Implement two stacks in an array

Monday 10 January 2022 10:01 AM Beirut Time · 536

Wajdi Alkayal

Create a data structure twoStacks that represents two stacks. Implementation of twoStacks should use only one array, i.e., both stacks should use the same array for storing elements. Following functions must be supported by twoStacks.
push1(int x) –> pushes x to first stack
push2(int x) –> pushes x to second stack
pop1() –> pops an element from first stack and return the popped element
pop2() –> pops an element from second stack and return the popped element
Implementation of twoStack should be space efficient.

Method 1 (Divide the space in two halves)
A simple way to implement two stacks is to divide the array in two halves and assign the half space to two stacks, i.e., use arr[0] to arr[n/2] for stack1, and arr[(n/2) + 1] to arr[n-1] for stack2 where arr[] is the array to be used to implement two stacks and size of array be n.
The problem with this method is inefficient use of array space. A stack push operation may result in stack overflow even if there is space available in arr[]. For example, say the array size is 6 and we push 3 elements to stack1 and do not push anything to second stack2. When we push 4th element to stack1, there will be overflow even if we have space for 3 more elements in array.

#include <iostream>
#include <stdlib.h>

using namespace std;

class twoStacks {
    int* arr;
    int size;
    int top1, top2;

public:
    // Constructor
    twoStacks(int n)
    {
        size = n;
        arr = new int[n];
        top1 = n / 2 + 1;
        top2 = n / 2;
    }

    // Method to push an element x to stack1
    void push1(int x)
    {
        // There is at least one empty
        // space for new element
        if (top1 > 0) {
            top1--;
            arr[top1] = x;
        }
        else {
            cout << "Stack Overflow"
                 << " By element :" << x << endl;
            return;
        }
    }

    // Method to push an element
    // x to stack2
    void push2(int x)
    {

        // There is at least one empty
        // space for new element
        if (top2 < size - 1) {
            top2++;
            arr[top2] = x;
        }
        else {
            cout << "Stack Overflow"
                 << " By element :" << x << endl;
            return;
        }
    }

    // Method to pop an element from first stack
    int pop1()
    {
        if (top1 <= size / 2) {
            int x = arr[top1];
            top1++;
            return x;
        }
        else {
            cout << "Stack UnderFlow";
            exit(1);
        }
    }

    // Method to pop an element
    // from second stack
    int pop2()
    {
        if (top2 >= size / 2 + 1) {
            int x = arr[top2];
            top2--;
            return x;
        }
        else {
            cout << "Stack UnderFlow";
            exit(1);
        }
    }
};

/* Driver program to test twStacks class */
int main()
{
    twoStacks ts(5);
    ts.push1(5);
    ts.push2(10);
    ts.push2(15);
    ts.push1(11);
    ts.push2(7);
    cout << "Popped element from stack1 is "
         << " : " << ts.pop1()
         << endl;
    ts.push2(40);
    cout << "\nPopped element from stack2 is "
         << ": " << ts.pop2()
         << endl;
    return 0;
}

Output:

Stack Overflow By element :7
Popped element from stack1 is  : 11
Stack Overflow By element :40
Popped element from stack2 is : 15

Complexity Analysis:

Time Complexity:
- Push operation : O(1)
- Pop operation : O(1)
Auxiliary Space: O(N).
Use of array to implement stack so. It is not the space-optimised method as explained above.

Method 2 (A space efficient implementation)
This method efficiently utilizes the available space. It doesn’t cause an overflow if there is space available in arr[]. The idea 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 (or shrink) 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.

C++
Java
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#include <iostream>
#include <stdlib.h>

using namespace std;

class twoStacks {
    int* arr;
    int size;
    int top1, 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 push1(int x)
    {
        // There is at least one empty space for new element
        if (top1 < top2 - 1) {
            top1++;
            arr[top1] = x;
        }
        else {
            cout << "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 {
            cout << "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 {
            cout << "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 {
            cout << "Stack UnderFlow";
            exit(1);
        }
    }
};

/* Driver program to test twStacks class */
int main()
{
    twoStacks ts(5);
    ts.push1(5);
    ts.push2(10);
    ts.push2(15);
    ts.push1(11);
    ts.push2(7);
    cout << "Popped element from stack1 is "
         << ts.pop1();
    ts.push2(40);
    cout << "\nPopped element from stack2 is "
         << ts.pop2();
    return 0;
}

Output:

Popped element from stack1 is 11
Popped element from stack2 is 40

Complexity Analysis:

Time Complexity:
- Push operation : O(1)
- Pop operation : O(1)
Auxiliary Space :O(N).
Use of array to implement stack so it is a space-optimized method.

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