Find maximum in stack in O(1) without using additional stack

Last Updated : 04 Apr, 2023

The task is to design a stack which can get the maximum value in the stack in O(1) time without using an additional stack.

Examples:

Input:
push(2)
findMax()
push(6)
findMax()
pop()
findMax()
Output:
2 inserted in stack
Maximum value in the stack: 2
6 inserted in stack
Maximum value in the stack: 6
Element popped
Maximum value in the stack: 2

Approach: Instead of pushing a single element to the stack, push a pair instead. The pair consists of the (value, localMax) where localMax is the maximum value upto that element.

When we insert a new element, if the new element is greater than the local maximum below it, we set the local maximum of a new element equal to the element itself.
Else, we set the local maximum of the new element equal to the local maximum of the element below it.
The local maximum of the top of the stack will be the overall maximum.
Now if we want to know the maximum at any given point, we ask the top of the stack for local maximum associated with it which can be done in O(1).

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;
 
struct Block {
 
    // A block has two elements
    // as components
    int value, localMax;
};
 
class Stack {
 
private:
    // Pointer of type block,
    // Will be useful later as the
    // size can be dynamically allocated
    struct Block* S;
    int size, top;
 
public:
    Stack(int);
    void push(int);
    void pop();
    void max();
};
 
Stack::Stack(int n)
{
 
    // Setting size of stack and
    // initial value of top
    size = n;
    S = new Block[n];
    top = -1;
}
 
// Function to push an element to the stack
void Stack::push(int n)
{
 
    // Doesn't allow pushing elements
    // if stack is full
    if (top == size - 1) {
        cout << "Stack is full" << endl;
    }
    else {
        top++;
 
        // If the inserted element is the first element
        // then it is the maximum element, since no other
        // elements is in the stack, so the localMax
        // of the first element is the element itself
        if (top == 0) {
            S[top].value = n;
            S[top].localMax = n;
        }
        else {
 
            // If the newly pushed element is
            // less than the localMax of element below it,
            // Then the over all maximum doesn't change
            // and hence, the localMax of the newly inserted
            // element is same as element below it
            if (S[top - 1].localMax > n) {
                S[top].value = n;
                S[top].localMax = S[top - 1].localMax;
            }
 
            // Newly inserted element is greater than the localMax
            // below it, hence the localMax of new element
            // is the element itself
            else {
                S[top].value = n;
                S[top].localMax = n;
            }
        }
 
        cout << n << " inserted in stack" << endl;
    }
}
 
// Function to remove an element
// from the top of the stack
void Stack::pop()
{
 
    // If stack is empty
    if (top == -1) {
        cout << "Stack is empty" << endl;
    }
 
    // Remove the element if the stack
    // is not empty
    else {
        top--;
        cout << "Element popped" << endl;
    }
}
 
// Function to find the maximum
// element from the stack
void Stack::max()
{
 
    // If stack is empty
    if (top == -1) {
        cout << "Stack is empty" << endl;
    }
    else {
 
        // The overall maximum is the local maximum
        // of the top element
        cout << "Maximum value in the stack: "
             << S[top].localMax << endl;
    }
}
 
// Driver code
int main()
{
 
    // Create stack of size 5
    Stack S1(5);
 
    S1.push(2);
    S1.max();
    S1.push(6);
    S1.max();
    S1.pop();
    S1.max();
 
    return 0;
}

Java

// Java implementation of the approach 
class GFG
{
     
static class Block 
{ 
 
    // A block has two elements 
    // as components 
    int value, localMax; 
}; 
 
static class Stack
{ 
 
    // Pointer of type block, 
    // Will be useful later as the 
    // size can be dynamically allocated 
    Block S[]; 
    int size, top; 
 
 
Stack(int n) 
{ 
 
    // Setting size of stack and 
    // initial value of top 
    size = n; 
    S = new Block[n]; 
    for(int i=0;i<n;i++)S[i]=new Block();
    top = -1; 
} 
 
// Function to push an element to the stack 
void push(int n) 
{ 
 
    // Doesn't allow pushing elements 
    // if stack is full 
    if (top == size - 1) 
    { 
        System.out.print( "Stack is full" ); 
    } 
    else
    { 
        top++; 
 
        // If the inserted element is the first element 
        // then it is the maximum element, since no other 
        // elements is in the stack, so the localMax 
        // of the first element is the element itself 
        if (top == 0) 
        { 
            S[top].value = n; 
            S[top].localMax = n; 
        } 
        else
        { 
 
            // If the newly pushed element is 
            // less than the localMax of element below it, 
            // Then the over all maximum doesn't change 
            // and hence, the localMax of the newly inserted 
            // element is same as element below it 
            if (S[top - 1].localMax > n) 
            { 
                S[top].value = n; 
                S[top].localMax = S[top - 1].localMax; 
            } 
 
            // Newly inserted element is greater than the localMax 
            // below it, hence the localMax of new element 
            // is the element itself 
            else
            { 
                S[top].value = n; 
                S[top].localMax = n; 
            } 
        } 
 
        System.out.println( n + " inserted in stack" ); 
    } 
} 
 
// Function to remove an element 
// from the top of the stack 
void pop() 
{ 
 
    // If stack is empty 
    if (top == -1) 
    { 
        System.out.println( "Stack is empty"); 
    } 
 
    // Remove the element if the stack 
    // is not empty 
    else
    { 
        top--; 
        System.out.println( "Element popped" ); 
    } 
} 
 
// Function to find the maximum 
// element from the stack 
void max() 
{ 
 
    // If stack is empty 
    if (top == -1) 
    { 
        System.out.println( "Stack is empty"); 
    } 
    else
    { 
 
        // The overall maximum is the local maximum 
        // of the top element 
        System.out.println( "Maximum value in the stack: "+ 
                                            S[top].localMax); 
    } 
} 
}
 
// Driver code 
public static void main(String args[])
{ 
 
    // Create stack of size 5 
    Stack S1=new Stack(5); 
 
    S1.push(2); 
    S1.max(); 
    S1.push(6); 
    S1.max(); 
    S1.pop(); 
    S1.max(); 
}
}
 
// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach
class Block:
     
    # A block has two elements
    # as components  (i.e. value and localMax)
    def __init__(self, value, localMax):
        self.value = value
        self.localMax = localMax
 
class Stack:
    def __init__(self, size):
         
        # Setting size of stack and
        # initial value of top
        self.stack = [None] * size
        self.size = size
        self.top = -1
 
    # Function to push an element
    # to the stack
    def push(self, value):
 
        # Don't allow pushing elements
        # if stack is full
        if self.top == self.size - 1:
            print("Stack is full")
        else:
            self.top += 1
             
            # If the inserted element is the first element
            # then it is the maximum element, since no other
            # elements is in the stack, so the localMax
            # of the first element is the element itself
            if self.top == 0:
                self.stack[self.top] = Block(value, value)
 
            else:
                 
                # If the newly pushed element is less 
                # than the localMax of element below it,
                # Then the over all maximum doesn't change
                # and hence, the localMax of the newly inserted
                # element is same as element below it
                if self.stack[self.top - 1].localMax > value:
                    self.stack[self.top] = Block(
                        value, self.stack[self.top - 1].localMax)
 
                # Newly inserted element is greater than 
                # the localMax below it, hence the localMax
                # of new element is the element itself
                else:
                    self.stack
                    self.stack[self.top] = Block(value, value)
                     
            print(value, "inserted in the stack")
             
    # Function to remove an element  
    # from the top of the stack          
    def pop(self):
         
        # If stack is empty
        if self.top == -1:
            print("Stack is empty")
 
        # Remove the element if the stack
        # is not empty
        else:
            self.top -= 1
            print("Element popped")
             
    # Function to find the maximum  
    # element from the stack  
    def max(self):
         
        # If stack is empty
        if self.top == -1:
            print("Stack is empty")
        else:
             
            # The overall maximum is the local maximum
            # of the top element
            print("Maximum value in the stack:", 
                  self.stack[self.top].localMax)
 
# Driver code
 
# Create stack of size 5
stack = Stack(5)
stack.push(2)
stack.max()
stack.push(6)
stack.max()
stack.pop()
stack.max()
 
# This code is contributed by girishthatte

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
public class Block 
{ 
 
    // A block has two elements 
    // as components 
    public int value, localMax; 
}; 
 
public class Stack
{ 
 
    // Pointer of type block, 
    // Will be useful later as the 
    // size can be dynamically allocated 
    public Block []S; 
    public int size, top; 
 
 
public Stack(int n) 
{ 
 
    // Setting size of stack and 
    // initial value of top 
    size = n; 
    S = new Block[n]; 
    for(int i = 0; i < n; i++)S[i] = new Block();
    top = -1; 
} 
 
// Function to push an element to the stack 
public void push(int n) 
{ 
 
    // Doesn't allow pushing elements 
    // if stack is full 
    if (top == size - 1) 
    { 
        Console.Write( "Stack is full" ); 
    } 
    else
    { 
        top++; 
 
        // If the inserted element is the first element 
        // then it is the maximum element, since no other 
        // elements is in the stack, so the localMax 
        // of the first element is the element itself 
        if (top == 0) 
        { 
            S[top].value = n; 
            S[top].localMax = n; 
        } 
        else
        { 
 
            // If the newly pushed element is 
            // less than the localMax of element below it, 
            // Then the over all maximum doesn't change 
            // and hence, the localMax of the newly inserted 
            // element is same as element below it 
            if (S[top - 1].localMax > n) 
            { 
                S[top].value = n; 
                S[top].localMax = S[top - 1].localMax; 
            } 
 
            // Newly inserted element is greater than the localMax 
            // below it, hence the localMax of new element 
            // is the element itself 
            else
            { 
                S[top].value = n; 
                S[top].localMax = n; 
            } 
        } 
 
        Console.WriteLine( n + " inserted in stack" ); 
    } 
} 
 
// Function to remove an element 
// from the top of the stack 
public void pop() 
{ 
 
    // If stack is empty 
    if (top == -1) 
    { 
        Console.WriteLine( "Stack is empty"); 
    } 
 
    // Remove the element if the stack 
    // is not empty 
    else
    { 
        top--; 
        Console.WriteLine( "Element popped" ); 
    } 
} 
 
// Function to find the maximum 
// element from the stack 
public void max() 
{ 
 
    // If stack is empty 
    if (top == -1) 
    { 
        Console.WriteLine( "Stack is empty"); 
    } 
    else
    { 
 
        // The overall maximum is the local maximum 
        // of the top element 
        Console.WriteLine( "Maximum value in the stack: "+ 
                                            S[top].localMax); 
    } 
} 
}
 
// Driver code 
public static void Main(String []args)
{ 
 
    // Create stack of size 5 
    Stack S1 = new Stack(5); 
 
    S1.push(2); 
    S1.max(); 
    S1.push(6); 
    S1.max(); 
    S1.pop(); 
    S1.max(); 
}
}
 
// This code contributed by Rajput-Ji

Javascript

<script>
 
/// JavaScript implementation of the approach
class Block{
 
    /// A block has two elements
    /// as components (i.e. value and localMax)
    constructor(value, localMax){
        this.value = value
        this.localMax = localMax
    }
}
 
class Stack{
    constructor(size){
         
        /// Setting size of stack and
        /// initial value of top
        this.stack = new Array(size).fill(null)
        this.size = size
        this.top = -1
    }
 
    // Function to push an element
    // to the stack
    push(value){
 
        // Don't allow pushing elements
        // if stack is full
        if(this.top == this.size - 1)
            document.write("Stack is full","</br>")
        else{
            this.top += 1
             
            // If the inserted element is the first element
            // then it is the maximum element, since no other
            // elements is in the stack, so the localMax
            // of the first element is the element itthis
            if(this.top == 0)
                this.stack[this.top] = new Block(value, value)
 
            else{
                 
                // If the newly pushed element is less
                // than the localMax of element below it,
                // Then the over all maximum doesn't change
                // and hence, the localMax of the newly inserted
                // element is same as element below it
                if(this.stack[this.top - 1].localMax > value){
                    this.stack[this.top] = new Block(
                        value, this.stack[this.top - 1].localMax)
                    }
 
                // Newly inserted element is greater than
                // the localMax below it, hence the localMax
                // of new element is the element itthis
                else{
                    this.stack[this.top] = new Block(value, value)
                }
            }
                     
            document.write(value, "inserted in the stack","</br>")
        }
    }
             
    // Function to remove an element
    // from the top of the stack        
    pop(){
         
        // If stack is empty
        if(this.top == -1)
            document.write("Stack is empty","</br>")
 
        // Remove the element if the stack
        // is not empty
        else{
            this.top -= 1
            document.write("Element popped","</br>")
        }
    }
             
    // Function to find the maximum
    // element from the stack
    max(){
         
        // If stack is empty
        if(this.top == -1)
            document.write("Stack is empty","</br>")
        else{
             
            // The overall maximum is the local maximum
            // of the top element
            document.write("Maximum value in the stack:",this.stack[this.top].localMax,"</br>")
        }
    }
}
 
// Driver code
 
// Create stack of size 5
let stack = new Stack(5)
stack.push(2)
stack.max()
stack.push(6)
stack.max()
stack.pop()
stack.max()
 
/// This code is contributed by shinjanpatra
 
</script>

Output:

2 inserted in stack
Maximum value in the stack: 2
6 inserted in stack
Maximum value in the stack: 6
Element popped
Maximum value in the stack: 2

Time Complexity : O(1)

Auxiliary Space: O(N)

Suggest improvement

How to efficiently implement k stacks in a single array?

Count the number of operations required to reduce the given number

Share your thoughts in the comments

Find maximum in stack in O(1) without using additional stack

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