Stacks, Queues, and Deques- Ed. 2, and 3.: Chapter 4- Ed. 4: Chapter 5

Stacks, Queues, and Deques

• Stacks- Stack ADT – interface- Stack implementation

• Queues- Queue ADT – interface- Queue implementation

• Deques- Deque ADT – interface- Deque implementation

• Sample case study application

Stacks

A stack is an interesting data structure with a lot of applications. In such a data structure, we are dealing with a collection of objects. Objects can be added to or removed from the collection, one at atime, as plates in a spring-loaded, cafeteria plate dispenser.

Stacks

Definition: A stack is a container of objects that are inserted and removed according to the last-in first-out (LIFO) principle. Two fundamental operations: push: Insert an object to the container. pop: Remove an object from the container.

push pop

The Stack Abstract Data Type

An array of integers is not a stack. However, if we enforce the LIFO principle, it becomes a stack. A stack S is an abstract data type (ADT) that supports following two fundamental methods: push(o): Insert object o at the top of the stack

Input: Object; Output: None. pop(): Remove from the stack and return the top object on

the stack; an error occurs if the stack is empty. Input: None; Output: Object

Other supporting methods:

size(): Return the number of objects in the stack. Input: None; Output: Integer

isEmpty(): Return a Boolean indicating if the stack is empty. Input: None; Output: Boolean

top(): Return the top object on the stack, without removingit; an error occurs if the stack is empty. Input: None; Output: Object

This table shows a series of stack operations and their effects.The stack is initially empty.

Operation Output S

push(5) - (5)

push(3) - (5,3)

pop() 3 (5)

push(7) - (5,7)

pop() 7 (5)

top() 5 (5)

pop() 5 ()

pop() “error” ()

isEmpty() true ()

push(9) - (9)

push(7) - (9,7)

push(3) - (9,7,3)

push(5) - (9,7,3,5)

size() 4 (9,7,3,5)

pop() 5 (9,7,3)

push(8) - (9,7,3,8)

pop() 8 (9,7,3)

pop() 3 (9,7)

5

53

5

The stack isnow empty.

A Stack Interface in Java

public interface Stack { public void push( Object element ); public Object pop() throws StackEmptyException; public int size(); public boolean isEmpty(); public Object top() throws StackEmptyException; } When we define an interface, we just indicate that a class which implements it should provide all the methods specified in it.

Here is the definition of the exception handler. public class StackEmptyException extends RuntimeException { public StackEmptyException( String err ) { super( err ); } }

A Simple Array-Based Implementation

To implement a stack with an array, we need:

1. An array of the given size 2. An index t for the top element (Normally, t is implemented

as a variable of type int. Initially, t is set to –1.)

S:

0 1 t N - 1… …

Algorithms: size(): return the number t + 1

S:

0 1 t N - 1… …

Algorithms: isEmpty():

return true if t < 0; otherwise, false.S:

0 1 t N - 1… …

t + 1 = 0 ort = -1 ort < 0

Algorithms: top():

if the stack is emptythrow a StackEmptyException

else return element S[t]

S:

0 1 t N - 1… …

push(o):if stack size is N

throw a StackFullExceptionelse

increase the index t by 1store the object to S[t]

It can be defined in the

same way forStackEmptyException.

S:

0 1 t N - 1… … t+1

push

pop():if stack is empty

throw a StackEmptyExceptionelse

save the element S[t] to a variable emake the element S[t] a null objectdecrease the index t by 1

return eS:

0 1 t N - 1… …

pop

Java code (a class implementing Interface Stack) public class ArrayStack implements Stack { public static final int CAPACITY = 1000; private int capacity; private Object S[]; private int top = -1; public ArrayStack() { this( CAPACITY ); } public ArrayStack( int cap ) { capacity = cap; S = new Object[ capacity ]; } public int size() { return ( top + 1 ); }

public boolean isEmpty() { return( top < 0 ); } public void push( Object obj ) throws StackFullException

{ if( size() == capacity ) throw new StackFullException( "Stack overflow" ); S[ ++top ] = obj; } public Object top() throws StackEmptyException { if( isEmpty() ) throw new StackEmptyException( "Stack is empty."

); return S[ top ]; }

public Object pop() throws StackEmptyException { Object elem; if( isEmpty() ) throw new StackEmptyException( "Stack is Empty." ); elem = S[ top ]; S[ top-- ] = null; return elem; } } public class StackFullException extends RuntimeException { public StackFullException( String err ) { super( err ); } }

Stacks in the Java Virtual Machine

Recall that programs written in Java must be compiled by a Java compiler. The product is a byte code, the instructions for Java virtual machines. To run the Java program, the Java virtual machine will execute the byte code instructions one by one. The Java virtual machine provides a stack called Java method stack for each running program. Data in the stack are organized into structures called frames. For each method, a frame is pushed into the stack. Within the frame are the local variables and arguments for the method.

PC=14main:

i = 5

PC=216cool:

j = 5k = 7

PC=320fool:

m = 7

Java stack

main() {int i = 5;...cool( i );...

}

cool( int j ) {int k = 7;...fool( k );...

}

fool( int m ) {...

}

14

216

320

Java program

Sample Case Study Application

We want to write a program to calculate the span of the stock’sprice on a given day. The span of the stock’s price on a given day: The maximumnumber of the consecutive days up to the current day (including thecurrent day) such that the stock price on each of those days has beenless than or equal to the price on the current day.

April 9

April 10

April 11

April 12

April 13

April 14

April 15

48.97 47.54 45.83 46.34 45.68 46.95 48.17

April 9

April 10

April 11

April 12

April 13

April 14

April 15

1 1 1 2 1 4 6

 

Exemplary data:

Price of 7-year bonds:

Spans:

Let us use some symbols to represent the problem: Stock Price:

April 9 April 10 April 11 April 12 April 13 April 14 April 15

p0 p1 p2 p3 p4 p5 p6

April 9 April 10 April 11 April 12 April 13 April 14 April 15

s0 s1 s2 s3 s4 s5 s6

The problem is how to calculate the span of a given day.

Analysis – property I:

Since price quotes start with a given day (day 0, which is April 9in this example), we have:

si i + 1

April 9

April 10

April 11

April 12

April 13

April 14

April 15

1 1 1 2 1 4 6

Spans:

si:

i + 1:

s0 = 1

0 + 1

s1 = 1

1 + 1

s2 = 1

2 + 1

s3 = 2

3 + 1

s4 = 1

4 + 1

s5 = 4

5 + 1

s6 = 6

6 + 1

Analysis – property II:

For a given day i, if the span is 1, it meanspi-1 > pi or (The price of day pi-1 is higher than that ofi = 0 day pi.)

If the span is 2, it means: pi-1 pi and pi-2 > pi .

If the span is 3, it means: pi-2 pi , pi-1 pi, and pi-3 > pi.

If the span is 4, it means: pi-3 pi , pi-2 pi, pi-1 pi, and

pi-4 > pi.

In general, if the span of pi is k, it means:pi-k-1 pi, pi-k-2 pi, … pi-1 pi, and pi-k > pi.

s0=1

s1=1

s2=1

s3=2

s4=1

s5=4

p0

s6=6

p1 p2 p3 p4 p5 p6

An Algorithm Without Using a Stack

Let us consider an algorithm without using a stack. To find out the span for a given day i: 1. Assign value 0 to a variable k2. Compare pi-k with pi. 3. If pi-k pi, increment the value in k and go to step 2.4. Step 2 and 3 are repeated until pi-k > pi or k > i. The span is the

value in the variable k.

To find out the spans for all the days, the procedure has to berepeated with i = 0, 1, …, n - 1. Therefore, the complete algorithmhas two loops: One for the days and another is to find the span fora given day. The spans are stored in an array s. for i = 0 to n -1 assign value 0 to a variable k repeat until pi-k > pi or k > i compare pi-k with pi. if pi-k pi

increment the value in k save the value of k to s[i] 

p1 p2 … … pi … … pn-1

i

k

for a certain i

s[i]

The running time of this algorithm is proportional to the squareof n, the number of days, namely, 

running time = O(n2)

(We say a quantity D is O(n2) if there exists a real constant c andan integer n0 1 such that D cn2 for every integer n n0.) Such an algorithm is called a quadratic-time algorithm.

Using a stack, however, we can cut down the running timedramatically.

An Algorithm Using a Stack

Note that the span si on a certain day i can be easily computedif we know the closest day preceding day i, such that the priceon that day is higher than the price on day i.If such a preceding day exists for a day i, let us denote it with h(i), and otherwise let us define h(i) = -1. Then, si = i – h(i).

Example: p0 p1 p2 p3 p4 p5 p6 48.97 47.54 45.83 46.34 45.68 46.95 48.17 h(0)

-1

h(1)

0

h(2)

1

h(3)

1

h(4)

3

h(5)

1

h(6)

0

si = i – h(i).

s0

1

s1

1

s2

1

s3

2

s4

1

s5

4

s6

6

The problem is how to compute h(i) efficiently?

Step 1:p0 = 48.97. h(0) = -1, s0 = 0 - h(0) = 0 – (-1) = 1

0 Day 0. It is possible that h(1) = 0.

Step 2:p1 = 47.54. Pop days with prices less than or equal to p1.At this point of time, we have only one element in the stack.It is 0 and p0 > p1. So h(1) = 0, s1 = 1 - h(1) = 1 – 0 = 1.

Day 1. It is possible that h(2) = 1.01

Step 3:p2 = 45.83. Pop days with prices less than or equal to p2.At this point of time, we have two elements in the stack.The top one is 1 and p1 > p2. So h(2) = 1, s2 = 2 - h(2) = 2 – 1 = 1.

Day 2. It is possible that h(3) = 2.12

0

Step 4:p3 = 46.34. Pop days with prices less than or equal to p3.The top one will be taken out since p3 > p2. The second one is 1 and p1 > p3. So h(3) = 1, s3 = 3 - h(3) = 3 – 1 = 2.

Day 3. It is possible that h(4) = 3.

013

Step 5:p4 = 45.68. Pop days with prices less than or equal to p4.The top one is 3 and p3 > p4. So h(4) = 3, s4 = 4 - h(3) = 4 – 3 = 1.

Day 4. It is possible that h(5) = 4.

0134

Step 6:p5 = 46.95. Pop days with prices less than or equal to p3.The top two will be taken out since p5 > p4 and p5 > p3. The third one is 1 and p1 > p5. So h(5) = 1, s5 = 5 - h(5) = 5 – 1 = 4.

Day 5. It is possible that h(6) = 5.

015

Step 7:p6 = 48.17. Pop days with prices less than or equal to p3.The top two will be taken out since p6 > p5 and p6 > p1. The third one is 0 and p0 > p6. So h(6) = 0, s6 = 6 - h(6) = 6 – 0 = 6.

Day 6. The price on day 6. The process stops.06

Java Implementation

public void computeDailyHighSpan( Quote Q[]) { int prevHigh; Stack D = new ArrayStack(); for( int i = 0; i < Q.length; i++ ) { while( !D.isEmpty() && Q[ i ].getPrice() >= (( Quote )D.top()).getPrice() ) D.pop(); if( D.isEmpty() ) prevHigh = -1; //prevHigh used to store h(i) else prevHigh = (( Quote )D.top()).getDay(); Q[ i ].setSpan( i - prevHigh ); D.push( Q[ i ]); } }

h(0)

h(i)

public class Quote { private int day, price, span; public Quote( int d, int p ) { setDay( d ); setPrice( p ); } public void setDay( int d ) { day = d; } public int getDay() { return day; } public void setPrice( int p ) { price = p; } public int getPrice() { return price; } public void setSpan( int s ) { span = s; } public int getSpan() { return span; } }

Note: What in the stack are the objects of class Quote rather than the days. Each object of class Quote has the day, stock price and the span.

public class DailyHighSpan { … …public static void main(String args[]) {

Quote Q[] = new Quote[10];System.out.println("Input:" + "\n");for (int i = 0; i < 10; i++) { Q[i] = new Quote(i,(int)(Math.random()*100)); System.out.print(Q[i].getPrice(); ); System.out.print('\t');

//System.out.print("-");}

System.out.println();try{computeDailyHighSpan(Q);}catch(EmptyStackException e) {}

}