27,28,29,30"12.Supposeyouhaveastringlike"Thisisasentence".Writeafunctionthatwouldseparatethesewords.
Andwillprintwholesentencewithspaces.13.Giventhreestringstr1,str2andstr3.Writeacomplementfunctiontofindthesmallestsub-sequencein
str1whichcontainsallthecharactersinstr2andbutnotthoseinstr3.14.GiventwostringsAandB,findwhetheranyanagramofstringAisasubstringofstringB.
Foreg:IfA=xyzandB=afdgzyxksldfmthentheprogramshouldreturntrue.
15. Given a string, find whether it contains any permutation of another string. For example, given"abcdefgh"and"ba",thefunctionshouldreturntrue,because"abcdefgh"hassubstring"ab",whichisapermutationofthegivenstring"ba".
16.Giveanalgorithmwhichremovestheoccurrenceof“a”by“bc”fromastring?Thealgorithmmustbe
in-place.17.Givenastring,"1010101010"inbase2convertitintostringwithbase4.Donotuseanextraspace.
CHAPTER15:ALGORITHMDESIGNTECHNIQUES
IntroductionInreallifewhenweareaskedtodosomework,wetrytocorrelateitwithourexperienceandthentrytosolveit.Similarly,whenwegetanewproblemtosolve.Wefirsttrytofindthesimilarityofthecurrentproblemwithsomeproblemsforwhichwealreadyknowthesolution.Thensolve thecurrentproblemandgetourdesiredresult.Thismethodprovidesfollowingbenefits:1)Itprovidesatemplateforsolvingawiderangeofproblems.2)Itprovidesustheideaofthesuitabledatastructurefortheproblem.3)Ithelpsusinanalysing,spaceandTimeComplexityofalgorithms.In the previous chapters,we have used various algorithms to solve different kind of problems. In thischapter,wewillreadaboutvarioustechniquesofsolvingalgorithmicproblems.VariousAlgorithmdesigntechniquesare:1)BruteForce2)GreedyAlgorithms3)Divide-and-Conquer,Decrease-and-Conquer4)DynamicProgramming5)Reduction/Transform-and-Conquer6)BacktrackingandBranch-and-Bound
BruteForceAlgorithmBruteForceisastraightforwardapproachofsolvingaproblembasedontheproblemstatement.Itisoneoftheeasiestapproachestosolveaparticularproblem.Itisusefulforsolvingsmallsizedatasetproblem.Someexamplesofbruteforcealgorithmsare:·Bubble-Sort·Selection-Sort·Sequentialsearchinanarray·Computingpow(a,n)bymultiplyinga,ntimes.·Convexhullproblem·Stringmatching·Exhaustivesearch:Travelingsalesman,Knapsack,andAssignmentproblems
GreedyAlgorithmIn greedy algorithm, solution is constructed through a sequence of steps.At each step, choice ismadewhich is locally optimal. Greedy algorithms are generally used to solve optimization problems. Wealwaystakethenextdatatobeprocesseddependinguponthedatasetwhichwehavealreadyprocessedandthenchoosethenextoptimumdatatobeprocessed.Greedyalgorithmsdoesnotalwaysgiveoptimumsolution.Someexamplesofbruteforcealgorithmsare:·Minimalspanningtree:Prim’salgorithm,Kruskal’salgorithm·Dijkstra’salgorithmforsingle-sourceshortestpathproblem·GreedyalgorithmfortheKnapsackproblem·Thecoinexchangeproblem·Huffmantreesforoptimalencoding
Divide-and-Conquer,Decrease-and-ConquerDivide-and-Conqueralgorithmsinvolvebasicthreesteps,firstsplittheproblemintoseveralsmallersub-problems,secondsolveeachsubproblemandthenfinallycombinethesubproblemsresultstoproducetheresult.In divide-and-conquer the size of the problem is reduced by a factor (half, one-third, etc.),While indecrease-and-conquerthesizeoftheproblemisreducedbyaconstant.Examplesofdivide-and-conqueralgorithms:·Merge-Sortalgorithm(usingrecursion)·Quicksortalgorithm(usingrecursion)·Computingthelengthofthelongestpathinabinarytree(usingrecursion)·ComputingFibonaccinumbers(usingrecursion)·Quick-hullExamplesofdecrease-and-conqueralgorithms:·Computingpow(a,n)bycalculatingpow(a,n/2)usingrecursion.·Binarysearchinasortedarray(usingrecursion)·SearchinginBST·Insertion-Sort·Graphtraversalalgorithms(DFSandBFS)·Topologicalsort·Warshall’salgorithm(usingrecursion)·Permutations(Minimalchangeapproach,Johnson-Trotteralgorithm)·Computingamedian,Topologicalsorting,Fake-coinproblem(Ternarysearch)
Considertheproblemofexponentiation:ComputexnBruteForce: n-1multiplicationsDivideandconquer: T(n)=2*T(n/2)+1=n-1Decreasebyone: T(n)=T(n-1)+1=n-1
Decreasebyconstantfactor:T(n)=T(n/a)+a-1=(a-1)n=nwhena=2
DynamicProgrammingWhilesolvingproblemsusingDivide-and-Conquermethod, theremaybeacasewhenrecursivelysub-problemscanresultinthesamecomputationbeingperformedmultipletimes.Thisproblemariseswhenthereareidenticalsub-problemsariserepeatedlyinarecursion.Dynamicprogrammingisusedtoavoidtherequirementofrepeatedcalculationofsamesub-problem.Inthismethod,weusuallystoretheresultofsub-problemsinatableandreferthattabletofindifwehavealreadycalculatedthesolutionofsub-problemsbeforecalculatingitagain.Dynamicprogrammingisabottomuptechniqueinwhichthesmallersub-problemsaresolvedfirstandtheresultofthesearesuedtofindthesolutionofthelargersub-problems.Examples:·Fibonaccinumberscomputedbyiteration.·Warshall’salgorithmfortransitiveclosureimplementedbyiterations·Floyd’salgorithmsforall-pairsshortestpaths
intfibonacci(intn)
if(n<=1)returnn;returnfibonacci(n-1)+fibonacci(n-2);
Usingdivideandconquerthesamesubproblemissolvedagainandagain,whichreducetheperformanceofthealgorithm.ThisalgorithmhasanexponentialTimeComplexityandlinearSpaceComplexity.intfibo(intn)
intfirst=0,second=1;inttemp,i;
if(n==0)
returnfirst;elseif(n==1)returnsecond;
for(i=2;i<=n;i++)temp=first+second;first=second;second=temp;returntemp;
Usingthisalgorithm,wewillgetFibonacciinlinearTimeComplexityandconstantSpaceComplexity.
Reduction/Transform-and-ConquerThesemethodsworksastwo-stageprocedure.First,theproblemistransformedintoaknownproblemforwhichweknowoptimalsolution.Inthesecondstage,theproblemissolved.Themostcommontypesoftransformationaresortofanarray.Forexample:Givenanarrayofnumbersfindsthetwoclosestnumber.Thebruteforcesolutionforthisproblemwilltakedistancebetweeneachelementinthearrayandwilltrytokeeptheminimumdistancepair,thisapproachwillhaveaTimeComplexityofO(n2)
Transformandconquer solution,willbe first sort thearray inO(nlogn) time and then find the closestnumberbyscanningthearrayinanotherO(n).WhichwillgivethetotalTimeComplexityofO(nlogn).Examples:·Gaussianelimination·HeapsandHeapsort
BacktrackingInreallife,letussupposesomeonegaveyoualockwithanumber(threedigitlock,numberrangefrom1to9).Youdonothavetheexactpasswordkeyforthelock.Youneedtotesteverycombinationuntilyougot therightone.Obviously,youneedtoteststartingfromsomethinglike“111”, then“112”andsoon.Moreover,youwillgetyourkeybeforeyoureach“999”.Therefore,whatyouaredoingisbacktracking.Supposethelockproducesomesound“click”correctdigitisselectedforanylevel.Ifwecanlistentothissoundsuchintelligence/heuristicswillhelpyoutoreachyourgoalmuchfaster.ThesefunctionsarecalledPruningfunctionorboundingfunctions.Backtracking is amethodbywhich solution is foundbyexhaustively searching through largebut finitenumberofstates,withsomepruningorboundingfunctionthatwillnarrowdownoursearch.Foralltheproblemslike(NPhardproblems)forwhichtheredoesnotexistanyothermethodweusebacktracking.Backtrackingproblemshavethefollowingcomponents:1.Initialstate2.Target/Goalstate3.Intermediatestates4.Pathfromtheinitialstatetothetarget/goalstate5.Operatorstogetfromonestatetoanother6.Pruningfunction(optional)The solvingprocessofbacktrackingalgorithmstartswith theconstructionof state’s tree,whosenodesrepresentsthestates.Therootnodeistheinitialstateandoneormoreleafnodewillbeourtargetstate.Eachedgeof the tree represents someoperation.Thesolution isobtainedby searching the treeuntil aTargetstateisfound.Backtrackingusesdepth-firstsearch:1)Storetheinitialstateinastack2)Whilethestackisnotempty,repeat:3)Readanodefromthestack.4)Whilethereareavailableoperators,do:
a.Applyanoperatortogenerateachildb.Ifthechildisagoalstate–returnsolutionc.Ifitisanewstate,andpruningfunctiondoesnotdiscarditpushthechildintothestack.
Therearethreemonksandthreedemonsatonesideofariver.Wewanttomoveallofthemtotheothersideusingasmallboat.Theboatcancarryonlytwopersonsatatime.Givenifonanyshorethenumberofdemonswillbemorethanmonksthentheywilleatthemonks.Howcanwemoveallofthesepeopletotheothersideoftheriversafely?Sameastheaboveproblemthereisafarmerwhohasagoat,acabbageandawolf.Ifthefarmerleaves,goatwithcabbage,goatwilleatthecabbage.Ifthefarmerleaveswolfalonewithgoat,wolfwillkillthegoat.Howcanthefarmermoveallhisbelongingstotheothersideoftheriver?
Youaregiventwojugs,a4-gallononeanda3-gallonone.Therearenomeasuringmarkersonjugs.Atapcanbeusedtofillthejugswithwater.Howcanyouget2gallonsofwaterinthe4-gallonjug?
Branch-and-boundBranchandboundmethodisusedwhenwecanevaluatecostofvisitingeachnodebyautilityfunctions.Ateachstep,wechoosethenodewithlowestcosttoproceedfurther.Branch-andboundalgorithmsareimplementedusingapriorityqueue.Inbranchandbound,wetraversethenodesinbreadth-firstmanner.
A*AlgorithmA*issortofanelaborationonbranch-and-bound.Inbranch-and-bound,ateachiterationweexpandtheshortestpaththatwehavefoundsofar.InA*,insteadofjustpickingthepathwiththeshortestlengthsofar,wepickthepathwiththeshortestestimatedtotallengthfromstart togoal,wherethetotallengthisestimatedaslengthtraversedsofarplusaheuristicestimateoftheremainingdistancefromthegoal.
Branch-and-boundwillalwaysfindanoptimalsolution,whichisshortestpath.A*willalwaysfindanoptimal solution if theheuristic is correct.Choosingagoodheuristic is themost important part ofA*algorithm.
ConclusionUsuallyagivenproblemcanbesolvedusinganumberofmethods,howeveritisnotwisetosettleforthefirstmethodthatcomestoourmind.Somemethodsresultinmuchmoreefficientsolutionsthanothers.For example the Fibonacci numbers calculated recursively ( decrease-and-conquer approach), andcomputedbyiterations(dynamicprogramming).InthefirstcasethecomplexityisO(2n),andintheothercasethecomplexityisO(n).Anotherexample,considersortingbasedontheInsertion-Sortandbasicbubblesort.Foralmostsortedfiles, Insertion-Sort will give almost linear complexity, while bubble sort sorting algorithms havequadraticcomplexity.Sothemostimportantquestionis,howtochoosethebestmethod?First,youshouldunderstandtheproblemstatement.Secondbyknowingvariousproblemsandtheirsolutions.
CHAPTER16:BRUTEFORCEALGORITHM
IntroductionBruteForceisastraightforwardapproachofsolvingaproblembasedontheproblemstatement.Itisoneoftheeasiestapproachestosolveaparticularproblem.Itisusefulforsolvingsmallsizedatasetproblem.Manytimes,otheralgorithmtechniquescanbeusedtogetabettersolutionofthesameproblem.Someexamplesofbruteforcealgorithmsare:·Bubble-Sort·Selection-Sort·Sequentialsearchinanarray·Computingpow(a,n)bymultiplyinga,ntimes.·Convexhullproblem·Stringmatching·Exhaustivesearch·Travelingsalesman·Knapsack·Assignmentproblems
ProblemsinBruteForceAlgorithm
Bubble-SortInBubble-Sort,adjacentelementsofthelistarecomparedandareexchangediftheyareoutoforder.//SortsagivenarraybyBubbleSort//Input:AnarrayAoforderableelements//Output:ArrayA[0..n-1]sortedinascendingorderAlgorithmBubbleSort(A[0..n-1])
sorted=falsewhile!sorteddosorted=trueforj=0ton-2doifA[j]>A[j+1]thenswapA[j]andA[j+1]sorted=false
TheTimeComplexityofthealgorithmisΘ(n2)
Selection-SortTheentiregivenlistofNelementsistraversedtofinditssmallestelementandexchangeitwiththefirstelement.Then, the list is traversed again to find the second element and exchanged itwith the secondelement.AfterN-1passes,thelistwillbefullysorted.//Sortsagivenarraybyselectionsort//Input:AnarrayA[0..n-1]oforderableelements//Output:ArrayA[0..n-1]sortedinascendingorderAlgorithmSelectionSort(A[0..n-1])
fori=0ton-2domin=iforj=i+1ton-1doifA[j]<A[min]min=jswapA[i]andA[min]
TheTimeComplexityofthealgorithmisΘ(n2)
SequentialSearchThealgorithmcomparesconsecutiveelementsofagivenlistwithagivensearchkeyworduntileitheramatchisfoundorthelistisexhausted.AlgorithmSequentialSearch(A[0..n],K)
i=0
WhileA[i]≠Kdoi=i+1ifi<nreturnielsereturn-1
WorstcaseTimeComplexityisΘ(n).
ComputingPOW(a,n)Computingan(a>0,andnisanonnegativeinteger)basedonthedefinitionofexponentiation.N-1multiplicationsarerequiredinbruteforcemethod.//Input:Arealnumberaandanintegern=0//Output:apowernAlgorithmPower(a,n)
result=1fori=1tondoresult=result*areturnresult
ThealgorithmrequiresΘ(n)
StringmatchingAbruteforcestringmatchingalgorithmtakestwoinputs,firsttextconsistsofncharactersandapatternconsistofmcharacter (m<=n).Thealgorithmstartsbycomparing thepatternwith thebeginningof thetext.Each character of the patters is compared to the corresponding character of the text.Comparisonstartsfromlefttorightuntilallthecharactersarematchedoramismatchisfound.Thesameprocessisrepeateduntilamatchisfound.Eachtimethecomparisonstartsonepositiontotheright.//Input:AnarrayT[0..n-1]ofncharactersrepresentingatext//anarrayP[0..m-1]ofmcharactersrepresentingapattern//Output:Thepositionofthefirstcharacterinthetextthatstartsthefirst//matchingsubstringifthesearchissuccessfuland-1otherwise.AlgorithmBruteForceStringMatch(T[0..n-1],P[0..m-1])
fori=0ton-mdoj=0whilej<mandP[j]=T[i+j]doj=j+1ifj=mthenreturnireturn-1
Intheworstcase,thealgorithmisO(mn).
Closest-PairBrute-ForceAlgorithmTheclosest-pairproblemistofindthetwoclosestpointsinasetofnpointsina2-dimensionalspace.Abruteforceimplementationofthisproblemcomputesthedistancebetweeneachpairofdistinctpointsandfindthesmallestdistancepair.//Findstwoclosestpointsbybruteforce//Input:AlistPofn>=2points//Output:TheclosestpairAlgorithmBruteForceClosestPair(P)
dmin=infinitefori=1ton-1doforj=i+1tondod= +ifd<dminthendmin=dimin=ijmin=jreturnimin,jmin
IntheTimeComplexityofthealgorithmisΘ(n2)
Convex-HullProblemConvex-hullofasetofpointsisthesmallestconvexpolygoncontainingallthepoints.Allthepointsofthesetwilllieontheconvexhullorinsidetheconvexhull.Illustratetherubber-bandinterpretationoftheconvexhull.Theconvex-hullofasetofpointsisasubsetofpointsinthegivensets.Howtofindthissubset?Answer:Therestofthepointsofthesetareallononeside.
Twopoints(x1,y1),(x2,y2)makethelineax+by=cWherea=y2-y1,b=x1-x2,andc=x1y2-y1x2Anddividestheplanebyax+by-c<0andax+by-c>0Soweneedtoonlycheckax+by-cfortherestofthepointsIfwefindallthepointsinthesetliesonesideofthelinewitheitherallhaveax+by-c<0orallthepointshaveax+by-c>0thenwewilladdthesepointstothedesiredconvexhullpointset.Foreachofn(n-1)/2pairsofdistinctpoints,oneneedstofindthesignofax+by-cineachoftheothern-2points.Whatistheworst-casecostofthealgorithm?O(n3)fori=0ton-1
forj=0ton-1if(xi,yi)!=(xj,yj)drawalinefrom(xi,yi)to(xj,yj)fork=0ton-1if(i!=kandj!=k)if(allotherpointslieonthesamesideoftheline(xi,yi)and(xj,yj))add(xi,yi)to(xj,yj)totheconvexhullset
ExhaustiveSearchExhaustivesearchisabruteforceapproachappliestocombinatorialproblems.In exhaustive search, we generate all the possible combinations. See if the combinations satisfy theproblemconstraintsandthenfindingthedesiredsolution.Examplesofexhaustivesearchare:·Travelingsalesmanproblem·Knapsackproblem·Assignmentproblem
TravelingSalesmanProblem(TSP)In the travelingsalesmanproblemweneed to find theshortest tour throughagivensetofNcities thatsalesmanvisitseachcityexactlyoncebeforereturningtothecitywherehestarted.Alternatively:FindingtheshortestHamiltoniancircuitinaweightedconnectedgraph.Acyclethatpassesthroughalltheverticesofthegraphexactlyonce.
TourswhereAisstartingcity:TourCostA→B→C→D→A1+3+6+5=15A→B→D→C→A1+4+6+8=19A→C→B→D→A8+3+4+5=20A→C→D→B→A8+6+4+1=19A→D→B→C→A5+4+3+8=20A→D→C→B→A5+6+3+1=15AlgorithmTSP
SelectacityMinTourCost=infiniteFor(Allpermutationsofcities)doIf(LengthOfPathSinglePermutation<MinTourCost)MinTourCost=LengthOfPath
Totalnumberofpossiblecombinations=(n-1)!Costforcalculatingthepath=Θ(n)Sothetotalcostforfindingtheshortestpath=Θ(n!)
KnapsackProblemGivenanitemwithcostC1,C2,...,Cn,andvolumeV1,V2,...,VnandknapsackofcapacityVmax,findthemostvaluable(max∑Cj)thatfitintheknapsack(∑Vj≤Vmax).The solution is one of all the subset of the set of object taking 1 to n objects at a time, so the TimeComplexitywillbeO(2n)AlgorithmKnapsackBruteForce
MaxProfit=0For(Allpermutationsofobjects)doCurrProfit=sumofobjectsselectedIf(MaxProfit<CurrProfit)
MaxProfit=CurrProfitStorethecurrentsetofobjectsselected
ConclusionBruteforceisthefirstalgorithmthatcomesintomindwhenweseesomeproblem.Theyarethesimplestalgorithms that are very easy to understand. However, these algorithms rarely provide an optimumsolution.Many caseswewill find other effective algorithm that ismore efficient than the brute forcemethod.Thisisthemostsimpletounderstandthekindofproblemsolvingtechnique.
CHAPTER17:GREEDYALGORITHM
IntroductionGreedyalgorithmsaregenerallyusedtosolveoptimizationproblems.Tofindthesolutionthatminimizesormaximizessomevalue(cost/profit/countetc.).In greedy algorithm, solution is constructed through a sequence of steps.At each step, choice ismadewhichislocallyoptimal.Wealwaystakethenextdatatobeprocesseddependinguponthedatasetwhichwehavealreadyprocessedandthenchoosethenextoptimumdatatobeprocessed.Greedyalgorithmsdoesnotalwaysgiveoptimumsolution.Forsomeproblems,greedyalgorithmgivesanoptimalsolution.Formost,theydonot,butcanbeusefulforfastapproximations.Greedyisastrategythatworkswellonoptimizationproblemswiththefollowingcharacteristics:1.Greedychoice:Aglobaloptimumcanbearrivedatbyselectingalocaloptimum.2. Optimal substructure: An optimal solution to the problem is made from optimal solutions of subproblems.Someexamplesofbruteforcealgorithmsare:Optimalsolutions:·Minimalspanningtree:
oPrim’salgorithm,oKruskal’salgorithm
·Dijkstra’salgorithmforsingle-sourceshortestpath·Huffmantreesforoptimalencoding·SchedulingproblemsApproximatesolutions:·GreedyalgorithmfortheKnapsackproblem·Coinexchangeproblem
ProblemsonGreedyAlgorithm
CoinexchangeproblemHowcanagivenamountofmoneyNbemadewiththeleastnumberofcoinsofgivenDoneminationsD=d1…dn?TheIndiancoinsystem5,10,20,25,50,100Supposewewanttogivechangeofacertainamountof40paisa.Wecanmakeasolutionbyrepeatedlychoosingacoin≤tothecurrentamount,resultinginanewamount.Thegreedysolutionalwayschoosethelargestvaluecoinwithoutexceedingthetotalamount.For40paisa:25,10,and5Theoptimalsolutionwillbe20,20Thegreedyalgorithmdidnotgiveusoptimalsolution,butitgaveafairapproximation.AlgorithmMAKE-CHANGE(N)
C=5,20,25,50,100//constant.S=//setthatwillholdthesolutionset.Value=NWHILEValue!=0x=largestiteminsetCsuchthatx<ValueIFnosuchitemTHENRETURN"NoSolution"S=S+xValue=Value-xRETURNS
MinimumSpanningTreeAspanningtreeofaconnectedgraphisatreecontainingallthevertices.A minimum spanning tree of a weighted graph is a spanning tree with the smallest sum of the edgeweights.
Prim’sAlgorithmPrim’salgorithmgrowsasingletreeT,oneedgeatatime,untilitbecomesaspanningtree.WeinitializeTwithzeroedges.Uwithsinglenode.WhereTisspanningtreeedgessetandUisspanningtreevertexset.Ateachstep,Prim’salgorithmaddsthesmallestvalueedgewithoneendpointinUandothernotinus.SinceeachedgeaddsonenewvertextoU,aftern−1additions,UcontainalltheverticesofthespanningtreeandTbecomesaspanningtree.//ReturnstheMSTbyPrim’sAlgorithm//Input:AweightedconnectedgraphG=(V,E)//Output:SetofedgescomprisingaMSTAlgorithmPrim(G)
T=LetrbeanyvertexinGU=rfori=1to|V|-1doe=minimum-weightedge(u,v)WithuinUandvinV-UU=U+vT=T+ereturnT
Prim’sAlgorithmusingapriorityqueue(minheap)togettheclosestfringevertexTimeComplexitywillbeO(mlogn)wherenverticesandmedgesoftheMST.
Kruskal’sAlgorithmKruskal’s Algorithm is used to create minimum spanning tree. Spanning tree is created by choosingsmallestweightedgethatdoesnotformacycle.Repeatthisprocessuntilalltheedgesfromtheoriginalsetisexhausted.Sorttheedgesinnon-decreasingorderofcost:c(e1)≤c(e2)≤···≤c(em).SetTtobetheemptytree.Addedgestotreeonebyoneifitdoesnotcreateacycle.(Ifthenewedgeformcyclethenignorethatedge.)//ReturnstheMSTbyKruskal’sAlgorithm//Input:AweightedconnectedgraphG=(V,E)//Output:SetofedgescomprisingaMSTAlgorithmKruskal(G)
SorttheedgesEbytheirweightsT=
while|T|+1<|V|doe=nextedgeinEifT+edoesnothaveacyclethenT=T+ereturnT
Kruskal’sAlgorithmisO(ElogV)usingefficientcycledetection.
Dijkstra’salgorithmforsingle-sourceshortestpathproblemDijkstra’salgorithmforsingle-sourceshortestpathproblemforweightededgeswithnonegativeweight.Itdeterminethelengthoftheshortestpathfromthesourcetoeachoftheothernodesofthegraph.GivenaweightedgraphG,weneedtofindshortestpathsfromthesourcevertexs toeachof theothervertices.
Thealgorithmstartsbykeepingtrackofthedistanceofeachnodeanditsparents.Allthedistanceissettoinfiniteinthebeginningaswedonotknowtheactualpathtothenodesandparentsofalltheverticesaresettonull.Alltheverticesareaddedtoapriorityqueue(minheapimplementation)Ateachstepalgorithmtakesonevertexfromthepriorityqueue(whichwillbethesourcevertexinthebeginning).Thenupdatethedistancearraycorrespondingtoalltheadjacentvertices.Whenthequeueisempty,thenwewillhavethedistanceandparentarrayfullypopulated.//SolvesSSSPbyDijkstra’sAlgorithm//Input:AweightedconnectedgraphG=(V,E)//withnonegativeweights,andsourcevertexv//Output:ThelengthandpathfromstoeveryvAlgorithmDijkstra(G,s)
foreachvinVdoD[v]=infinite//UnknowndistanceP[v]=null//unknownpreviousnodeaddvtoPQ//addingallnodestopriorityqueue
D[source]=0//Distancefromsourcetosource
while(Qisnotempty)u=vertexfromPQwithsmallestD[u]removeufromPQforeachvadjacentfromudoalt=D[u]+length(u,v)ifalt<D[v]thenD[v]=altP[v]=uReturnD[],P[]
TimeComplexitywillbeO(|E|log|V|).Note:Dijkstra’salgorithmdoesnotworkforgraphswithnegativeedgesweight.Note:Dijkstra’salgorithmisapplicabletobothundirectedanddirectedgraphs.
HuffmantreesforoptimalencodingCodingisanassignmentofbitstringsofalphabetcharacters.Therearetwotypesofencoding:·Fixed-lengthencoding(eg.,ASCII)·Variable-lengthencoding(eg.,Huffmancode)Variable length encoding canonlyworkonprefix free encoding.Whichmeans that no codeword is aprefixofanothercodeword.Huffmancodesarethebestprefixfreecode.Anybinarytreewithedgeslabelledas0and1willproduceaprefixfreecodeofcharactersassignedtoitsleafnodes.Huffman’s algorithm is used to construct a binary treewhose leaf value is assigned a code,which isoptimal for the compressionof thewhole text need to be processed.For example, themost frequentlyoccurringwordswillgetthesmallestcodesothatthefinalencodedtextiscompressed.Initializenone-nodetreeswithwordsandthetreeweightswiththeirfrequencies.Jointhetwobinarytreewithsmallestweightintooneandtheweightofthenewformedtreeasthesumofweightofthetwosmalltrees.RepeattheaboveprocessN-1timesandwhenthereisjustonebigtreeleftyouaredone.Markedgesleadingtoleftandrightsubtreeswith0’sand1’s,respectively.Word FrequencyApple 30Banana 25Mango 21Orange 14
Pineapple 10
Word Value CodeApple 30 11Banana 25 10Mango 21 01Orange 14 001Pineapple 10 000ItisclearthatmorefrequencywordsgetssmallerHuffman’scode.//Computesoptimalprefixcode.//Input:ArrayWofcharacterprobabilities//Output:TheHuffmantree.AlgorithmHuffman(C[0..n-1],W[0..n-1])
PQ=//priorityqueuefori=0ton-1doT.char=C[i]T.weight=W[i]addTtopriorityqueuePQ
fori=0ton-2doL=removeminfromPQR=removeminfromPQT=nodewithchildrenLandRT.weight=L.weight+R.weightaddTtopriorityqueuePQreturnT
TheTimeComplexityisO(nlogn).
ActivitySelectionProblemSuppose that activities require exclusiveuseof common resources, andyouwant to schedule asmanyactivitiesaspossible.LetS=a1,...,anbeasetofnactivities.Eachactivityaineedstheresourceduringatimestartingatsiandfinishingbeforefi,i.e.,during[si,fi).Theoptimizationproblemistoselectthenon-overlappinglargestsetofactivitiesfromS.WeassumethatactivitiesS=a1,...,anaresortedinfinishtimef1≤f2≤...fn-1≤fn(thiscanbedoneinΘ(nlogn)).ExampleConsidertheseactivities:
I 1 2 3 4 5 6 7 8 9 10 11
S[i] 1 3 0 5 3 5 6 8 8 2 11
F[i] 4 5 6 7 8 9 10 11 12 13 14
Hereisagraphicrepresentation:
Wechoseanactivitythatstartfirst,andthenlookforthenextactivitythatstartsafteritisfinished.Thiscouldresultina4,a7,a8,butthissolutionisnotoptimal.Anoptimal solution is a1, a3, a6, a8. (Itmaximizes theobjective functionof a numberof activities
scheduled.)Anotheroneisa2,a5,a7,a9.(Optimalsolutionsarenotnecessarilyunique.)Howdowefind(oneof)theseoptimalsolutions?Letusconsideritasadynamicprogrammingproblem...Wearetryingtooptimizethenumberofactivities.Letusbegreedy!
·Themoretimeleftafterrunninganactivity,themoreactivitieswecanfitin.·Ifwechoosethefirstactivitytofinish,themoretimewillbeleft.·Sinceactivitiesaresortedbyfinishtime,wewillalwaysstartwitha1.·Thenwecansolvethesinglesubproblemofactivityschedulinginthisremainingtime.
AlgorithmActivitySelection(S[],F[],N)
SortS[]andF[]inincreasingorderoffinishingtimeA=a1K=1Form=2toNdoIfS[m]>=F[k]A=A+amK=mReturnA
KnapsackProblemAthiefentersastoreandseesanumberofitemswiththeircostandweightmentioned.HisKnapsackcanholdamaxweight.Whatshouldhestealtomaximizeprofit?
FractionalKnapsackproblemAthiefcantakeafractionofanitem(theyaredivisiblesubstances,likegoldpowder).
The fractional knapsack problem has a greedy solution one should first sort the items in term of costdensityagainstweight.Thenfillupasmuchof themostvaluablesubstancebyweightasonecanhold,thenasmuchofthenextmostvaluablesubstance,etc.UntilWisreached.
Item A B CCost 300 190 180Weight 3 2 2Cost/weight 100 95 90Foraknapsackofcapacityof4kg.Theoptimumsolutionoftheabovewilltake3kgofAand1kgofB.AlgorithmFractionalKnapsack(W[],C[],Wk)
Fori=1tondoX[i]=0Weight=0//UseMaxheapH=BuildMaxHeap(C/W)WhileWeight<Wkdoi=H.GetMax()If(Weight+W[i]<=Wk)doX[i]=1Weight=Weight+W[i]ElseX[i]=(Wk–Weight)/W[i]Weight=WkReturnX
0/1KnapsackProblemAthiefcanonlytakeorleavetheitem.Hecannottakeafraction.Agreedy strategy same as above could result in empty space, reducing the overall cost density of theknapsack.Intheaboveexample,afterchoosingobjectAthereisnoplaceforBorCsothereleavesemptyspaceof1kg.Moreover,theresultofthegreedysolutionisnotoptimal.The optimal solutionwill bewhenwe take objectB andC.This problem can be solved by dynamicprogrammingthatwewillseeinthecomingchapter.
CHAPTER18:DIVIDE-AND-CONQUER,DECREASE-AND-CONQUER
IntroductionDivide-and-Conqueralgorithmsworksby recursivelybreakingdownaproblem into twoormore sub-problems (divide), until these sub problems become simple enough so that can be solved directly(conquer).Thesolutionofthesesubproblemsisthencombinedtogiveasolutionoftheoriginalproblem.Divide-and-Conqueralgorithmsinvolvebasicthreesteps1.Dividetheproblemintosmallerproblems.2.Conquerbysolvingtheseproblems.3.Combinetheseresultstogether.In divide-and-conquer the size of the problem is reduced by a factor (half, one-third etc.),While indecrease-and-conquerthesizeoftheproblemisreducedbyaconstant.
Examplesofdivide-and-conqueralgorithms:·Merge-Sortalgorithm(recursion)·Quicksortalgorithm(recursion)·Computingthelengthofthelongestpathinabinarytree(recursion)·ComputingFibonaccinumbers(recursion)·ConvexHullExamplesofdecrease-and-conqueralgorithms:·ComputingPOW(a,n)bycalculatingPOW(a,n/2)usingrecursion·Binarysearchinasortedarray(recursion)·SearchinginBST·Insertion-Sort·Graphtraversalalgorithms(DFSandBFS)·Topologicalsort·Warshall’salgorithm(recursion)·Permutations(Minimalchangeapproach,Johnson-Trotteralgorithm)
·Fake-coinproblem(Ternarysearch)·Computingamedian
GeneralDivide-and-ConquerRecurrenceT(n)=aT(n/b)+f(n)·Wherea≥1andb>1.·"n"isthesizeofaproblem.·"a"isanumberofsub-problemintherecursion.·“n/b”isthesizeofeachsub-problem.·"f(n)"isthecostofthedivisionoftheproblemintosubproblemormergeoftheresultsofsub-problem
togettheresult.
MasterTheoremThemastertheoremsolvesrecurrencerelationsoftheform:T(n)=aT(n/b)+f(n)Itispossibletodetermineanasymptotictightboundinthesethreecases:Case1:when )andconstantϵ>1,thenthefinalTimeComplexitywillbe:
Case2:when )andconstantk≥0,thenthefinalTimeComplexitywillbe:
)Case3:when andconstantϵ>1,ThenthefinalTimeComplexitywillbe:T(n)=Ɵ(f(n))
ModifiedMastertheorem:Thisisashortcuttosolvingthesameproblemeasilyandfast.IftherecurrencerelationisintheformofT(n)=aT(n/b)+dxs
Example1:TakeanexampleofMerge-Sort,T(n)=2T(n/2)+nSol:-Logba=log22=1
)Case2appliesand )T(n)=Ɵ(nlog(n))Example2:BinarySearchT(n)=T(n/2)+O(1)Logba=log21=0
)Case2appliesand )T(n)=Ɵ(log(n))Example3:BinarytreetraversalT(n)=2T(n/2)+O(1 )Logba=log22=1
)Case1appliesandT(n)=Ɵ(n)
ProblemsonDivide-and-ConquerAlgorithm
Merge-Sortalgorithm
//Sortsagivenarraybymergesort//Input:AnarrayAoforderableelements//Output:ArrayA[0..n−1]inascendingorder
AlgorithmMergesort(A[0..n−1])ifn≤1thenreturn;copyA[0..⌊n/2⌋−1]toB[0..⌊n/2⌋−1]copyA[⌊n/2⌋..n−1]toC[0..⌈n/2⌉−1]Mergesort(B)Mergesort(C)Merge(B,C,A)
//Mergestwosortedarraysintoonearray//Input:SortedarraysBandC//Output:SortedarrayAAlgorithmMerge(B[0..p−1],C[0..q−1],A[0..p+q−1])
i=0j=0fork=0top+q−1doifi<pand(j=qorB[i]≤C[j])thenA[k]=B[i]i=i+1elseA[k]=C[j]j=j+1
TimeComplexity:O(nlogn),SpaceComplexity:O(n)TheTimeComplexityofMerge-SortisO(nlogn)inall3cases(worst,averageandbest)asMerge-Sortalwaysdividesthearrayintotwohalvesandtakelineartimetomergetwohalves.Itrequirestheequalamountofadditionalspaceastheunsortedlist.Hence,itisnotatallrecommendedforsearchinglargeunsortedlists.
Quick-Sort
//Sortsasubarraybyquicksort//Input:AnsubarrayofA//Output:ArrayA[l..r]inascendingorderAlgorithmQuicksort(A[l..r])
ifl<rthenp←Partition(A[l..r])//pisindexofpivotQuicksort(A[l..p−1])Quicksort(A[p+1..r])
//PartitionsasubarrayusingA[..]aspivot
//Input:SubarrayofA//Output:FinalpositionofpivotAlgorithmPartition(A[],left,right)
pivot=A[left]lower=leftupper=rightwhilelower<upperwhileA[lower]<=pivotlower=lower+1whileA[upper]>pivotupper=upper–1iflower<upperthenswapA[lower]andA[upper]swapA[lower]andA[upper]//upperisthepivotpositionreturnupper
WorstCaseTimeComplexity:O(n2)BestCaseTimeComplexity:O(nlogn)AverageTimeComplexity:O(nlogn)SpaceComplexity:O(nlogn)ThespacerequiredbyQuick-Sortisveryless,onlyO(nlogn)additionalspaceisrequired.Quicksortisnotastablesortingtechnique,soitmightchangetheoccurrenceoftwosimilarelementsinthelistwhilesorting.
ExternalSortingExternalsortingisalsodoneusingdivideandconqueralgorithm.
BinarySearchWegetthemiddlepointfromthesortedarrayandstartcomparingwiththedesiredvalue.Note:Binarysearchrequiresthearraytobesortedotherwisebinarysearchcannotbeapplied.//Searchesavalueinasortedarrayusingbinarysearch//Input:AnsortedarrayAandakeyK//Output:TheindexofKor−1AlgorithmBinarySearch(A[0..N−1],N,K)
low=0high=N-1whilelow<=highdomid=⌊(low+high)/2⌋ifK=A[mid]thenreturnmidelseifA[mid]<Klow=mid+1elsehigh=mid-1return−1
//Searchesavalueinasortedarrayusingbinarysearch//Input:AnsortedarrayAandakeyK//Output:TheindexofKor−1AlgorithmBinarySearch(A[],low,high,K)
Iflow>highreturn-1mid=⌊(low+high)/2⌋ifK=A[mid]thenreturnmidelseifA[mid]<KreturnBinarySearch(A[],mid+1,high,K)elsereturnBinarySearch(A[],low,mid-1,K)
TimeComplexity:O(logn).Ifyounoticetheaboveprograms,youseethatwealwaystakehalfinputandthrowingouttheotherhalf.SotherecurrencerelationforbinarysearchisT(n)=T(n/2)+c.Usingadivideandconquermastertheorem,wegetT(n)=O(logn).SpaceComplexity:O(1)
Powerfunction//ComputeNthpowerofXusingdivodeandconquerusingrecursion//Input:ValueXandpowerN//Output:Power(X,N)
AlgorithmPower(X,N)IfN=0Return1ElseifN%2==0Value=Power(X,N/2)ReturnValue*ValueElseValue=Power(X,N/2)ReturnValue*Value*X
ConvexHullSortpointsbyX-coordinates.DividepointsintoequalhalvesAandB.RecursivelycomputeHAandHB.MergeHAandHBtoobtainCH
LowerTangent(HA,HB)A=rightmostpointofHAB=leftmostpointofHBWhileabisnotalowertangentforHAandHBdo
WhileabisnotalowertangenttoHAdoa=a−1(moveaclockwise)WhileabisnotalowertangenttoHBdob=b+1(movebcounterclockwise)
ReturnabSimilarlyfinduppertangentandcombinethetwohulls.
InitialsortingtakesO(nlogn)timeRecurrencerelationT(N)=2T(N/2)+O(N)Where,O(N)timefortangentcomputationinsidemergingFinalTimeComplexitywillbeT(N)=O(nlogn).
ClosestPairGivenNpointsin2-dimensionalplane,findtwopointswhosemutualdistanceissmallest.
Abruteforcealgorithmtakeseverypointandfinditsdistancewithalltheotherpointsintheplane.Thenkeeptrackoftheminimumdistancepointsandminimumdistance.TheclosestpairwillbefoundinO(n2)time.Letussupposethereisaverticalline,whichdividethegraphintotwoseparateparts(letuscallitleftandrightpart).Thebruteforcealgorithm,wewillnoticethatwearecomparingallthepointsinthelefthalfwiththepointsintherighthalf.Thisisthepointwherewearedoingsomeextrawork.
Tofindtheminimumweneedtoconsideronlythreecases:1)Closestpairintherighthalf2)Closestpairinthelefthalf.3)Closestpairintheboundaryregionofthetwohalves.(Grey)EverytimewewilldividethespaceSintotwopartsS1andS2byaverticalline.RecursivelywewillcomputetheclosestpairinbothS1andS2.LetuscallminimumdistanceinspaceS1asδ1andminimumdistanceinspaceS2asδ2.Wewillfindδ=min(δ1,δ2)Nowwewillfindtheclosestpairintheboundaryregion.BytakingonepointeachfromS1andS2intheboundaryrangeofδwidthonbothsides.Thecandidatepairofpoint(p,q)wherepЄS1andqЄS2.
Wecanfindthepoints,whichlieinthisregioninlineartimeO(N)byjustscanningthroughallthepoints,andfindingthatallpointslieinthisregion.NowwecansorttheminincreasingorderinY-axisinjustO(nlogn)time.Thenscanthroughthemandgettheminimuminjustoneextralinearpass.Closestpaircannotbefarapartfromeachother.Letuslookintothenextfigure.
Thenthequestionishowmanypointsweneedtocompare.WeneedtocomparethepointssortedinY-axisonlyintherangeofδ.Therefore,thenumberofpointswillcomedowntoonly6points.
Bydoingthis,wearegettingequation.T(N)=2T(N/2)+N+NlogN+6N=O(n(logn)2)Canweoptimizethisfurther?YesInitially,whenwearesortingthepointsinXcoordinatewearesortingtheminYcoordinatetoo.When we divide the problem, then we traverse through the Y coordinate list too, and construct thecorrespondingYcoordinatelistforbothS1andS2.Thenpassthatlisttothem.SincewehavetheYcoordinatelistpassedtoafunctiontheδregionpointscanbefoundsortedintheYcoordinatesinjustonesinglepassinjustO(N)time.T(N)=2T(N/2)+N+N+6N=O(nlogn)//Findsclosestpairofpoints//Input:Asetofnpointssortedbycoordinates//Output:DistancebetweenclosestpairAlgorithmClosestPair(P)
ifn<2thenreturn∞elseifn=2thenreturndistancebetweenpairelsem=medianvalueforxcoordinateδ1=ClosestPair(pointswithx<m)δ2=ClosestPair(pointswithx>m)δ=min(δ1,δ2)δ3=processpointswithm−δ<x<m+δ
returnmin(δ,δ3)Firstpre-processthepointsbysortingtheminXandYcoordinates.Usetwoseparateliststokeepthesesortedpoints.Beforerecursivelysolvingsub-problempassthesortedlistforthatsub-problem.
CHAPTER19:DYNAMICPROGRAMMING
IntroductionWhilesolvingproblemsusingDivide-and-Conquermethod, theremaybeacasewhenrecursivelysub-problemscanresultinthesamecomputationbeingperformedmultipletimes.Thisproblemariseswhenthereareidenticalsub-problemsariserepeatedlyinarecursion.Dynamicprogrammingisusedtoavoidtherequirementofrepeatedcalculationofsamesub-problem.Inthismethod,weusuallystoretheresultofsub-problemsinsomedatastructure(likeatable)andreferittofindifwehavealreadycalculatedthesolutionofsub-problemsbeforecalculatingitagain.Dynamicprogrammingisappliedtosolveproblemswiththefollowingproperties:1.OptimalSubstructure:Anoptimalsolutionconstructedfromtheoptimalsolutionsofitssub-problems.2.OverlappingSubproblems:Whilecalculatingtheoptimalsolutionofsubproblemssamecomputationisrepeatedagainandagain.Examples:1.Fibonaccinumberscomputedbyiteration.2.Assembly-lineScheduling3.Matrix-chainMultiplication4.0/1KnapsackProblem5.LongestCommonSubsequence6.OptimalBinaryTree7.Warshall’salgorithmfortransitiveclosureimplementedbyiterations8.Floyd’salgorithmsforall-pairsshortestpaths9.OptimalPolygonTriangulation10.Floyd-Warshall’sAlgorithmStepsforsolving/recognizingifDPapplies.1.OptimalSubstructure:Trytofindifthereisarecursiverelationbetweenproblemandsub-problem.2.Writerecursiverelationoftheproblem.(ObserveOverlappingSubproblemsatthisstep.)3.Computethevalueofsubproblemsinabottomupfashionandstorethisvalueinsometable.4.Constructtheoptimalsolutionfromthevaluestoredinstep3.5.Repeatstep3and4untilyougetyoursolution.
ProblemsonDynamicprogrammingAlgorithm
Fibonaccinumbersintfibonacci(intn)
if(n<=1)returnn;returnfibonacci(n-1)+fibonacci(n-2);
Usingdivideandconquersamesub-problemissolvedagainandagain,whichreducetheperformanceofthealgorithm.ThisalgorithmhasanexponentialTimeComplexity.SameproblemofFibonaccicanbesolvedinlineartimeifwesorttheresultsofsubproblems.intfibonacci(intn)
intfirst=0,second=1;inttemp,i;
if(n==0)returnfirst;elseif(n==1)returnsecond;
for(i=2;i<=n;i++)temp=first+second;first=second;second=temp;returntemp;
Usingthisalgorithm,wewillgetFibonacciinlinearTimeComplexityandconstantSpaceComplexity.
Assembly-lineScheduling
Weconsider theproblemofcalculating the leastamountof timenecessary tobuildacarwhenusingamanufacturingchainwithtwoassemblinglines,asshowninthefigureTheproblemvariables:·e[i]:entrytimeinassemblylinei·x[i]:exittimefromassemblylinei·a[i,j]:TimerequiredatstationS[i,j](assemblylinei,stagej)·t[i,j]:TimerequiredtotransitfromstationS[i,j]totheotherassemblylineYourprogrammustcalculate:·Theleastamountoftimeneededtobuildacar·Thelistofstationstotraverseinordertoassembleacarasfastaspossible.Themanufacturingchainwillhavenomorethan50stations.If we want to solve this problem in the brute force approach, there will be in total 2n DifferentcombinationssotheTimeComplexitywillbeO(2n)Step1:CharacterizingthestructureoftheoptimalsolutionTocalculatethefastestassemblytime,weonlyneedtoknowthefastesttimetoS1;nandthefastesttimetoS2;n,includingtheassemblytimeforthenthpart.Thenwechoosebetweenthetwoexitingpointsbytakingintoconsiderationtheextratimerequired,x1andx2.TocomputethefastesttimetoS1;nweonlyneedtoknowthefastesttimetoS1;n1andtoS2;n1.Thenthereareonlytwochoices...Step2:Arecursivedefinitionofthevaluestobecomputed
Step3:Computingthefastesttimefinally,computef*asStep4:Computingthefastestpathcomputeasli[j]asthechoicemadeforfi[j](whetherthefirstorthesecondtermgivestheminimum).Also,computethechoiceforf*asl*.FASTEST-WAY(a,t,e,x,n)
f1[1]←e1+a1,1f2[1]←e2+a2,1forj←2tondoiff1[j-1]+a1,j≤f2[j-1]+t2,j-1+a1,jthenf1[j]←f1[j-1]+a1,jl1[j]←1elsef1[j]←f2[j-1]+t2,j-1+a1,jl1[j]←2iff2[j-1]+a2,j≤f1[j-1]+t1,j-1+a2,jthenf2[j]←f2[j-1]+a2,jl2[j]←2elsef2[j]∞f1[j-1]+t1,j-1+a2,jl2[j]←1iff1[n]+x1≤f2[n]+x2
thenf*=f1[n]+x1l*=1elsef*=f2[n]+x2l*=2
MatrixchainmultiplicationSameproblemisalsoknownasMatrixChainOrderingProblemorOptimal-parenthesizationofmatrixproblem.Givenasequenceofmatrices,M=M1,…,Mn.Thegoalofthisproblemistofindthemostefficientwaytomultiplythesematrices.Theguildisnottoperformtheactualmultiplication,buttodecidethesequenceofthematrixmultiplications,sothattheresultwillbecalculatedinminimaloperations.Tocomputetheproductof twomatricesofdimensionspXqandqXr,pqrnumberofoperationswillberequired.Matrixmultiplicationoperationsareassociativeinnature.Therefore,matrixmultiplicationcanbedoneinmanyways.Forexample,M1,M2,M3andM4,canbefullyparenthesizedas:(M1·(M2·(M3·M4)))(M1·((M2·M3)·M4))((M1·M2)·(M3·M4))(((M1·M2)·M3)·M4)((M1·(M2·M3))·M4)Forexample,LetM1dimensionsare10×100,M2dimensionsare100×10,andM3dimensionsare10×50.
((M1·M2)·M3)=(10*100*10)+(10*10*50)=15000(M1·(M2·M3)=(100*10*50)+(10*100*50)=100000Therefore, in thisproblemweneed toparenthesize thematrix chain so that totalmultiplicationcost isminimized.GivenasequenceofnmatricesM1,M2,…Mn.Andtheirdimensionsarep0,p1,p2,…,pn.WherematrixAi has dimension pi − 1 × pi for 1 ≤ i ≤ n.Determine the order ofmultiplication thatminimizesthetotalnumberofmultiplications.If you try to solve this problem using the brute - force method, then you will find all possibleparenthesization. Thenwill compute the cost ofmultiplication. Thenwill pick the best solution. Thisapproachwillbeexponentialinnature.Thereisaninsufficiencyinthebruteforceapproach.TakeanexampleofM1,M2,…,Mn.Whenyouhavecalculated that ((M1·M2) ·M3) is better than (M1· (M2·M3) so there is no point of calculating thencombinationsof(M1·(M2·M3)with(M4,M5….Mn).Optimalsubstructure:AssumethatM(1,N)istheoptimumcostofproductionoftheM1,…,Mn.Anarrayp[]torecordthedimensionsofthematrices.P[0]=rowoftheM1p[i]=colofMi1<=i<=NForsomekM(1,N)=M(1,K)+M(K+1,N)+p0*pk*pnIfM(1,N)isminimalthenbothM(1,K)&M(K+1,N)areminimal.Otherwise,ifthereissomeM’(1,K)istherewhosecostislessthanM(1..K),thenM(1..N)can'tbeminimalandthereisamoreoptimalsolutionpossible.Forsomegeneraliandj.M(i,j)=M(i,K)+M(K+1,j)+pi-1*pk*pjRecurrencerelation:
M(i,j)=OverlappingSubproblems:
Directlycallingrecursivefunctionwillleadtocalculationofsamesub-problemmultipletimes.Thiswillleadtoexponentialsolution.AlgorithmMatrixChainMultiplication(p[])
fori:=1tonM[i,i]:=0;forl=2ton//listhemovinglinefori=1ton−l+1j=i+l−1;M[i,j]=i≤k<j
TimeComplexitywillO(n3ConstructingoptimalparenthesisSolutionUse another table s[1..n, 1..n]. Each entry s[i, j] records the value of k such that the optimalparenthesizationofMiMi+1...MjsplitstheproductbetweenMkandMk+1.AlgorithmMatrixChainMultiplication(p[])fori:=1ton
M[i,i]:=0;forl=2ton//listhemovingline
fori=1ton−l+1j=i+l−1;M[i,j]=i≤k<jS[i,j]=i≤k<j
AlgorithmMatrixChainMultiplication(p[])
fori:=1ton
M[i,i]:=0;forl=2ton//listhemovinglinefori=1ton−l+1j=i+l−1;fork=itojif((M[i,j]=(S[i,j]=k
AlgorithmPrintOptimalParenthesis(s[],i,j)
Ifi=jPrintAiElsePrint“(”PrintOptimalParenthesis(s[],i,s[i,j])PrintOptimalParenthesis(s[],s[i,j],j)Print“)”
LongestCommonSubsequenceLetX=x1,x2,….,xmisasequenceofcharacters.AndY=y1,y2,…,ynisanothersequence.ZisasubsequenceofXifitcanbedrivenbydeletingsomeelementsofX.ZisasubsequenceofYifitcanbedrivenbydeletingsomeelementsformY.ZisLCSofitissubsequencetobothXandY,andthereisnosubsequencewhoselengthisgreaterthanZ.OptimalSubstructure:LetX=<x1,x2,...,xm>andY=<y1,y2,...,yn>betwosequences,andletZ=<z1,z2,...,zk>beaLCSofXandY.·Ifxm=yn,thenzk=xm=yn⇒Zk−1isaLCSofXm−1andYn−1·Ifxm!=yn,then:
ozk!=xm⇒ZisanLCSofXm−1andY.ozk!=yn⇒ZisanLCSofXandYn−1.
RecurrencerelationLetc[i,j]bethelengthofthelongestcommonsubsequencebetweenX=x1,x2,….,xiandY=y1,y2,…,yj.Thenc[n,m]containsthelengthofanLCSofXandY
AlgorithmLCS(X[],m,Y[],n)
fori=1tomc[i,0]=0forj=1tonc[0,j]=0;fori=1tomforj=1tonifX[i]==Y[j]c[i,j]=c[i-1,j-1]+1b[i,j]=elseifc[i-1,j]≥c[i,j-1]c[i,j]=c[i-1,j]b[i,j]=↑elsec[i,j]=c[i,j-1]b[i,j]=←
AlgorithmPrintLCS(b[],X[],i,j)
ifi=0returnifj=0returnifb[i,j]=PrintLCS(b[],X[],i−1,j−1)printX[i]elseifb[i,j]=↑PrintLCS(b[],X[],i−1,j)elsePrintLCS(b[],X[],i,j−1)
CoinExchangingproblemHowcanagivenamountofmoneyNbemadewiththeleastnumberofcoinsofgivenDoneminationsD=d1…dn?Forexample,Indiancoinsystem5,10,20,25,50,100.Supposewewanttogivechangeofacertainamountof40paisa.Wecanmakeasolutionbyrepeatedlychoosingacoin≤tothecurrentamount,resultinginanewamount.Thegreedysolutionalwayschoosethelargestcoinvaluepossible.For40paisa:25,10,and5Thisishowbillionsofpeoplearoundtheglobedochangeeveryday.Thatisanapproximatesolutionoftheproblem.Butthisisnottheoptimalway,theoptimalsolutionfortheaboveproblemis20,20
Step(I):Characterizethestructureofacoin-changesolution.DefineC[j]tobetheminimumnumberofcoinsweneedtomakeachangeforjcents.If we knew that an optimal solution for the problem of making change for j cents used a coin ofDoneminationdi,wewouldhave:C[j]=1+C[j−di]Strep(II):Recursivelydefinesthevalueofanoptimalsolution.
Step(III):Computevaluesinabottom-upfashion.AlgorithmCoinExchange(n,d[],k)
C[0]=0forj=1tondoC[j]=infinitefori=1tokdoifj<diand1+C[j−di]<C[j]thenC[j]=1+C[j−di]returnC
Complexity:O(nk)Step(iv):ConstructanoptimalsolutionWeuseanadditionalarrayDone[1..n],whereDone[j]istheDoneminationofacoinusedinanoptimalsolution.AlgorithmCoinExchange(n,d[],k)
C[0]=0forj=1tondoC[j]=infinitefori=1tokdoifj<diand1+C[j−di]<C[j]thenC[j]=1+C[j−di]Done[j]=direturnC
AlgorithmPrintCoins(Done[],j)
ifj>0PrintCoins(Done,j−Done[j])printDone[j]
CHAPTER20:BACKTRACKINGANDBRANCH-AND-BOUND
IntroductionSupposethelockproducesomesound“click”correctdigitisselectedforanylevel.Youjustwillfindthefirstdigit, thenfindtheseconddigit, thenfindthethirddigitanddone.Thiswillbeagreedyalgorithmandyouwillfindthesolutionveryquickly.However,letussupposethelockissomeoldoneanditcreatessamesoundnotonlyatthecorrectdigitbut at some other digits. Therefore,when you are trying to find the digit of the first ring, then itmayproductsoundatmultipleinstances.Soatthispointyouarenotdirectlygoingstraighttothesolution,butyouneedtotestvariousstatesandincasethosestatesarenotthesolutionyouarelookingfor,thenyouneedtobacktrackonestepatatimeandfindthenextsolution.Sure,thisintelligence/heuristicsofclicksound will help you to reach your goal much faster. These functions are called Pruning function orboundingfunctions.
ProblemsonBacktrackingAlgorithm
NQueensProblemThere areN queens given, you need to arrange them in aNxN chessboard, such that no queen shouldattacheachother.voidprint(int*Q,intn)
for(inti=0;i<n;i++)printf("%d",Q[i]);
intFeasible(int*Q,intk)
for(inti=0;i<k;i++)if(Q[k]==Q[i]||abs(Q[i]-Q[k])==abs(i-k))return0;return1;
voidNQueens(int*Q,intk,intn)
if(k==n)print(Q,n);return;for(inti=0;i<n;i++)Q[k]=i;if(Feasible(Q,k))NQueens(Q,k+1,n);
intmain()
intQ[8];NQueens(Q,0,8);return0;
TowerofHanoiTheTowerofHanoipuzzle,disksneedtobemovedfromonepillartoanothersuchthatanylargediskcannotrestaboveanysmalldisk.Thisisafamouspuzzleintheprogrammingworld,itsoriginscanbetrackedbacktoIndia."There isastoryaboutanIndian temple inKashiViswanathanwhichcontainsa largeroomwith threetimeworn posts in it surrounded by 64 golden disks. Brahmin priests, acting out the command of anancientHinduprophecy, have beenmoving these disks, in accordancewith the immutable rules of theBrahma thecreatorofuniverse, since thebeginningof time.Thepuzzle is thereforealsoknownas theTowerofBrahmapuzzle.Accordingtotheprophecy,whenthelastmoveofthepuzzlewillbecompleted,theworldwillend.";););)
voidtowers(intnum,charfrom,charto,chartemp)if(num<1)return;towers(num-1,from,temp,to);printf("\nMovedisk%dfrompeg%ctopeg%c",num,from,to);towers(num-1,temp,to,from);
intmain()
intnum=10;printf("ThesequenceofmovesinvolvedintheTowerofHanoiare:\n");towers(num,'A','C','B');return0;
CHAPTER21:COMPLEXITYTHEORYANDNPCOMPLETENESS
IntroductionComputational complexity is the measurement of how much resources are required to solve someproblem.Therearetwotypesofresources:1.Time:howmanystepsittakestosolveaproblem2.Space:howmuchmemoryittakestosolveaproblem.
Decisionproblem
MuchofComplexity theorydealswithdecisionproblems.Adecisionproblemalwayshasayesornoanswer.Manyproblemscanbeconvertedtoadecisionproblemthathaveansweredasyesorno.Forexample:1.Searching:Theproblemofsearchingelementcanbeadecisionproblemifweasktofindifaparticular
numberisthereinthelist.2.Sortingof list and find if the list is sortedyou canmakeadecisionproblem is the list is sorted in
increasingorderornot?3.Graphcolouringalgorithms:thisisalsocanbeconvertedtoadecisionproblem.Canwedothegraph
colouringbyusingXnumberofcolours?4.Hamiltoniancycle:Is thereisapathfromall thenodes,eachnodeisvisitedexactlyonceandcome
backtothestartingnodewithoutbreaking?
ComplexityClassesProblemsaredividedintomanyclassessuchthathowdifficulttosolvethemorhowdifficulttofindifthegivensolutioniscorrectornot.
ClassPproblemsTheclassPconsistsofasetofproblemsthatcanbesolvedinpolynomialtime.ThecomplexityofaPproblemisO(nk)Wherenisinputsizeandkissomeconstant(itcannotdependonn).Class PDefinition: The class P contains all decision problems forwhich a Turingmachine algorithmleadstothe“yes/no”answerinadefinitenumberofstepsboundedbyapolynomialfunction.Forexample:Givenasequencea1,a2,a3….an.FindifanumberXisinthisarray.Wecansearch,thenumberXinthisarrayinlineartime(polynomialtime)Anotherexample:Givenasequencea1,a2,a3….an.Ifweareaskedtosortthesequence.WecansortandarrayinpolynomialtimeusingBubble-Sort,thisisalsolineartime.Note:O(logn)isalsopolynomial.Anyalgorithmthathascomplexitylessthansomepolynomialfunctionisalsopolynomial.SomeproblemofPclassis:1.Shortestpath2.Minimumspanningtree3.Maximumproblem.4.Maxflowgraphproblem.5.Convexhull
ClassNPproblemsSet of problems forwhich there is a polynomial time checking algorithm.Given a solution ifwe cancheckinapolynomialtimeifthatsolutioniscorrectornotthen,theproblemisNPproblem.ClassNPDefinition:TheclassNPcontainsalldecisionproblemsforwhich,givenasolution,thereexistsapolynomialtime“proof”or“certificate”thatcanverifyifthesolutionistheright“yes/no”answer
Note:Thereisnoguaranteethatyouwillbeabletosolvethisprobleminpolynomialtime.However,ifaproblemisanNPproblem,thenyoucanverifyananswerinpolynomialtime.NPdoesnotmeans“non-polynomial”.Actually, it is“Non-DeterministicPolynomial” typeofproblem.They are the kind of problems that can be solved in polynomial time by a Non-Deterministic Turingmachine.Ateachpoint,allthepossibilitiesareexecutedinparallel.Iftherearenpossiblechoices,thenallncaseswillbeexecutedinparallel.Wedonothavenon-deterministiccomputers.DonotconfuseitwithparallelcomputingbecausethenumberofCPUislimitedinparallelcomputingitmaybe16coreor32core,butitcannotbeN-Core.InshortNPproblemsarethoseproblemsforwhich,ifasolutionisgiven.Wecanverifythatsolution(ifitiscorrectornot)inpolynomialtime.
BooleanSatisfiabilityproblemABooleanformulaissatisfiedifthereexistsomeassignmentofthevalues0and1toitsvariablesthatcausesittoevaluateto1.
ThereareintotalNDifferentBooleanVariablesA1,A2…AN.ThereareanMnumberofbrackets.EachbrackethasKvariables.There isNvariable so thenumberof solutionswill be2n In addition, to verify if the solutions reallyevaluatetheequationto1willtaketotal2n*kmstepsGivensolutionofthisproblemyoucanfindiftheformulasatisfiesornotinKMsteps.
HamiltoniancycleHamiltoniancycle isapath fromall thenodesofagraph,eachnode isvisitedexactlyonce and comebacktothestartingnodewithoutbreaking.IsanNPproblem,ifyouhaveasolutiontoit,thenyoujustneedtoseeifallthenodesarethereinthepath
andyoucamebacktowhereyoustartedandyouaredone?Thecheckingisdoneinlineartimeandyouaredone.DeterminingwhetheradirectedgraphhasaHamiltoniancycledoesnothaveapolynomialtimealgorithm.O(n!)However, ifsomeonehavegivenyouasequenceofvertices,determiningwhether thatsequenceformsaHamiltoniancyclecanbedone inpolynomial time(Linear time).HamiltoniancyclesareinNP
CliqueProblemIn a graph given is there is a clique of size K or more. A clique is a subset of nodes that are fullyconnectedtoeachother.ThisproblemisNPproblem.Givenasetofnodes,youcanveryeasilyfindoutwhetheritisacliqueornot.Forexample:
PrimeNumberFindingPrimenumberisNP.Givenasolution,itiseasytofindifitisaPrimeornotinpolynomialtime.Findingprimenumbersisimportantascryptographyheavilyusesprimenumbers.intisPrime(intn)
intanswer=(n>1)?1:0;
for(inti=2;i*i<=n;++i)if(n%i==0)answer=0;break;returnanswer;
Checkingwill happen until the square root of number so the TimeComplexitywill beO(√n).Hence,primenumberfindingisanNPproblemaswecanverifythesolutioninpolynomialtime.
Graphtheoryhavewonderfulsetofproblems•Shortestpathalgorithms?•LongestpathisNPcomplete.•Euleriantoursisapolynomialtimeproblem.•HamiltoniantoursisaNPcomplete
Classco-NPSet of problems forwhich there is a polynomial time checking algorithm.Given a solution, ifwe cancheckinapolynomialtimeifthatsolutionisincorrecttheproblemisco-NPproblem.Class co-NP Definition: The class co-NP contains all decision problems such that there exists apolynomialtimeproofthatcanverifyiftheproblemdoesnothavetheright“yes/no”answer.
ClassPisSubsetofClassNPAllproblemsthatarePalsoareNP(P NP).ProblemsetPisasubsetofproblemsetNP.SearchingIfwehavesomenumbersequencea1,a2,a3….an.WealreadyknowthatsearchinganumberXinsidethisarrayisoftypeP.IfitisgiventhatnumberXisinsidethissequence,thenwecanverifybylookingintoeveryentryagainandfindiftheansweriscorrectinpolynomialtime(lineartime.)SortingAnotherexampleofsortinganumbersequence,ifitisgiventhatthearrayb1,b2,b3..bnisasortedthenwecanloopthroughthisgivenarrayandfindifthelistisreallysortedinpolynomialtime(lineartimeagain.)
NP–Hard:AproblemisNP-HardifalltheproblemsinNPcanbereducedtoitinpolynomialtime.
NP–CompleteProblemsSetofproblemisNP-CompleteifitisanNPproblemandanNP-Hardproblem.Itshouldfollowboththeproperties:1)Itssolutionscanbeverifiedinapolynomialtime.2)AllproblemsofNParereducedtoNPcompleteproblemsinpolynomialtime.YoucanalwaysreduceanyNPproblemintoNP-Completeinpolynomialtime.Moreover,whenyougettheanswertotheproblem,thenyoucanverifythissolutioninpolynomialtime.
AnyNPproblemispolynomialreducedtoNP-Completeproblem, ifwecanfindasolutiontoasingleNP-Completeproblem inpolynomial time, thenwecansolveall theNPproblems inpolynomial time.However,sofarnooneisabletofindanysolutionofNP-Completeprobleminpolynomialtime.P≠NP
Reduction
Itisaprocessoftransformationofoneproblemintoanotherproblem.Thetransformationtimeshouldbepolynomial.IfaproblemAistransformedintoBandweknowthesolutionofBinpolynomialtime,thenAcanalsobesolvedinpolynomialtime.Forexample,QuadraticEquationSolver:WehaveaQuadraticEquationSolver,whichsolvesequationoftheformax2+bx+c=0.IttakesInputa,b,candgenerateoutputr1,r2.Nowtry tosolvea linearequation2x+4=0.Usingreductionsecondequationcanbe transformedto thefirstequation.2x+4=0x2+2x+4=0ATLAS:Wehaveanatlasandweneedtocolourmapssothatnotwocountrieshavethesamecolour.Letussupposebelowisthevariouscountries.Inaddition,differentpatternrepresentsdifferentcolour.
Wecanseethatsameproblemofatlascolouringcanbereducedtographcolouringandifweknowthesolutionofgraphcolouringthensamesolutioncanworkforatlascolouringtoo.Whereeachnodeofthegraphrepresentsonecountryandtheadjacentcountryrelationisrepresentedbytheedgesbetweennodes.
Thesortingproblemreduces(≤)toConvexHullproblem.SATreduces(≤)to3SAT
TravelingSalesmanProblem(TSP)
Thetravelingsalesmanproblemtriestofindtheshortesttourthroughagivensetofncitiesthatvisitseachcityexactlyoncebeforereturningtothecitywhereitstarted.Alternatively find the shortestHamiltonian circuit in aweighted connected graph.A cycle that passesthroughalltheverticesofthegraphexactlyonce.AlgorithmTSPSelectacityMinTourCost=infiniteFor(Allpermutationsofcities)do
If(LengthOfPathSinglePermutation<MinTourCost)MinTourCost=LengthOfPath
Totalnumberofpossiblecombinations=(n-1)!Costforcalculatingthepath:Θ(n)Sothetotalcostforfindingtheshortestpath:Θ(n!)It is anNP-Hard problem there is no efficient algorithm to find its solution. Even if some solution isgiven, it is equally hard to verify that this is a correct solution or not. However, some approximatealgorithmscanbeusedtofindagoodsolution.Wewillnotalwaysget thebestsolution,butwillgetagoodsolution.Our approximate algorithm is based on the minimum spanning tree problem. In which we have toconstructatreefromagraphsuchthateverynodeisconnectedbyedgesofthegraphandthetotalsumofthecostofalltheedgesitminimum.
Intheabovediagram,wehaveagroupofcities(eachcityisrepresentedbyacircle.)Whicharelocatedinthegridandthedistancebetweenthecitiesissameaspertheactualdistance.Andthereisapathfromeachcitytoanothercitywhichisastraightpathfromonetoanother.
Wehavemadeaminimumspanningtreefortheabovecitygraph.Whatwewant toprove that the shortestpath inaTSPwill alwaysbegreater than the lengthofMST.Since inMST all nodes are connected to the next node,which is also theminimumdistance from thegroupofnode.Therefore,tomakeitapathwithoutrepeatingthenodesweneedtogodirectlyfromonenodetootherwithoutfollowingMST.Atthatpoint,whenwearenotfollowingMSTwearechoosinganedge,whichisgrater,thentheedgesprovidedbyMST.SoTSPpathwillalwaysbegreaterthanorequaltoMSTpath.
Nowletustakeapathfromstartingnodeandtraverseeachnodeonthewaygivenaboveandthencomebacktothestartingnode.Thetotalcostofthepathis2MST.Theonlydifferenceisthatwearevisitingmanynodesmultipletimes.
NowletuschangeourtraversalalgorithmsothatitwillbecomeTSPinourtraversal,wedidnotvisitanalreadyvisitednodewewillskipthemandwillvisitthenextunvisitednode.Inthisalgorithm,wewillreach thenextnodebyas shorterpath. (The sumof the lengthof all the edgesof apolygon is alwaysgreaterthanasingleedge.)Ultimately,wewillgettheTSPanditspathlengthisnomorethantwicetheoptimalsolution.Therefore,theproposedalgorithmgivesagoodresult.
EndNoteNobody has come upwith such a polynomial-time algorithm to solve aNP-Complete problem.Manyimportantalgorithmsdependsuponit.However,atthesametimenobodyhasproventhatnopolynomialtime algorithm is possible.There is amillionUSdollars for anyonewho cando either solve anyNPCompleteprobleminpolynomial time.Thewholeeconomyof theworldwill fallasmostof thebanksdependsonpublickeyencryptionwillbeeasytobreakifP=NPsolutionisfound.
CHAPTER22:INTERVIEWSTRATEGY
IntroductionSuccessintechinterviewdependsonsomanyfactors,yournon-technicalskills,yourtechnicalskills,etc.Abovealltheinterviewersshouldbeconvincedthattheywouldenjoyworkingwithyou.
ResumeAgoodresumeisonethatcommunicateyourskillsandaccomplishmentsinaclearandeffectiveway.Agoodresumeformathasthefollowingattributes:1. Multiple Columns: Multiple columns make it easier for someone to skim your company name,
positions,collage,andotherkeyfacts.2.Brief:Interviewerisgoingtospendabout30Secreadingyourresume.Youshouldjustfocusonthe
highlights.Onepageisallyouneed,butifyouare10+yearsofexperience,thenyoucanjustifytwopages.
3.NoJunk:Noobjective,Nooath,Summarysection/Keyskills sectionmaybe fine, ifyour resume isshortandconcisethenyoudonotneedasummarysection.
4.UseTables:Youcanusetables,butitshouldnotwastespace.5.Highlights:highlightsshouldbeshort.Keepyourhighlightstoone-liner.6. Neat: Keep your resume neat and clean. Use appropriate Fonts and Formatting. Bold to represent
highlightsandmaybeitalicsinsomeplaces.
NontechnicalquestionsPrepareforvariousnon-technicalquestions.Thefirst thingtodois toprepareanswersofanyquestionthatisrelatedtoyourresume.Theinterviewerisgoingtolookintoitandaskafewquestionstogetanideaaboutyou.Sogothroughallthepast/currentjobandprojectsandmakesureyouknowaboutthemandyourrole.Thesequestionsmaybelike:1.WhatwasthemostchallengingactivityyouhavedoneinprojectABC?2.WhatdidyoulearnfromprojectABC?3.Whatareyourresponsibilitiesinthecurrentjob?4.Whatwasthemostinterestingthingyouhavedoneinyourcurrentjob?5.Whichcourseinuniversitydidyoulikemostandwhy?
TechnicalquestionsSolving a technical question is not just about knowing the algorithms and designing a good softwaresystem.Theinterviewerwantstoknowyouapproachtowardsanygivenproblem.Manypeoplemakemistakes,astheydonotaskclarifyingquestionsaboutagivenproblem.Theyassumemanythingsandbeginworkingwith that.Well the truth is the interviewer toactuallyexpectyoutoaskconstraintsquestions.Thereislotofdatathatismissingthatyouneedtocollectfromyourinterviewerbeforebeginningtosolveaproblem.Forexample,Letussupposetheintervieweraskyoutogiveabestsortingalgorithm.Some interviewee will directly jump to Quick-SortO(nlogn). Oops, mistake you need to ask manyquestionsbeforebeginningtosolvethisproblem.Questions:1.Whatkindofdatawearegoingtosort?Aretheyintegers?2.Howmuchdataarewegoingtosort?3.Whatexactlyisthisdataabout?4.Whatkindofdata-structureusedtoholdthisdata?5.Canwemodifythegivendata-structure?Andmore…?Answer:1.Yes,theyareintegers.2.Maybethousands.3.Theystoreaperson’sage.4.Dataaregivenintheformofsomearray.5.No,youcannotmodifythedatastructureprovided.Okfromthefirstanswer,wewilldeducethatthedataisinteger.Thedataisnotsobigitjustcontainsafewthousandentries.Thethirdanswerisinterestingfromthiswededucethattherangeofdatais1-150.Dataisprovidedinanarray.Fromfifthsanswerwededucethatwehavetocreateourowndatastructureandwecannotmodifythearrayprovided.Sofinally,weconclude,wecanjustusebucketsorttosortthedata.Therangeisjust1-150soweneedjust151-capacityintegralarray.DataisunderthousandssowedonothavetoworryaboutdataoverflowandwegetthesolutioninlineartimeO(N).
CHAPTER23:SYSTEMDESIGN
SystemDesignThesectionwewilllookintoquestionsinwhichintervieweraskstodesignahigh-levelarchitectureofanysoftwaresystem.Note:-ThisisanadvancechapterItmaybethattheuserisnotabletounderstanditcompletely.Iwouldsuggest that give it some time read the chapter and try to read online.Themore time you give to thischapterthebetterunderstandingyouwillget.Itmayalsohelpifyougivemultipleroundsofreading.Therearetwokindsofquestionsinthisandwhichwillbeaskeddependsonthetypeofcompanies.Thefirstkindofquestions is todesign somekindofelevator system,valetparkingsystem,etc. In this, theinterviewerjustwantstotesthowwellyouareabletodesignasystem,especiallyhowwellyourclassesareinteracting.The Second kind of systemdesign problems ismore interesting, inwhich the interviewer asks you todesignsomekindofwebsiteorsomekindofserviceorsomeAPIinterface.Forexample,designgooglesearchengineordesignsomefeatureofFacebooklikehowfriendsmappingisdoneonFacebook,designa web-based game that allows 4 people play poker etc. They are interesting one and in this, theinterviewercanaskaboutscalabilityaspect.Nowcomesaquestiontoourmind,howwouldyoudesigngooglesearchenginein10-15minutes?Well,theanswerisyoucannot.Ittookmanydaysifnotyearsbyagroupofasmartengineertodesigngooglesearchengine.TheinterviewerisexpectingaHigher-levelarchitectureofthesystemthatcanaddressthegivenUse-Cases and Constraints of the problem in hand. There is no single right solution. The sameproblemcanbesolvedinanumberofways.Themostimportantthingisthatyoushouldbeabletojustifyyoursolution.
SystemDesignProcessLetuslookintoa5Stepsapproachforsolvingsystemdesignproblems:1.UseCasesGeneration2.ConstraintsandAnalysis3.BasicDesign4.Bottlenecks5.Scalability
UseCasesJust like algorithm design problems, the system design questions are alsomost likelyweakly defined.Thereissomuchinformationthatismissingandwithoutthis,thedesignisimpossible.Sofirstthinginthedesignprocessistogatherallthepossibleusecases.Youshouldaskquestionstotheinterviewertofindtheusecaseofthesystem.Theinterviewerwantstoseeyourrequirementgatheringcapability.Sameasalgorithmquestionsneverassumethings,whicharenotstated.
ConstraintsandAnalysisThis is thestepinwhichyouwilldefinevariousconstraintsof thesystemandthenanalyse them.Yoursystemdesignwilldependontheanalysisthatyoudointhisstep.Inthisstep,youneedtofindanswerstoquestionslike.Howmanyuserswillbeusingthesystem?Whatkindofdatathatwearegoingtostore?Etc.
BasicDesignInthisstep,youwilldesignthemostbasicdesignofthesystem.Drawyourmaincomponentsandmakeconnectionsbetweenthem.Inthisstep,youneedtodesignasystemwiththesuppositionthatthereisnomemorylimitationandalldatacanfitinonesinglemachine.Youshouldbeabletojustifyyouridea.Inthisstep,youneedtohandlealltheuse-cases.
BottlenecksAnalysisIn this step, you will find the one or more bottlenecks on the basic design you had proposed. The“Scalability Theory” given belowwill help to identify the bottlenecks. You need to know the belowtheory which experts had developed over time. In this step, you will consider how much data yourproposedsystemcanhandle,memorylimitationsetc.
ScalabilityInthisstep,youwillremoveallthebottlenecksofthesystemandyouaredone.Theremaybemultipleiterationsbetween“Bottlenecksanalysis”and“Scalability”untilwereachourfinalsolution.Wewillbereading various concepts like Vertical scaling, Horizontal scaling, Load-Balancer, Redundancy andCachinginthischapter.“ScalabilityTheory”givenbelowwillhelpyoutounderstandtheseconcepts.
ScalabilityTheoryIn this section,wewill be designing a genericweb server,whichwill be handling a large number ofrequests.YoucanimagineitassomesortofwebsitelikeFacebookinwhichlargenumberofusersareaccessingit.
VerticalscalingVerticalscalingmeansthatyouscalebyaddingmoreresources(HigherspeedCPU,MoreRAMetc.)Toyourexistingmachine.Verticalscalinghasitsownlimititcanhelpyoutohandlemoreload,butuntilitslimitisreached,thenwehavetogoforhorizontalscaling.
HorizontalscalingHorizontalscalingmeansthatyouscalebyaddingmoremachinestoyourpoolofresources.Distributetherequestbydistributingtherequestamongmorethanonewebserver.Indoingthisweneedtohavealoadbalancer,whichwilldistributetherequestamongtheservers.
LoadBalancer(Applicationlayer)Loadbalancerhastodecidewhichservershouldservethenextrequest.Sodistributingtheloadcanbemadeusingdifferentstrategies:1.RoundRobbin:Roundrobinisthewayofdistributingrequestsinasequentialfashion.Therequestis
sent to theserver1 then thenext request issent toserver2andsoonuntilwereach theendof theserverlist.Thenwhenwereachtheend,itissentagaintoserver1.Roundrobinhasaproblemthataserver,which isalreadybusy,maygetanother request.Round robinalsohasaproblemwithstickysessions.Wewantthatarequesttobesenttothesameserverthenexttime.
2.Anotherapproachistoselectservercorrespondstothehashvalueofthedata.Findthehashvalueofthedata,modthehashvaluebythenumberofservers.Assignthejobtoamachinewhosevaluewegotaftermod.Sticksessionproblemisalreadysolvedinhashvalueapproach.However,theproblemofuneven load distribution is there, there is possible to have amore load sent to a server, which isalreadybusy.
3.May be the load balancer know, howmuch load each server has or how busy each the server is.
Moreover,willsendthenextrequesttotheleastbusyserver.4.Theservercanbeaspecializedoneservingimagesomeservingvideoandsomeservingotherdata.
ProblemsofLoadBalancingConsideracustomerwhohadselectedsomeitemsinhisbuycartonAmazon.Whenheselectsanotheritemthenitshouldbeaddedtothesamecartsoitshouldbesenttothesameserver.Inaddition,theuserprofile that is saved to one server and if the user request reaches the other server his profilewill beemptythisisalsonotagoodidea.This problem can be solved bymaking the load balancer decide that a particular user requestwouldalwaysgotothesameserver.Theuserprofileandcartdetailsshouldbesavedinsomedatabase.Sticksession:samesessionsshouldlendtothesameserver.Howtogetthisdone.ThefirstapproachisthatwestoretheIPaddressreturnedbyloadbalancingintoacookieandthenusethisIPaddressinthesubsequentrequests.However,thisrevealstheIPaddressoftheservertotheworldthatwedonotwant.Therefore,anothersolutionisthatweusesomesessionidthatisanumberthattheloadbalancerknowsthatbelongstowhichserver.Bythis,wearepreventingourserversbeingexposedtotheouterworldandpreventitfrombeingattacked.
LoadBalancer(Databaselayer)1.ThemostbasicapproachisRoundRobin.Dataisdistributedinacircularfashion.First,datagotothe
first database, the secondwill go to the second database and so one.Each database server had anequalload.However,ithasadisadvantagethatthedatalookupiscomplex.Andneedalargelookuptable.
2.Anotherapproachistodividethedatainsuchawaythatallthedatawillgotothefirstmachineuntilit
reaches its maximum capacity. When maximum capacity is reached, then data goes to the secondmachineandsoon.Thisapproachhasanadvantagethatonlytherequirednumberofmachinesisused.However,ithasadisadvantagethatthedatalookupiscomplex.Andneedalargelookuptable.
3.Anotherapproachistoselectdatabasecorrespondstothehashvalueofthedata.Findthehashvalueof
thedata.Modthehashvaluebythenumberofdatabases.Thedataarethenstoredinthedatabasevalue
wegotaftermodulus.Forhasavalueapproachwedonotrequireanylookuptable.Wecanfindthedatabase, which is storing the data, by finding the hash value. However, the problem of unevendistributionofdataisthere,thereispossibletohaveamoredatasenttoadatabase,whichhasalreadyreached itsmaximumcapacity. In thiscase,weneed to findabetter load-balancingkeyor split thedatafromthedatabaseintoanumberofdatabases.
4. In thehashvalue,baseddistributionofdata there isno relationbetween thedata that is stored ina
particulardatabase.Informationaboutthedatacanbeusedtomakethedatabaseaccessiblefaster.Forexample, in social networking likeFacebook, if someonewho lives in India ismore likely tohavefriendsfromIndia.AndsomeonewholivesintheUSAismorelikelytohavefriendsintheUSA.
5.Perhapslocationaware(approach4)andthehashvaluebased(approach3)distributionofdatamaybe
thebestapproachtokeepthedatasothatitcantakeadvantageofboththeapproaches.CountrycodeanduserIDcanbeusedtogetthelocationofthedatabase.
RedundancyThereisoneprobleminoursystem,thereisaredundancyintheserversbutourloadbalancerisnowoursinglepointof failure.Weadda secondary loadbalancer incase theprimary loadbalancerdies, thensecondaryloadbalancerbecomesprimaryandthenalltherequestswillbehandledbyit.
Raid (Redundant Array of Inexpensive Disk): Raid is a technology to create redundancy in thedatabases.Multipleharddrivesareusedtoreplicatedata,therebyprovingredundancy.
CachingAcacheisasimplekey-valuestoreanditshouldresideasabufferinglayerbetweenyourapplicationandyourdatastorage.Wheneveryourapplicationwants toread, it first tries toretrieve thedatafromyourcache.Onlyifdataisnotpresentinthecache,thenonlyittriestogetthedatafromthemaindatabase.Cachingimprovesapplicationperformancebystoringportionofdatainmemoryforlow-latencyaccess.Weneeddatabasesandaccesstothedatabaseisslow,soweusemultipletypesofcachingtomakeoursystemfaster.Databaseservers itselfdoescachingsodo theotherentities inbetween theuserand thedatabase.
Memcached:Itisaserversoftware,whatitdoesisitkeptwhateveryouaccessinmemory.Itcanrunonthesameserverasthewebserveroritcanrunonaseparatemachinealltogether.Redis:ItisadataserverbasedonNoSQL,whichisakey-valuedatastore.Dataisstoredasthevaluewith respect tocorrespondingkey.Thisdata is later retrievedby theuseof thekey.Redis isused forcachingitisbesttostorethewholeobjectasoneinstancesothatthedatacanbeaccessedinparallelanddataexpirationwillflushoutthewholeobject.Thereisaproblemsinceramisfinite,thenthecachewillgetfull.Theexpiredobjectwillberemovedsoeverythingthatisaccessedthenitsexpirywillberesetandifthereisanobjectthatisnotusedforsometimethenitwasdeleted.Cacheismoreimportantwhenthewebsitethatwearedesigningismorereadheavierthanawrite.
AcompletewebserverimplementationThesummaryoftheabovesystem.1.TheWeb-Serversofscalablewebserviceishiddenbehindaloadbalancer.Theloadbalancerevenly
distributesloadacrossalltheservers.2.Theusershouldget thesameresult fromweb-server regardlesswhichserver isactuallyserving the
request.Therefore,everyservershouldbeidenticaltoeachother.Serversshouldnotcontainanydatalikesessioninformationoruserprofile.
3.Sessionneedtobestoredinacentralizeddatastore(DB)whichisaccessibletoalltheservers.Datacanbestoredinsomeexternaldatabase.Redundancyinthedatabaseisprovidedbyraidtechnology.
4.Thedatabaseisslow,soweneedacache.In-memorybasedcachelikeRedisorMemcached.5.However,thecachehasaproblemofexpiring.Whenatablechanges,thenthecacheisoutdated.6.ForMemcachedtherearetwooptions:
a.WecansavequeriestotheDBb.Wecansavethewholeobjectthatwillkeepusclosetoweb-server.
7.CDN(Contentdeliverynetworks)canbeusedtoprovideapre-processedwebpage.
Belowdiagramwillgiveyouacompletepictureofthewholesystem.
DesignsimplifiedFacebookDesignsimplifiedFacebookwherepeoplecanaddotherpeopleasfriends.Inaddition,wherepeoplecanpostmessagesandthatmessagesarevisibleontheirfriend’spage.Thedesignshouldbesuchthatitcanhandle10millionofpeople.Theremaybe,onanaverage100friendseachpersonhas.Everydayeachpersonpostssome10messagesonanaverage.
UseCase1.Ausercancreatetheirownprofile.2.Ausercanaddotheruserstohisfriendlist.3.Userscanpostmessagestotheirtimeline.4.Thesystemshoulddisplaypostsoffriendstothedisplayboard/timeline.5.Peoplecanlikeapost.6.Peoplecansharetheirfriendsposttotheirowndisplayboard/timeline.
Constraints1.Considerawholenetworkofpeopleasrepresentedbyagraph.Eachpersonisanodeandeachfriend
relationshipisanedgeofthegraph.2.Totalnumberofdistinctusers/nodes:10million3.Totalnumberofdistinctfriend’srelationship/edgesinthegraph:100*10million4.Numberofmessagespostedbyasingleuserperday:105.Totalnumberofmessagespostedbythewholenetworkperday:10*10million
BasicDesignOursystemarchitectureisdividedintotwoparts:1.First,thewebserverthatwillhandlealltheincomingrequests.2.Theseconddatabase,whichwillstoretheentireperson'sprofile,theirfriendrelationsandposts.
First,threerequirementscreatingaprofile,addingfriends,postingmessagesarewrittensomeinformationtothedatabase.Whilethelastoperationisreadingdatafromthedatabase.Thesystemwilllooklikethis:1.Eachuserwillhaveaprofile.2.Therewillbealistoffriendsineachuserprofile.
3.Eachuserwillhavetheirownhomepagewherehispostswillbevisible.Auser can like anypostof their friendand that likeswill reflecton the actualmessage sharedbyhisfriend.Ifausersharessomepost,thenthispostwillbeaddedtotheuserhomepageandalltheotherfriendsoftheuserwillseethispostasanewpost.
BottleneckA number of requests posted per day is 100million.Approximate some 1000 request are posted persecond.Therewillbeanunevendistributionofloadsothesystemthatwewilldesignshouldbeabletohandleafewthousandrequestsperseconds.
ScalabilitySincethereis,aheavyloadweneedhorizontalscalingmanywebserverswillbehandlingtherequests.Indoingthisweneedtohavealoadbalancer,whichwilldistributetherequestamongtheservers.
Thisapproachgivesusaflexibilitythatwhentheloadincreases,wecanaddmorewebserverstohandletheincreasedload.Theseweb servers are responsible forhandlingnewpost addedby theuser.Theyare responsible forgenerating various user homepage and timeline pages. In our diagram, the client is the web browser,whichisrenderingthepagefortheuser.Weneedtostoredataaboutuserprofile,Usersfriendlist,User-generatedposts,Userlikestatuestotheposts.Letusfindouthowmuchstorageweneedtostoreallthisdata.Thetotalnumberofusers10million.Letus suppose each user is usingFacebook for 5 to 6 years, so the total number of posts that a user hadproducedinthiswholetimeisapproximately20,000millionor20billion.Letussupposeeachmessageconsistsof100wordsor500characters.Letusassumeeachcharactertake2bytes.
Totalmemoryrequired=20*500*2billionbytes.
=20,000billionbytes=20,000GB=20TB
1gigabyte(GB)=1billionbytes1000gigabytes(GB)=1TerabytesMost of thememory is taken from the posts and the user profile and friend list will take nominal ascomparedwith the posts.We can use a relational database like SQL to store this data. Facebook andtwitterareusingarelationaldatabasetostoretheirdata.
Responsiveness is key for social networking site. Databases have their own cache to increase theirperformance.Stilldatabaseaccess isslowasdatabasesarestoredonharddrivesand theyareslowerthan RAM. Database performance can be increased by replication of the database. Requests can bedistributedbetweenthevariouscopiesofthedatabases.Also,therewillbemorereadsthenwritesinthedatabasesotherecanbemultipleslaveDBwhichareusedforreadingandtherecanbefewmasterDBforwriting.Stilldatabaseaccessisslowtowewillusesomecachingmechanism likeMemcached inbetweenapplication serveranddatabase.Highlypopularusersandtheirhomepagewillalwaysremaininthecache.Theremaybethecasewhenthereplicationnolongersolvestheperformanceproblem.Inaddition,weneedtodosomeGeo-locationbasedoptimizationinoursolution.Again,lookforacompletediagraminthescalabilitytheorysection.If it were asked in the interview how you would store the data in the database. The schema of thedatabasecanlooklike:TableUsers
·UserId·FirstName
TablePosts·PostId
·LastName·Email·Password·Gender·Birthday·Relationship
·AuthorId·DateofCreation·Content
TableFriends·RelationId·FirstFriendId·SecondFriendId
TableLikes·Id·PostId·UserId
DesignFacebookFriendssuggestionfunctionDesignasystemtoimplementafriendsuggestionfunctionalityofFacebook,withmillionsofusers.Thealgorithmshouldsuggestallthefriendsoftheimmediatefriendsasaproposedlisttoaddasfriends.
UseCaseThesystemshouldsuggestfriendsofthefriendsassuggestednewfriends.
ConstraintMillionsofuser’slotofdatawithbillionsofrelations.
BasicDesignForgetaboutmillionsofusers. Justconsider thereareonlya fewpersonsand theyareconnectedwitheach other as friends.Consider that people are represented by vertices of graphs and their friendshiprelationisrepresentedbyedges.Sincethereareonlyafewpeople,thenwecankeepeverythinginmemoryandfindthefriendsuggestionusingBreadthFirstTraversal.Wejustneedtofindthenodes,whicharejust2degreesapartfromthestartingnode.
BottleneckSincetherearemillionsofusers,wecannothaveeverythinginmemory.Sincetherearemillionsofusers,wecannotkeepthedataononemachine.Onefriends’profilesmaylieonmanydifferentmachines.
ScalabilitySincetherearemillionsofusers,theiruserprofileisdistributedamongmanydifferentdatabaseservers.User profiles can be distributed depending uponGeo-Location. The Indian users profilewill lie in aserverlocatedinIndiaandUScitizen'sprofilelieintheserverlocatedintheUS.EachuserwillhavecorrespondingUserIdassociatedwiththem.SomeportionofIDcanbeusedtogetGeolocationoftheuser.Anotherportionofuseridcanfindtheuserprofileonthatserver.Theuserprofileisnotthatfrequentlyupdatedsothereismorereadthanwrite.Sosinglemasterwriter-multipleslavereaderarchitectureismostsuitableforthisapplication.Theapplicationservercanprocessthedata;itcandotheoptimizationtoquerylessfromthedatabasebyaccumulatinguserlisttobeprocessed.classsystem
privatemap<int,int>personIdToMachineIdMap;privatemap<int,Machine>machineIdToMachineMap;
MachinegetMachine(intmachineId);getPerson(intpersonId)intmachienId=personIdToMachienIdMap[personId];Machinem=machineIdToMachineMap[machienId];returnm.getPersonWithId(personId);
Optimization:Reduced the number of jumpsby first finding the list of friendswhose profile is on thesamemachine.Thensendthefindnextdegreefriendsquerythatwillreturnthelistofnextlevelfriends.Bydoing,thisworkisdistributedamongvariousmachines.Finally,theresultofthevariousquerieswillbemerged,andthenfriendlistissuggested.Betterresult:Youcancalculatethedegreeofthefriendswiththefriendlist.Thepersonwhoisafriendofmanyofmyfriendsismorelikelytobemyfriendthanthepersonwhoisjustafriendofoneofmyfriends.WeneedtokeeptrackofthefriendreferencecountsbykeepingHash-Tableforthefriendlistandmakethecount1wheneverwefindanewpersonotherwiseincreasethecountby1.Ifwewanttotakeadvantageofcaching,thenweneedtoaddsomedatabasecacheinbetween.
Therecanbemultiplewebservers,whichwillbequeryingthedatabases.Alsomultipleuserswhoareaccessing their Facebook profile and each one of them is proposedwith new friends list so the finalarchitecture is again sameas theoneproposed in the completeweb server implemented in scalabilitytheory.
DesignashorteningservicelikeBitly
UseCaseBasicusecase:1.ShorteningtakesaURLandreturnsashortURL.2.RedirectiontakesashortURLandredirectstotheoriginalURL.3.CustomURL.4.Highavailabilityofthesystem.Additionalusecases:1.Analytics2.Automaticlinkexpiration.3.Manuallinkremoval.4.SpecificcompanyURL.5.UIorjustAPIRequirementAnalysis/MathFirst,weneedtofindtheusagepattern.You can directly ask this data from the interviewer or you can derive it using some data that theinterviewerprovides.Letussuppose that the interviewer tells that therewillbe1billion requestspermonth.Inaddition,outofthese10%times,itisanewrequestand90%ofthetime,itisaredirectionofthealreadyshortenedURL.Letuswritedownthedatathatweget.1.1BNrequestspermonth2.10%arefornewURL/shorteningand90%areforredirection.3.NewURLspermonthis100MLN4.Requestspersecond1BN/(30*24*3600)=385.Roughly,youcanassumeit400requestsperseconds.5.TotalnumberofURLsstoredin5years.
5*12*100MLN=6BNURLsin5years.6.LetussupposethespacerequiredbyeachURLis500bytes.7.LetussupposethespacerequiredbyeachHashcodeforcorrespondingURLsis6bytelong.8.Totaldataweneedtostoreinfiveyears.3TBsforalltheURLsand36gbforhashes
6,000,000,000*500bytes=3terabytes6,000,000,000*6bytes=36gigabytes
9.Newdatawriterequestspersecond:40*(500+6):20k
BasicdesignWebserver:providethewebsitefortheBitlyservicewhereuserscangeneratetheshortURL.ApplicationServer:providesthefollowingservices:1.Shorteningservice2.Redirectionservice3.Key=HashFunction(URL)
DatabaseServer:1.KeeptrackofhashtoURLmapping.2.WorkslikeahugeHash-Tablestoresthenewmappingandretrievesoldmappinggivenkey.Bottleneck1.Trafficisnotmuch2.Datastoragecanbeaproblem.
ScalabilityApplicationServer:1.Startwiththesinglemachine.2.Testhowfarittakesup.3.Wedoaverticalscalingforsometime.4.Addloadbalancerandaclusterofmachinestohandlespikesandtoincreaseavailability.DataStorage:1.Billionsofobjects2.Eachobjectissmall3.Thereisnorelationshipbetweenobjects4.Readsaremorethanwrite.5.3TBsofURLsand36GBofhash.MySQL:1.Widelyused2.Amaturetechnology3.Cleanscalingparadigms(master/slave,master/master)4.UsedbyFacebook,google,twitteretc.5.Indexlookupisveryfast.Mappings:<Hash,URL>1.UseonlyMySQLtablewithtwofields.2.Createauniqueindexonthehashwewanttoholditinmemorytospeeduplookups.3.VerticalscalingofMySQLforawhile4.Partitionofdataintomanypartitions5.Master-slave(readfromslaveandwritetomaster.)6.Eventually,partitionthedatabytakingthefirstcharacterofthehashmodthenumberofpartitions.
StockQueryServerImplement a stockquery service that canprovide an interface toget stockprice information likeopenprice,closeprice,highestprice, lowestpriceetc.Youshouldprovideaninterfacethatwillbeusedtoenterthesedataandinterfacetoreadthisdata.
UseCaseTherewillbetwointerfacestothissystem.1)Firstinterfacetoadddailystockpriceinformationtothesystem.2)Secondinterfacetoreadstockpriceinformationgivingthedateandstockidasinput.
ConstraintsLetussupposethesystemwillbeusedbythousandsofusers.Foreachstock,therewillbeonlyonewriteoperationperday.However,foreverystocktherewillbemultiplereadoperationsperday.Therefore,theapplicationismorereadheavythenwriteheavy.Thesolutionshouldbeflexibleenoughsothatifnewdatafieldsneedtobeaddedtothestocktheycaneasilybeaddedtothesystem.Thesolutionprovidedshouldbesecure.
BasicDesignWecanuseadatabaselikeSQLtostorestockdata.Clientcanaccessthedatabaseusingthewebserverinterface.Belowdiagramwillshowthebasicarchitecture.
Intheabovearchitecture,theusercanaccessthedatabaseusingwebservice.Anynumberofflexibilitycanbeprovidedfortheuse.Forexample,whatisthemaxpriceofsomestockin6monthsetc.?Atthesametime,theuserdoesnothaveaccesstothedatatheyshouldnothave.Wecanprovidedifferentaccessofreadandwritedependingonthenormalusersoradministrator.Well-definedrollingback,backingupdataandsecurityfeaturesareprovidedbytheSQLdatabase.Theabovearchitectureiseasilyextendabletousewithawebsiteorsomemobileapplication.
ScalabilitySincewehave1000’sofusers,thenhavingasinglewebserverandasingledatabaseisnotextendable.We need to distribute data among N number of Databases, which sit behind some load balancer. In
addition,multipleNnumberofwebserverwhichwillsitbehindsomeloadbalancer.Eachoftheloadbalancersneedstobeprovidedwithsomeredundancy,astheywillbeasinglepointoffailure.Finally,the solutionwill look likebelowdiagram. (Fordetails, see scalability theoryexplainedearlier in thischapter)
DesignabasicsearchengineDatabaseYouaregivenmillionsofURLs;howwouldyoucreateyourdatabase.So thatgivenaquery stringofwords,howtofindtheURLs,whichcontainallthewordsofthequerystring.Thewordscancomeinanyorder.
UseCase1)WearegivenalistofmillionsofURLs.2) The user of the systemwill provide query string. In addition, we need to return the URLs, which
containallthewordsofthequerystring.3)Itissomekindofsearchenginesowecanpre-processthedataandmakeourdatabase.
RequirementAnalysisIntherequirementstep,youneedtofindouthowmanyusersaregoingtousethissearchengine.Inourcase,letussupposetherearenotmanyuserswhoaregoingtouseoursystemsothesourmainconcernisadatabase.MaybewehaveNnumberofmachinesthatcanbeusedtofastourdatapre-processing.
BasicDesignInthisstep,wewillmakethebasicdesignsoletusmakeaworkingsystemwithjustafewURLs.HowwouldyoufindtherequiredURLfromthegivenURLs,whichcontainsallthewordsoftheinputquerystring?WecanmakeaHash-Tableinwhichwordsarethekeysanddocumentidsarevalues.“Hello”->url1,url2,url3“World”->url2,url4,url5Tosearch thedocument,whichcontains“helloworld”,wecanfind the intersectionof the twolists. Inaddition,url2istheresult.
BottleneckIn this step,wewill look back to our original problem inwhich pre-processing ofmillions ofURLs.TheremaybeanumberofdifferentwordssoitmaynotbepossibletokeepthewholeHash-Tableonasinglemachine.Therefore,weneedtodividetheHash-Tableandkeepitonaseparatemachine.WeneedtoretrievetheURLsthatmatchagivenwordefficiently.Sothatwecanfindtheintersection.Pre-processingallthemillions,URLsbysinglemachinewillbeslow.Weneedtofindawaytoparallelprocesspre-processingstep.
ScalabilityLetuslookintotheproblemofkeepingtheHash-Tableindifferentdatabases.Onesolutionistodividethewordsalphabetically.Wecanmaketablescorrespondingtoeachword.Eachdatabasecontainstablesofwordsundersomerange.Forexample,DB1containsallthewords,whichstartwithalphabet“a”,andDB2containsallthewords,whichstartswiththealphabets“b”andsoon.Dataisstoredinthedatabase.
When a database reaches itsmaximum capacity, the data is stored to nextmachine and a tree kind ofstructurecanbemade.Finding the listofURLscorresponding tosomeword iseasy,wecango to thecorresponding database and find the table and get all the data of that table. Finally, we can take theintersectionoftheresultofvariouswords.Inaddition,theresultwillbegivenasoutput.ProcessingofthemillionsofURLswithasinglemachineisslow.Therefore,wecandivideabunchofURLprocessingamonganNnumberofmachines,eachURLprocessingisindependentofeachotherandthefinalHash-TableoftheURLscanbefinallycombined.ThisapproachofprocessingindependentdataandfinallycombiningtheirresultisusedinMapReduce.MapReduce:AMapReduce divides the input dataset into independent chunks,which are processed inparallel.Thentheiroutputiscombinedtogettheresult.
DesignabasicsearchengineCachingGivenasearchenginedatabaseimplementationwhichsupportsQuerySearch()functionwhichwillreturnthe best list of URLs based on thewords of the query. This time you need to design theweb serverimplementationof this such that thereareNnumberof thewebservers that are responding to theuserqueries.Anywebservercanbepickedat random.QuerySearch() isaheavyoperationsoyouneed todesign,cachingforthissystemsothatdatabaseaccessisreduced.
UseCase1.TheuserofthesystemwillprovidequerystringandsystemwillrespondwiththeproperlistofURLs
correspondingtohisrequest.2.Giventhat,thedatabaseoperationsareveryheavyweneedtominimizethembycachingthequeriesat
thewebserver.3.Weneedtokeepthefrequentqueriesinthecacheandstalequeriesneedtoberemovedfromthecache.4.Weneedtohavesomeproperrefreshmechanismforeachquery.
BasicDesignLetuslustforgetaboutanNnumberofmachinesandassumethatQuerySearch()operationhappensonasinglemachine.Nowwewouldliketocachequeries.Eachquerywillconsistofsomestring.Inaddition,theresultofaqueryisalistofURLs.WeneedtohaveaquickcachelookupsothatwecangettheresultoftheQueryfromthecacheif it ispresentthere.Also,needtohavesomeproperrefreshmechanismforeachquery.TheHash-Tableismosteffectivetokeepthecache.Byusingahash,atablelookupisfast.However,ifthecacheisfilledhowyouwouldremovetheleastuseddatafromthecache.Alinkedlistcanbeusedtoremovetheolddata.Youcankeepadoublelinkedlisttomanagetheolddataremoval.Wheneveradataisaccessed,itcanbemovedtothefrontofthelinkedlistandtheremovalcanhappenfromtheendofthelinkedlist.Takingadvantageofboththesolutions,wecankeepthecacheinalinkedlistandadditsreferencetotheHash-Table.Now the last problem of how to remove the data upon the expiry of it. For example,most frequentlyaccessedqueryresultwillalwaysremaininthelinkedlisteventhoughthatresultischangedandifitisaccessedagainfromthedatabasethenitwillgivesomeupdatedresult.Forthis,weneedtohavesomeTTL(timetolive)associatedwitheachquerydependingupontheresultofURLswegetfromeachquery.Forexample,someweatherorcurrentnewsrelatedqueriesshouldhaveaTTLofdays.Ontheotherhand,somehistoricdatashouldhavealongTTL.TheTTLcanbederivedfromhowfrequentlytheURLsarechanginginthequeryresult.
BottleneckThereareNdifferentwebservers.Inaddition,anyparticularquerycanbeservedbyanyserver.
Dataaccessshouldbefast.
ScalabilityThevarioussolutionsthatwecanthinkaboutare:Approach1:Serverscanhavetheirowncache.IfsomequeryissenttoMachine1,itwillcatchitinitscachewhenthesamequeryissenttoitagainitwillreturnitfromitscache.However,ifsomequeryissenttoMachine1first, itwillcacheitandif thesamequeryissent toMachine2, itwillagaindoadatabaselookupandcache it to its own cache. This implementation is suboptimal as it is doingmore number of databaselookupsthanwhatisactuallyrequired.Approach2:Anotherapproachisthateachmachinestoresidenticalcache.Wheneversomedatabaseaccesshappenedthen thesamecache isupdatedbyall thewebservers.Thisapproachhasadrawback thatwheneveradataisupdatedincachesamecacheupdateisfiredtoalltheNwebservers.Anotherdisadvantageisthatallthecachestoresthesamedatasowearewastingpreciouscachespace.Approach3:Inthisapproach,wewilldivideourcachesuchthateachwebserverholdsadifferentpartofthecache.Whenaqueryreachtosomewebserveritknowswhichwebserveractuallyholdsthecacheforthisqueryoratleastknowsthatwhichserverissupposedtokeepacacheoftheparticularquery.Todothisweneedahash-basedapproach.Wefindtheserver,whichservesthequery,byjustfindingthehash(query)percentageN.Whenaqueryrequestcometosomewebserver,itwillfindthewebservercorrespondingtothisquerybyapplyingtheformula.ItwillasktheQuerySearch()functiontothatparticularserver.Thatserverwillinturnwillquerythedatabaseifrequiredorprovidetheresultfromitsowncache.Now, regarding thecacheexpirationandoldcacheremoval.Wearekeeping theTTLcorresponding toeach query so there can be a thread runningwhich looks for the expired data and remove it from thecache.Inaddition,combinationoflinkedlistandHash-Tableisusedtokeepthecachetogetridoflessaccesseddatawhenthecacheisalmostfull.As a further improvement,we can think of some sort ofGeo location awarewebserver selection andcachepolicysothatqueryrelatedtoIndiaismoresupposedtobedoneinIndiaandqueryrelatedtochinaissupposedtocomemorefromchina.
DuplicateintegerinmillionsofdocumentsGivenmillionsofdocumentswithalldistinctnumbers,findthenumber,whichoccursmultipletimes.
BasicDesignConsidertherearejustafewnumbers,andwewanttofindtheduplicatenumbers.Thefirstapproach is tofindkeepasorted listof thenumbersandsee if thenextreadnumbermatcheswithsomenumberinthelist.Anotherbetter approach is to findahashvaluecorresponding tonumber andadd that thenumber to aHash-Table.
ConstraintMillionsofdocumentsandthereisnorangeofnumbersowecannotkeepeverythinginmemory.
ScalabilityWecanfindthehashvalueforalltheintegersandthenaddthatintegertoitscorrespondinghashvaluefileordatabase.Ifthereissomeduplicate,thentheywillfallinthesamefile.Inthefirstpass,variousfilesarecreatedandintegersaredistributed.In thesecondpassall, thedataof the individual filescanbe loaded intomemoryandsorted to find ifthereissomeduplicatevalue.Wecanusethesametechniqueexplainedabovetoprocessthevariousdocumentsofintegerinparallelbydifferentmachinesandthencombinetheiroutputtogetourresultfaster.
Zomato
UseCase1.Givenalocation,thelistofhotelsinthatlocalityneedstobedisplayed.2.Givenahotelnamethathotel’srating,review,andmenuneedtobedisplayed3.Thereshouldbesomeoptiontofindifadeliveryoptionisthereinthehotel.4.Thereshouldbesomeoptiontoselectahotelonveg/non-vegcategory.5.Thereshouldbesomeoptiontoselecthotels,whichservealcohol.6.Theusershouldbeabletoaddreviews,addpersonalratingstothehotels.7.Theuserhassomeaccountorcanaccessasguest.8.Users/Adminshouldbeabletoaddnewhotelstothesystem.
Constraints1.Anumberofqueriespersecondsuppose100queriespersecond.2.Therearemorereadsthanwrites.3.90%ofthetimethereisreadoperationandonly10%ofthetimethereisawriteoperation.4.100*60*60*24=8,640,000
BasicDesign
ThescalabilitywillbesameasthatoftheexamplesexplainedaboveinthecaseofbasicFacebook.Sameconceptofredundancy,loadbalancing,scalabilityetc.
AbstractDesignEachhotelhassomehotelidassociatedwithit.1.DataofthehotelcanbeName,Address,Rating,ReviewList,Veg-Nonveg,andAlcoholetc.2.Theregionisafieldintheaddress.3.Search:Whenauserdoesaregionbased,queryalltheentriesofthehotelsinthatregionneedtobe
displayedtotheuser.4.Search:Usershouldbeabletosearchspecifichotel.5.AddReview/Rating:Whenusersassessahotel,thenheshouldbeabletoaddreviewsandratingfor
thehotel.6.ObviouslytheimagesarestoredinCDN
ApplicationServicelayer1.Startwithfewermachines2.Allloadbalancer+aclusterofmachinesovertime.3.Trafficspikehandling.4.Highavailability.DataStorageLayer:1.Thousandsofhotels.2.Therearenorelationshipsbetweentheobject.3.Readsaremorethanwriting.4.RelationaldatabaseoptionisMySQL5.Widelyused.6.Clearscalingparadigms.(Master-Masterreplication,Master-Slavereplication)7.Indexlookupsareveryfast.Oneoptimizationthatwecanassignanidtohotelstheidcanbederivedfromthelocalitysothatitwouldbeeasytofindhotelsinthatlocality.
YouTube
Scenarios1.Usershavesomeprofileaccordingtowhichcontentisshown.2.ContentthumbnailsareshownwhentheuseropenstheYouTubewebpage.3.Whentheuserclicksonsomethumbnail,thenthatvideoisplayedonflashplayer.
Constraints1.Millionsofusersaregoingtousethisservice.2.200millionvideorequestsservedperday.3.Morereadsthanwrites.
Design1.YouTubeissupposedtoservehugenumberofvideosforwhichithasvideoservingclusters.Asingle
videocanbeservedfrommultipleserversinclustersandfrommultipleclusterstherebydistributingthediskreadwhichincreasestheperformanceofthesystem.
2. Themost popular videos are served fromCDN.CDN ismore close to the user,which reduce theresponsetime.Thisalsoreducetheloadtothevideoservingclusters.
3.Therestofthemetadataofthevideoisservedfromotherservers,astheuserisnotmuchinterestedinthemetadata.
4.Therestoftheapplicationwillbesame,asitwillhaveanapplicationserver,databaseservers,loadbalancer,cachingetc.
5.There ismorereadthanwriteso themasterserver topologywillbeused.Therefore, therecanbeasinglemasterforwritingandmultipleslavesforreading.
6.Master data is replicated to slaves. Since slaves are same asmaster then themaster is down, thenslavescanbepromotedtomakeasmaster.
7.Pagetobedisplayedtotheuserdependsonhissubscribedpages,Historyetc.8.InformationcanbecachedintheMemcachedimplementednearthedatabaseloadbalancer.
DesignIRCTC
Scenario
1.Theusershouldbeabletoquerytrainsbetweentwostations.2.Thequeryshouldbebasedonthedate,quota,fromandtothestation.3.Theusershouldbeabletoseetheavailabilityinthetrainlistretrievedfromtheabovequery.4.Theusershouldbeabletobookticketsfortheavailableseats.
Constraints1. Therewill be a huge number of people requesting the service. Let us suppose 0.01 percent of the
populationusetheservicedailyonce.2.Geo-Redundancyshouldbeprovided.3.Morereadquery/requestthenwritesrequests.
Design1.Thebasicarchitecturefromthescalabilitytheorytopiccanbeusedheretoo.2.Thereshouldbeahugenumberofservers,whichareservingtheusers.3.Thereshouldbemultipleserversatmultiplegeo locations toprovidegeo-redundancy.Thedatabase
shouldbereplicatedatthesemultiplegeolocations.Theremaybemultipleserversinoneparticularzonetoo.
4.Thereisahugenumberofreadquery,theusergenerallydoesalargenumberofquerytofindtheseathewantstobook.Therewillbemorereadsthenwritesomaster-slave.
5.Queriescanbecached;littleolddataisok.6.Allthesearchwillbeservedbyslaveservers.7.Whenwebookaticketthentransactiongoesdirectlytothemasterserver.Locksontrainnumbercanbe
takentopreventraceconditions.Oncealockisacquiredthenonlyyoucanbookaticket.Somecountscanbeusedtoavoidunnecessarylockingandsomecountercanbeusedforthis.
8.Eachstationhasaquotaintrainandseatsareallocatedfromthatquota.9.Eachphysicaltrainwillhavetwotrainidsonewhentransgofromsourcetodestinationstationandone
whenitcomesback.Sointhesystem,therewillbetwotrainids.Keepingthesetwoseparateidswillmakethequeryeasiertoimplement.
10.When final charting is done then each seat is swapped for the empty slots andwe try to find therequestfromsourcestationwhosedestinationisalsointhatslot.Thefirstfitisallottedthatseat.
11.Aloadbalancerisusedtodistributetraffic.12.Theremaybemultiplebookingserverswhichaskforabookingtoken to themasterserver.Master
serverallocateatokenforthatserverandreserveitforsometime.Whenalltheuserinformationisfilledandpaymentisdonethenonlyitallocatesrealseatdependinguponuserpreference.
13.Slaveserverwillhandleuserrequesttilltheend.Finalbookingrequestwiththeuserpaymentandhiscompleteinformationwillgotothemasterserverandthecorrespondingticketwillbebooked.
AlarmClockHowwouldyoudesignanalarmclock?
UseCaseAlarmclockshouldhaveallthefunctionsofclock.Shouldbeabletoshowtime.TheUsercansetthealarmtimeTheUsercanresetthealarm.TheUsercansetthealarm.
ConstrainsTheGranularityofalarmcanbe15min.
TestCaseSetthealarmatsometime6:00AMandsetit.TheAlarmshouldworkat6:00AMStopthealarm,thenalarmshouldstopringing.
DesignTherecanbeaclockclass,whichmanagestimeandshoetimetothescreen.IthasfunctionsgetTime()andsetTime()Alarm Clock extends Clock, and have some more function like startAlarm(), stopAlarm(),setAlarmTime(),ring()
ImplementationAtimerentrywillrunforgranularityof15min.Alternatively,onemindependsoncustomerrequirement.Itwillcheckifcurrenttimeisequaltoalarmtime.Iftrueandthealarmisonthenring.IfstartAlarmiscalled,itwillsetalarmonIfstopalarmiscalled,itwillsetalarmoff.
DesignforElevatorofabuilding
ScenariosAtypicallifthasbuttons(Elevatorbuttons)insidethecabintolettheuserwhogotthelifttoselecthis/herdesired floor.Similarly,each floorhasbuttons (Floorbuttons) tocall the lift togo floorsaboveandafloorbelowrespectively.Thebuttonsilluminateindicatingtherequestisaccepted.Inaddition,thebuttonstopsilluminatingwhentheliftreachestherequestedfloor.Usecases:User•Pressesthefloorbuttontocallthelift•PressestheelevatorbuttontomovetothedesiredfloorFloorButton&ElevatorButton•Illuminateswhenpressedbyuser•PlacesanelevatorrequestwhenpressedElevator•Movesup/downasperinstruction•Opens/closesthedoor
DesignEachbuttonpressresultsinanelevatorrequestwhichhastobeserved.Eachoftheserequestsistrackedatacentralizedplace.ElevatorRequests,theclassthatstores,elevatorrequestscanusedifferentalgotoscheduletheelevatorrequests.Theelevatorismanagedbyacontrollerclass,whichwecallElevatorController.Elevatorcontrollerclassprovideinstructionstotheelevator.Elevatorcontrollerreadsthenextelevatorrequesttobeprocessedandserved.
Thebuttonisanabstractclassdefiningcommonbehaviourlikeilluminate,doNotIlluminate.FloorButton,ElevatorButtonextendsButtontypeanddefineplaceRequest()methodwhichisinvokedwhenabuttonispressed.Whenafloorbuttonorelevatorbuttonispressesarequestsisaddedtoacommonqueue.ElevatorControllerreadsthenextrequestandinstructnextactiontotheelevator.Howcanweextendthistomultipleelevators?Inthesingleelevatorscenario,thereisasingleelevatorandanelevatorcontrollerandacommonserverwhere the floor requests and the elevator button request are stored.Which are processed as per theschedulingalgorithm.Toextend this tomultiple elevator scenarios therewill still be single elevator controller.Floorbasedrequests can be served by any elevator whereas elevator button requests will be served only by theelevatortowhomthebuttonbelongs.FloorButton's placeRequest() adds a request to the common queue,which is accessed by the elevatorcontroller thereby assigning the request to one of the elevators. ElevatorButton's placeRequest adds arequesttotheelevatordirectlyasitissupposedtoserveit.Elevatorcontrollerwillberunningvariousalgorithmslikeshortestseeketc.todecidewhichliftissupposedtohandlewhichrequest.
ValetparkingsystemDesignavaletparkingsystem.
UseCaseTherequirementsofthevaletparkingsystemshouldbe:1.GivenaParkinglothavingafixednumberofslots2.Whereacarcanentertheslotifthereisafreeslotandthenitwillbegiventhedirectionofthefreeslot.3.Whenexitingthecarhastopaythefeesforthedurationofthetimeitusesparking.
Constraints1.Parkingslotscomeinmultiplesizes-small,midandlarge2.Threetypesofvehicles,small,mid,large3.Asmallvehiclecanparkinasmall,medium,orlargespot4.Amediumvehiclecanparkinamediumorlargespot5.Alargevehiclecanparkonlyinalargespot
Design&ImplementationTheparkinglotwillhavethefollowinginterfaceparkingLot
Map<int,Space>unreservedMap;Map<int,Space>reservedMap;
reserveSpace(Space)//Itwillfindifthereisspaceintheunreservedmap//Ifyes,thenwewillpickthatelementand//putintothereservedmapwiththecurrenttimevalue.
unreserveSpace(Space)//Itwillfindtheentryinreservemap.Ifvaluefoundthen//wewillpickthatElementandputintotheunreservedmap.//Andreturnthechargeunitswiththecurrenttimevalue.
OOdesignforaMcDonaldsshopLetusstartwiththedescriptionofhowtheMcDonaldsshopworks.1.InaMcDonaldsshop,theCustomerselectstheburgeranddirectlyplacestheorderwiththecashier.2. In a McDonalds shop, the Customer waits for the order ready notification. Customer upon being
notifiedthattheorderisreadycollectstheburgerhimself.Therearethreedifferentactorsinourscenarioandbelowisthelistofactionstheydo.Customer1.Paysthecashtothecashierandplaceshisorder,getatokennumberandreceipt2.Waitsfortheintimationthatorderforhistokenisready3.Uponintimation/notification,hecollectstheburgerandenjoyshisdrinkCashier1.Takesanorderandpaymentfromthecustomer2.Uponpayment,createsanorderandplacesitintotheorderqueue3.ProvidetokenandreceipttothecustomerCook1.Getsthenextorderfromthequeue2.Preparestheburger3.Placestheburgerinthecompletedorderqueue4.Placesanotificationthatorderfortokenisready
ObjectorienteddesignforaRestaurantLetusdescribehowtherestaurantworks.1.Inarestaurant,thewaitertakesorderfromthecustomer.2.Thewaiterwaitsfortheordertobereadyandoncereadyservesthedishestothecustomer.ThesearethedifferentactorsinthemodelandIhavelistedthedifferentactionsagainsteachactorCustomer1.Selectsthedishfromthemenuandcalluponawaiter2.Placestheorder3.Enjoyshismealoncethedishisservedonhisplate4.Askforthebill5.PaysfortheservicesWaiter1.Respondstothecustomerscallonthetablesheiswaiting2.Takesthecustomer'sorder3.Placestheorderinthependingorderqueue4.Waitsfortheorderreadynotifications5.Oncenotificationisreceived,collectsthedishandservesthedishtothecorrespondingcustomer6.Receivesthebillrequestfromcustomer7.AskstheCashiertopreparethebill8.GivesthebilltothecustomerandacceptsthepaymentCashier1.Acceptsthepreparedbillrequestfromthewaiterforthegivenorderdetails2.Preparesthebillsandhandsitovertothewaiter3.AcceptsthecashfromthewaitertowardstheorderCook1.Getsthenextorderfromthependingorderqueue2.Preparesthedishandpushtheordertofinishedorderqueue3.Sendsanotificationthattheorderisready
ClassdiagramfortheRestaurant.
ObjectorienteddesignforaLibrarysystemA library has a set of books,which the users can borrow for a certain period and return.Usersmaychoosetorenewthereturndateiftheyfeeltheyneedmoretimetoreadthebook.
Thetypicaluseractionswiththisonlinelibrarywouldbe·Signin/register·Searchbooks·Borrowbooks·Renewbooks·Returnbooks·Viewhisprofile
Theonline librarymustkeep trackof thedifferentbooks in the librarycurrentlyavailable forusers toborrowandthebooksalreadyborrowedbyusers.Putitsimplytheinventoryshouldbemanaged.Thevariouscomponentsofthesystem:1.User2.Librarian3.Library4.Book5.Transection6.EventManagerThebelowclassdiagram,whichdepictshowthesecomponentsinter-operates.
TheUsereitherinteractswiththeLibrarian,theuserrequest,returnorrenewsabook.TheLibrarianwill
searchforthebookifthebookisavailableintheLibrarythenissueittotheuser.ATransectionwillbecreated and added to theEventManager.EventManagerwill support add transaction and send returnrequestinterface.Oncethebookisoverduethentheeventmanagerwillsendanindicationtothestudentthatthebookneedstobereturned.WhenthebookisrenewedthenthelibrarystateisnotchangedbuttheTransectiondetailisrenewedattheEventManager.
Suggestashortestpath
UseCaseTheuserhadsomecoordinatebysearchingthecoordinatefromthename.Showthewholemapconsideringthecoordinateasitscentre.Suggesttheshortestpathbetweentwopoints.
ConstraintsAllpathsarepositiveincost.For simplicity, I am considering all paths are for vehicle only, no pedestrian (pedestrian canwalk ineitherdirectioneveninone-wayroad.)
DesignThewholecitymapisstoredasagraphingoogle.Weneedtofindthemapbylookingintotheobjects,whichareinthedistanceshownbythebrowser.Thesamepathisstoredasdirectedgraph.Inaddition,thegraphthatneedstoberendereddependsonthezoomlevel.Thepreferredalgorithmisa*forthisapplicationtogettheshortestpath.Weight=h(x,y)+g(x,y)
Exercise1.DesignasystemtoimplementsocialnetworkinglikeFacebook,withmillionsofusers.Howwouldyou
findtheconnectionbetweentwopeople?2.Autocompleteinwww.booking.com.Designautocompletefeatureforwww.booking.com.3. Instagram, Instagram is anonlinemobile-basedphoto sharing,video sharing service,whichenables
userstotakepictures,andvideouploadthemtotheserverandsharethemonsocialnetworkingsiteslikeFacebookorTwitter.Note:-CDNisusedtostoreactiveimages.
4.MonolithicWebsite,assumeyouhaveamonolithicwebsiteandyouareaskedtoarchitectthewebsite
Hint:-Discusswholescalabilitytheorysectionhere.5. TripAdvisor: URL's are parsed; content is collected from various services, and then applied to a
template.6.Cinchcast:Liveaudiostreamingforbusinesstodoconferences.7.BlogTalkRadio:Audiosocialnetwork8.Clientbasedrecommendationfeature:Howwouldyoudesignaclientbasedrecommendationfeature
(basedoncustomerhistory)on theproductdetailpage?DesignCustomerswhovieweditemAalsoviewitemBanditemCinanonlineshoppingportal.
9.Car rentingsystem:Designacar rentingsystem, includingreservingacar,checking inandchecking
out.Considerallthecases:reserveacar,thencheckoutsuccessfully;reserveacar,butthecarissoldoutbeforeyoucheckout.TestCases:a)Trytoreserveacarformorethanonepersonb)Trytoreserveacarthatissoldoutc)Verify thecheckoutprocess.After checkingout aparticular, you shouldbeable to reserve it for
anothercustomer.d)Trytoreservethesamecarfordifferentcustomersindifferentdates
10.Onlinecabbookingsystem(likeUber)
AdminModulea)Adminshouldbeabletoaddnewdriver/taxidetails.b) Should be able to calculate the amount that needs to be paid to the drivers.Monthly,weekly or
daily.
UserModulea)Shouldbeabletochoosefromandtolocation.b)AvailableTaxiestype,alongwithfaredetails.
c)SelectaTaxityped)Bookthetaxi.e)Aconfirmationmessageforthebooking.
DriverModulea)AdrivershouldbeabletoregisterasadrivertoUber.b)Whenajobisdisplayedtothedriver,heshouldbeabletoacceptthejob.c)Whenthedriverreachestothecustomerthenheshouldbeabletostartatrip.d)Whenthedriverhadtakenthecustomertothedesiredlocationthenheshouldstopthetrip.e)Thedrivershouldcollectthefarebasedontheamountdisplayedintheapp.f)Thedrivershouldbeabletogivecustomerfeedback.
Note:Justassume2minutesisequalto1KM.
11.Onlineteachingsystem
a)Inanonlineteachingsystem,therearennumberofteachersandeachoneteachesonlyonesubjecttoanynumberofstudents.
b)Astudentcanjointoanynumberofteacherstolearnthosesubjects.c)Eachstudentcangiveonepreferencethroughwhichhecangetupdatesaboutthesubjectorclass
timingsetc.d)ThosepreferencescanbethroughSMSorTwitter/FacebookorEmailetc.e)Designabovesystemanddrawthediagramforabove.
12.CustomerOrderBookingSystem
AdminModulea)Shouldbeabletoadd/edit/deleteitem,alongwithquantity,price,andunit.b)Shouldbeabletoseeallorders.
CustomerModulea)Shouldbeabletoenterhis/herdetailsforshipping,alongwillbasicinformationlikename,email,
contactetc.b)Canchooseitem,quantityc)Automaticallypayablepriceshouldbegeneratedasperselecteditemandquantity.d)Shouldbeabletoconfirmtheorder.e) After confirmation can see order confirmation report along with order number, which will be,
systemgenerated.13.OnlineMovieBookingSystem
AdminModulea)Shouldbeable toenterallmovies,whichhavebeenreleased,andabout torelease innextweek
withallpossibledetailsliketheatrelocation,price,showtimingsandseats.b)Shouldbeabletodeletemovies,whicharenolongerinthetheatre.c)Canseeanumberofbookedticketsandremainingticketsforsingletheatreorforalltheatre.
UserModule
a)User should be able to check all ongoingmovies in theatre alongwith locations, availability ofseats,price,andshowtimings
b)Theusershouldbeabletocheckallupcomingmoviesfornextweektoo.c)Allmoviesthosearerunningontheatreshouldbeavailableforbooking(oneticketormorethanone
ticketcanbebooked).d)Afterbookingusershouldseetheconfirmationmessageofbooking.
14.Design an onlineAuction system (similar to e-bay). Functionalities include enlisting a product for
auctionbybidowner,placingthebidforaproductbybidders,Bidwinnerselection,Notificationofbidwinneretc.).
APPENDIX
AppendixAAlgorithms TimeComplexityBinarySearchinasortedarrayofNelements O(logN)ReversingastringofNelements O(N)LinearsearchinanunsortedarrayofNelements O(N)ComparetwostringswithlengthsL1andL2 O(min(L1,L2))ComputingtheNthFibonaccinumberusingdynamicprogramming O(N)CheckingifastringofNcharactersisapalindrome O(N)FindingastringinanotherstringusingtheAho-Corasickalgorithm O(N)SortinganarrayofNelementsusingMerge-Sort/Quick-Sort/Heap-Sort O(N∗logN)SortinganarrayofNelementsusingBubble-Sort O(N!)Twonestedloopsfrom1toN O(N!)TheKnapsackproblemofNelementswithcapacityM O(N∗N)Findingastringinanotherstring–thenaiveapproach O(L1∗L2)Threenestedloopsfrom1toN O(N3)Twenty-eightnestedloops…yougettheidea O(N28)Stack Addingavaluetothetopofastack O(1)Removingthevalueatthetopofastack O(1)Reversingastack O(N)Queue Addingavaluetoendofthequeue O(1)Removingthevalueatthefrontofthequeue O(1)Reversingaqueue O()Heap Addingavaluetotheheap O(logN)Removingthevalueatthetopoftheheap O(logN)Hash Addingavaluetoahash O(1)Checkingifavalueisinahash O(1)
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