Part-B1 Stacks

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Abstract Data Types (ADTs)

An abstract data type (ADT) is an abstraction of a data structureAn ADT specifies: Data stored Operations on

the data Error conditions

associated with operations

Example: ADT modeling a students record The data stored are

Student name, id No., as1, as2,as3, exam

The operations supported are

int averageAs(as1,as2,as3) Int finalMark(as1, as2,as3,

exam) ) Error conditions:

Calculate the final mark for absent student

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The Stack ADT (§4.2)The Stack ADT stores arbitrary objectsInsertions and deletions follow the last-in first-out schemeThink of a spring-loaded plate dispenserMain stack operations:

push(object): inserts an element

object pop(): removes and returns the last inserted element

Auxiliary stack operations:

object top(): returns the last inserted element without removing it

integer size(): returns the number of elements stored

boolean isEmpty(): indicates whether no elements are stored

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Stack Interface in Java

Java interface corresponding to our Stack ADTRequires the definition of class EmptyStackException

public interface Stack {

public int size();

public boolean isEmpty();

public Object top()throws EmptyStackException;

public void push(Object o);

public Object pop() throws EmptyStackException;

}

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ExceptionsAttempting the execution of an operation of ADT may sometimes cause an error condition, called an exceptionExceptions are said to be “thrown” by an operation that cannot be executed

In the Stack ADT, operations pop and top cannot be performed if the stack is emptyAttempting the execution of pop or top on an empty stack throws an EmptyStackException

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Applications of Stacks

Direct applications Undo sequence in a text editor Chain of method calls in the Java

Virtual Machine

Indirect applications Auxiliary data structure for algorithms Component of other data structures

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Array-based Stack (Implementation)

A simple way of implementing the Stack ADT uses an arrayWe add elements from left to rightA variable t keeps track of the index of the top element

S0 1 2 t

…

Algorithm size()return t + 1

Algorithm pop()if isEmpty() then

throw EmptyStackException else

t t 1return S[t + 1]

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Array-based Stack (cont.)The array storing the stack elements may become fullA push operation will then throw a FullStackException

Limitation of the array-based implementation

Not intrinsic to the Stack ADT

S0 1 2 t

…

Algorithm push(o)if t = S.length 1 then

throw FullStackException else

t t + 1S[t] o

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Array-based Stack (Cont.)

A Stack might be emptytop returns the element at the top of the Stack, but does not remove the top element. When the Stack is empty, an error occurs.

S0 1 2 t

…

Algorithm isEmpty() if t<0 then return true else return false

Algorithm top()if isEmpty() then

throw EmptyStackException return S[t ]

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Performance and Limitations for array-based Stack

Performance Let n be the number of elements in the stack The space used is O(n) Each operation runs in time O(1)

Limitations The maximum size of the stack must be

defined a priori and cannot be changed Trying to push a new element into a full stack

causes an implementation-specific exception

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Array-based Stack in Javapublic class ArrayStack implements Stack{// holds the stack elements

private Object S[ ];// index to top elementprivate int top = -1;// constructor

public ArrayStack(int capacity) { S = new Object[capacity]);

}

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Array-based Stack in Javapublic Object pop() throws EmptyStackException{

if isEmpty()throw new EmptyStackException

(“Empty stack: cannot pop”);Object temp = S[top];// facilitates garbage collection S[top] = null;top = top – 1;return temp;

} public int size() { return (top + 1);

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Array-based Stack in Java public boolean isEmpty() { return (top < 0); }public void push(Object obj) throws FullStackException { if (size() == capacity) throw new FullStackException("Stack overflow."); S[++top] = obj; } public Object top() throws EmptyStackException { if (isEmpty()) throw new EmptyStackException("Stack is empty."); return S[top]; }

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Parentheses Matching

An expression, i.e.,[(2+3)*x+5]*2. Each “(”, “{”, or “[” must be paired with a matching “)”, “}”, or “[” correct: ( )(( )){([( )])} correct: ((( )(( )){([( )])} incorrect: )(( )){([( )])} incorrect: ({[ ])} incorrect: (

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Parentheses Matching Algorithm

Algorithm ParenMatch(X,n):Input: An array X of n tokens, each of which is either a grouping symbol, avariable, an arithmetic operator, or a numberOutput: true if and only if all the grouping symbols in X matchLet S be an empty stackfor i=0 to n-1 do

if X[i] is an opening grouping symbol thenS.push(X[i])

else if X[i] is a closing grouping symbol thenif S.isEmpty() then

return false {nothing to match with} if S.pop() does not match the type of X[i] then

return false {wrong type} if S.isEmpty() then

return true {every symbol matched} else

return false {some symbols were never matched}

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Parentheses Matching Example 1

Input: () (() [()])

i X[i]

Operation Stack Output

0 ( Push ( (

1 ) Pop ( Test if ( and X[i] match? YES

2 ( Push ( (

3 ( Push ( ((

4 ) Pop (Test if ( and X[i] match? YES

(

5 [ Push [ ([

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Parentheses Matching Example 1

Input: () (() [()])

i X[i]

Operation Stack

Output

6 ( Push ( ([(

7 ) Pop (Test if ( and X[i] match? YES

([

8 ] Pop [Test if [ and X[i] match? YES

(

9 ) Pop (Test if ( and X[i] match? YES

Test if stack is Empty? YES TRUE

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Parentheses Matching Example 2

Input: ( () [] ]()

i X[i]

Operation Stack Output

0 ( Push ( (

1 ( Push ( ((

2 ) Pop (Test if ( and X[i] match? YES

(

3 [ Push [ ([

4 ] Pop [Test if [ and X[i] match? YES

(

5 ] Pop (Test if ( and X[i] match ? NO FASLE

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Computing Spans (not in book)

We show how to use a stack as an auxiliary data structure in an algorithmGiven an array X, the span S[i] of X[i] is the maximum number of consecutive elements X[j] immediately preceding X[i] and such that X[j] X[j+1] Spans have applications to financial analysis

E.g., stock at 52-week high

6 3 4 5 2

1 1 2 3 1

X

S

01234567

0 1 2 3 4

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Quadratic AlgorithmAlgorithm spans1(X, n)

Input array X of n integersOutput array S of spans of X #S new array of n integers nfor i 0 to n 1 do n

s 1 nwhile s i X[i s] X[i-s+1] 1 2 … (n 1)

s s 1 1 2 … (n 1)S[i] s n

return S 1X[]= 1,2,3, …, n-1, n.

Algorithm spans1 runs in O(n2) time

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Computing Spans with a Stack

We push the n elements in the stackWe sacn the elements in the stack in the reverse order.

We find the last element x[j] with x[j]<=x[i] and x[j-1]>x[i] and set s[i]=i-j+1.

If no such a j exists, s[i]=i+1.

Compute the s[i] for the remaining i’s as follows:

for (i=n-2; i>=0; i--) if (s[i+1]>1 & s[i]==0)

then s[i]=s[i+1]-1

01234567

0 1 2 3 4 5 6 7

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Example:

i= 0, 1, 2, 3, 4, 5, 6, 7 X[i]= 6, 3, 4, 1, 2, 3, 5, 4S[i]= 1, 1, 2, 1, 2, 3, 4, 1.

4 5 53 32 21 1 4 4 43 3 36 6 6 6Stack Stack Stack Stack Stack

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Linear AlgorithmAlgorithm spans2(X, n) #

S new array of n integers nA new empty stack 1for i 0 to n 1 doA.push(i) n

i=n-1 ; j=n-1; 1 while (i<=0) do while (A.isEmpty()

X[A.top()] X[j] ) do n

j A.pop() n if A.isEmpty() then n

S[i] i 1 1 else

S[i] i j +1 n i=j-1; n

return S 1

Each index of the array

Is pushed into the stack exactly one

Is popped from the stack at most once

The statements in the two while loops are executed at most n times Algorithm spans2 runs in O(n) time for (i=n-2; i>=0; i--)

if (s[i+1]>1 & s[i]==0) then s[i]=s[i+1]-1

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Summary of Stack

Understand the definition of Stack (basic)

Applications1. Parentheses (moderate)2. Computing Spans (not required)

Implementation of Stack is not required. You should be able to use stack to write

programs to solve problems

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Week 3: Tutorial Arrangement

Week 3 Tutorial (Feb. 6, 2006, 6:30p.m. – 7:20p.m) will be in MMW2478 (CS lab, second floor, Lift 18, on the left hand side after you go through the tunnel.)

City Uni.

Lift 18

Tunnel connecting city U and Festival Walk

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Remarks

Emphasize the concept of ADT.More examples about ADTDelete the Span application examples. Add Queue part (perhaps to week 3)Teach slowly.