Explain how to implement two stacks in one array A[1..n] in such a way that neither stack overflows unless the total number of elements in both stacks together is n. The PUSH and POP operations should run in O(1) time.

Answer:

->The purpose is to start two stacks from two extreme corners of arr[]/stack1 starts from the leftmost element, the first element in stack1 is pushed at index 0.

->The stack2 starts from the rightmost corner, the first element in stack2 is pushed at index(n-1). Both stacks grow in opposite direction.

->To check for overflow, all we need to check is for space between top elements of both stacks. This check is highlighted in the below code:

Programming Code:

import java.io.*;

import java.util.*;

class twoStacks

{

int *arr;

int size;

int tops1,top2;

public:

  twoStacks(int n) //constructor

{

size=n;

arr= new int[n];

top1=-1;

top2= size;

}

//Method to push an element x to stack1

void push(intx)

{

//There is atleast one empty space for new element

if(top1<top2-1)

{

top1++;

arr[top1]=x;

}

else

{

System.out.println("Stack Overflow");

exit(1);

}

//Method to push an element x to stack2

void push2(int x)

{

//There is at least one empty space for new element

if(top1<top2-1)

{

top2--;

arr[top2]=x;

}

else

{

System.out.println("Stack Overflow");

exit(1);

}

}

//Method to pop an element from first stack

int pop1()

{

if(top1>=0)

{

int x=arr[top1];

top1..;

return x;

}

else

{

System.out.println("Stack underFlow");

exit(1);

}

}

//Method to pop an element from second stack

int pop2()

{

if(top2<size)

{

int x=arr[top2];

top2++;

return x;

}

else

{

System.out.println("Stack underFlow");

exit(1);

}

}

}