Find maximum in a stack in O(1) time and O(1) extra space

Given a stack of integers. The task is to design a special stack such that maximum element can be found in O(1) time and O(1) extra space.

Examples:

Given Stack :
2
5
1
64   --> Maximum

So Output must be 64 when getMax() is called.

Below are the different functions designed to push and pop elements from the stack.
Push(x) : Inserts x at the top of stack.

  • If stack is empty, insert x into the stack and make maxEle equal to x.
  • If stack is not empty, compare x with maxEle. Two cases arise:
    • If x is less than or equal to maxEle, simply insert x.
    • If x is greater than maxEle, insert (2*x – maxEle) into the stack and make maxEle equal to x. For example, let previous maxEle was 3. Now we want to insert 4. We update maxEle as 4 and insert 2*4 – 3 = 5 into the stack.

Pop() : Removes an element from top of stack.

  • Remove element from top. Let the removed element be y. Two cases arise:
    • If y is less than or equal to maxEle, the maximum element in the stack is still maxEle.
    • If y is greater than maxEle, the maximum element now becomes (2*maxEle – y), so update (maxEle = 2*maxEle – y). This is where we retrieve previous maximum from current maximum and its value in stack. For example, let the element to be removed be 5 and maxEle be 4. We remove 5 and update maxEle as 2*4 – 5 = 3.

Important Points:



  • Stack doesn’t hold actual value of an element if it is maximum so far.
  • Actual maximum element is always stored in maxEle

Illustration

Push(x)

  • Number to be Inserted: 3, Stack is empty, so insert 3 into stack and maxEle = 3.
  • Number to be Inserted: 5, Stack is not empty, 5> maxEle, insert (2*5-3) into stack and maxEle = 5.
  • Number to be Inserted: 2, Stack is not empty, 2< maxEle, insert 2 into stack and maxEle = 5.
  • Number to be Inserted: 1, Stack is not empty, 1< maxEle, insert 1 into stack and maxEle = 5.

Pop()

  • Initially the maximum element maxEle in the stack is 5.
  • Number removed: 1, Since 1 is less than maxEle just pop 1. maxEle=5.
  • Number removed: 2, 2<maxEle, so number removed is 2 and maxEle is still equal to 5.
  • Number removed: 7, 7> maxEle, original number is maxEle which is 5 and new maxEle = 2*5 – 7 = 3.
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// C++ program to implement a stack that supports
// getMaximum() in O(1) time and O(1) extra space.
#include <bits/stdc++.h>
using namespace std;
  
// A user defined stack that supports getMax() in
// addition to push() and pop()
struct MyStack {
    stack<int> s;
    int maxEle;
  
    // Prints maximum element of MyStack
    void getMax()
    {
        if (s.empty())
            cout << "Stack is empty\n";
  
        // variable maxEle stores the maximum element
        // in the stack.
        else
            cout << "Maximum Element in the stack is: "
                 << maxEle << "\n";
    }
  
    // Prints top element of MyStack
    void peek()
    {
        if (s.empty()) {
            cout << "Stack is empty ";
            return;
        }
  
        int t = s.top(); // Top element.
  
        cout << "Top Most Element is: ";
  
        // If t < maxEle means maxEle stores
        // value of t.
        (t > maxEle) ? cout << maxEle : cout << t;
    }
  
    // Remove the top element from MyStack
    void pop()
    {
        if (s.empty()) {
            cout << "Stack is empty\n";
            return;
        }
  
        cout << "Top Most Element Removed: ";
        int t = s.top();
        s.pop();
  
        // Maximum will change as the maximum element
        // of the stack is being removed.
        if (t > maxEle) {
            cout << maxEle << "\n";
            maxEle = 2 * maxEle - t;
        }
  
        else
            cout << t << "\n";
    }
  
    // Removes top element from MyStack
    void push(int x)
    {
        // Insert new number into the stack
        if (s.empty()) {
            maxEle = x;
            s.push(x);
            cout << "Number Inserted: " << x << "\n";
            return;
        }
  
        // If new number is less than maxEle
        if (x > maxEle) {
            s.push(2 * x - maxEle);
            maxEle = x;
        }
  
        else
            s.push(x);
  
        cout << "Number Inserted: " << x << "\n";
    }
};
  
// Driver Code
int main()
{
    MyStack s;
    s.push(3);
    s.push(5);
    s.getMax();
    s.push(7);
    s.push(19);
    s.getMax();
    s.pop();
    s.getMax();
    s.pop();
    s.peek();
  
    return 0;
}

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Output:


Number Inserted: 3
Number Inserted: 5
Maximum Element in the stack is: 5
Number Inserted: 7
Number Inserted: 19
Maximum Element in the stack is: 19
Top Most Element Removed: 19
Maximum Element in the stack is: 7
Top Most Element Removed: 7
Top Most Element is: 5






          
          
          
          

            

    

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