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 the 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.]

How does the Above Equation work?  

While pushing x into stack if x is greater than maxEle we push modified value = 2 * x-maxEle.  
So, x > val 
=> x-val>0 
=> adding x to both sides 2x – val >x (this is our updated maxEle  from next steps). 
Means the modified value is greater than maxEle.
After popped out the maxEle should be updated 
so maxEle = 2*maxEle – y (top value) 
= 2*maxEle – (2*x- previousMax) 
= 2*maxEle – 2*maxEle (maxELe = x in next steps) + previousMax   
= previousMax.

 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.

Implementation:

C++

// 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 greater 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;
}

Java

// Java program to implement a stack that supports
// getMaximum() in O(1) time and O(1) extra space.
import java.util.*;

class GFG
{

// A user defined stack that supports getMax() in
// addition to push() and pop()
static class MyStack
{
    Stack<Integer> s = new Stack<Integer>();
    int maxEle;

    // Prints maximum element of MyStack
    void getMax()
    {
        if (s.empty())
            System.out.print("Stack is empty\n");

        // variable maxEle stores the maximum element
        // in the stack.
        else
            System.out.print("Maximum Element in" +
                        "the stack is: "+maxEle + "\n");

    }

    // Prints top element of MyStack
    void peek()
    {
        if (s.empty())
        {

            System.out.print("Stack is empty ");
            return;
        }

        int t = s.peek(); // Top element.

        System.out.print("Top Most Element is: ");

        // If t > maxEle means maxEle stores
        // value of maximum Element.
        if(t > maxEle)
            System.out.print(maxEle);
        else
            System.out.print(t);
    }

    // Remove the top element from MyStack
    void pop()
    {
        if (s.empty())
        {
            System.out.print("Stack is empty\n");
            return;
        }

        System.out.print("Top Most Element Removed: ");
        int t = s.peek();
        s.pop();

        // Maximum will change as the maximum element
        // of the stack is being removed.
        if (t > maxEle)
        {
            System.out.print(maxEle + "\n");
            maxEle = 2 * maxEle - t;
        }

        else
            System.out.print(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);
            System.out.print("Number Inserted: " + x + "\n");
            return;
        }

        // If new number is Greater than maxEle
        if (x > maxEle)
        {
            s.push(2 * x - maxEle);
            maxEle = x;
        }

        else
            s.push(x);

        System.out.print("Number Inserted: " + x + "\n");
    }
};

// Driver Code
public static void main(String[] args)
{
    MyStack s = new MyStack();
    s.push(3);
    s.push(5);
    s.getMax();
    s.push(7);
    s.push(19);
    s.getMax();
    s.pop();
    s.getMax();
    s.pop();
    s.peek();
    }
}

/* This code contributed by PrinciRaj1992 */

Python 3

# Class to make a Node
class Node:
    #Constructor which assign argument to nade's value
    def __init__(self, value):
        self.value = value
        self.next = None

    # This method returns the string representation of the object.
    def __str__(self):
        return "Node({})".format(self.value)

    #  __repr__ is same as __str__
    __repr__ = __str__


class Stack:
    # Stack Constructor initialise top of stack and counter.
    def __init__(self):
        self.top = None
        self.count = 0
        self.maximum = None

    #This method returns the string representation of the object (stack).
    def __str__(self):
        temp=self.top
        out=[]
        while temp:
            out.append(str(temp.value))
            temp=temp.next
        out='\n'.join(out)
        return ('Top {} \n\nStack :\n{}'.format(self.top,out))

    #  __repr__ is same as __str__
    __repr__=__str__

    #This method is used to get minimum element of stack
    def getMax(self):
        if self.top is None:
            return "Stack is empty"
        else:
            print("Maximum Element in the stack is: {}" .format(self.maximum))



    # Method to check if Stack is Empty or not
    def isEmpty(self):
        # If top equals to None then stack is empty
        if self.top == None:
            return True
        else:
        # If top not equal to None then stack is empty
            return False

    # This method returns length of stack
    def __len__(self):
        self.count = 0
        tempNode = self.top
        while tempNode:
            tempNode = tempNode.next
            self.count+=1
        return self.count

    # This method returns top of stack
    def peek(self):
        if self.top is None:
            print ("Stack is empty")
        else:
            if self.top.value > self.maximum:
                print("Top Most Element is: {}" .format(self.maximum))
            else:
                print("Top Most Element is: {}" .format(self.top.value))

    #This method is used to add node to stack
    def push(self,value):
        if self.top is None:
            self.top = Node(value)
            self.maximum = value

        elif value > self.maximum :
            temp = (2 * value) - self.maximum
            new_node = Node(temp)
            new_node.next = self.top
            self.top = new_node
            self.maximum = value
        else:
            new_node = Node(value)
            new_node.next = self.top
            self.top = new_node
        print("Number Inserted: {}" .format(value))

    #This method is used to pop top of stack
    def pop(self):
        if self.top is None:
            print( "Stack is empty")
        else:
            removedNode = self.top.value
            self.top = self.top.next
            if removedNode > self.maximum:
                print ("Top Most Element Removed :{} " .format(self.maximum))
                self.maximum = ( ( 2 * self.maximum ) - removedNode )
            else:
                print ("Top Most Element Removed : {}" .format(removedNode))




 # Driver program to test above class
stack = Stack()

stack.push(3)
stack.push(5)
stack.getMax()
stack.push(7)
stack.push(19)
stack.getMax()
stack.pop()
stack.getMax()
stack.pop()
stack.peek()

# This code is contributed by Blinkii

C#

// C# program to implement a stack that supports
// getMaximum() in O(1) time and O(1) extra space.
using System;
using System.Collections.Generic;

class GFG
{

// A user defined stack that supports getMax() in
// addition to push() and pop()
public class MyStack
{
    public Stack<int> s = new Stack<int>();
    public int maxEle;

    // Prints maximum element of MyStack
    public void getMax()
    {
        if (s.Count == 0)
            Console.Write("Stack is empty\n");

        // variable maxEle stores the maximum element
        // in the stack.
        else
            Console.Write("Maximum Element in" +
                        "the stack is: "+maxEle + "\n");

    }

    // Prints top element of MyStack
    public void peek()
    {
        if (s.Count == 0)
        {

            Console.Write("Stack is empty ");
            return;
        }

        int t = s.Peek(); // Top element.

        Console.Write("Top Most Element is: ");

        // If t < maxEle means maxEle stores
        // value of t.
        if(t > maxEle)
            Console.Write(maxEle);
        else
            Console.Write(t);
    }

    // Remove the top element from MyStack
    public void pop()
    {
        if (s.Count == 0)
        {
            Console.Write("Stack is empty\n");
            return;
        }

        Console.Write("Top Most Element Removed: ");
        int t = s.Peek();
        s.Pop();

        // Maximum will change as the maximum element
        // of the stack is being removed.
        if (t > maxEle)
        {
            Console.Write(maxEle + "\n");
            maxEle = 2 * maxEle - t;
        }

        else
            Console.Write(t + "\n");
    }

    // Removes top element from MyStack
    public void push(int x)
    {
        // Insert new number into the stack
        if (s.Count == 0)
        {
            maxEle = x;
            s.Push(x);
            Console.Write("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);

        Console.Write("Number Inserted: " + x + "\n");
    }
};

// Driver Code
public static void Main(String[] args)
{
    MyStack s = new MyStack();
    s.push(3);
    s.push(5);
    s.getMax();
    s.push(7);
    s.push(19);
    s.getMax();
    s.pop();
    s.getMax();
    s.pop();
    s.peek();
}
}

// This code is contributed by Princi Singh

Javascript

<script>
    // Javascript program to implement a stack that supports
    // getMaximum() in O(1) time and O(1) extra space.

    let s = [];
    let maxEle;

    // Prints maximum element of MyStack
    function getMax()
    {
        if (s.length == 0)
            document.write("Stack is empty" + "</br>");

        // variable maxEle stores the maximum element
        // in the stack.
        else
            document.write("Maximum Element in " +
                        "the stack is: "+maxEle + "</br>");

    }

    // Prints top element of MyStack
    function peek()
    {
        if (s.length == 0)
        {

            document.write("Stack is empty ");
            return;
        }

        let t = s[s.length - 1]; // Top element.

        document.write("Top Most Element is: ");

        // If t < maxEle means maxEle stores
        // value of t.
        if(t > maxEle)
            document.write(maxEle);
        else
            document.write(t);
    }

    // Remove the top element from MyStack
    function pop()
    {
        if (s.length == 0)
        {
            document.write("Stack is empty" + "</br>");
            return;
        }

        document.write("Top Most Element Removed: ");
        let t = s[s.length - 1];
        s.pop();

        // Maximum will change as the maximum element
        // of the stack is being removed.
        if (t > maxEle)
        {
            document.write(maxEle + "</br>");
            maxEle = 2 * maxEle - t;
        }

        else
            document.write(t + "</br>");
    }

    // Removes top element from MyStack
    function push(x)
    {
        // Insert new number into the stack
        if (s.length == 0)
        {
            maxEle = x;
            s.push(x);
            document.write("Number Inserted: " + x + "</br>");
            return;
        }

        // If new number is less than maxEle
        if (x > maxEle)
        {
            s.push(2 * x - maxEle);
            maxEle = x;
        }

        else
            s.push(x);

        document.write("Number Inserted: " + x + "</br>");
    }

    push(3);
    push(5);
    getMax();
    push(7);
    push(19);
    getMax();
    pop();
    getMax();
    pop();
    peek();

    //  This code is contributed by rameshtravel07.
</script>

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

Complexity Analysis:

• Time Complexity: O(1) 
• Auxiliary Space: O(1)