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Delete all even elements from a stack

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  • Difficulty Level : Easy
  • Last Updated : 18 Aug, 2021
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Given a stack with n elements, the task is to remove all the elements of the stack without affecting the order of elements.
Examples: 
 

Input : s = 16 <- 15 <- 29 <- 24 <- 19 (TOP) 
Output: 19 29 15 
19 29 15 is the order of odd elements in which 
they will be popped from the given stack.
Input : s = 1 <- 2 <- 3 <- 4 <- 5 (TOP) 
Output: 5 3 1 
 

 

Approach: 
 

  1. Create a temporary stack temp and start popping the elements of the given stack s.
  2. For every popped element say val, if val % 2 == 1 then push it to temp.
  3. At the end of step 2, temp will contain all the odd elements from s but in reverse order.
  4. Now, to get the original order, pop every element from temp and push it to s.

Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <stack>
#include <iostream>
#include <stdio.h>
 
using namespace std;
 
// Utility function to print
// the contents of a stack
static void printStack(stack<int> s)
{
    while (!s.empty())
    {
        cout << s.top() << " ";
        s.pop();
    }
}
 
// Function to delete all even
// elements from the stack
static void deleteEven(stack<int> s)
{
    stack<int> temp;
    // While stack is not empty
    while (!s.empty())
    {
        int val = s.top();
        s.pop();
 
        // If value is odd then push
        // it to the temporary stack
        if (val % 2 == 1)
            temp.push(val);
    }
 
    // Transfer the contents of the temporary stack
    // to the original stack in order to get
    // the original order of the elements
    while (!temp.empty())
    {
        s.push(temp.top());
        temp.pop();
    }
 
    // Print the modified stack content
    printStack(s);
}
 
// Driver Code
int main()
{
    stack<int> s;
    s.push(16);
    s.push(15);
    s.push(29);
    s.push(24);
    s.push(19);
    deleteEven(s);
    return 0;
}
  
// This code is contributed by Vivekkumar Singh

Java




// Java implementation of the approach
import java.util.Stack;
class GFG {
 
    // Utility function to print
    // the contents of a stack
    static void printStack(Stack<Integer> stack)
    {
        while (!stack.isEmpty())
            System.out.print(stack.pop() + " ");
    }
 
    // Function to delete all even
    // elements from the stack
    static void deleteEven(Stack<Integer> stack)
    {
        Stack<Integer> temp = new Stack<>();
 
        // While stack is not empty
        while (!stack.isEmpty()) {
            int val = stack.pop();
 
            // If value is odd then push
            // it to the temporary stack
            if (val % 2 == 1)
                temp.push(val);
        }
 
        // Transfer the contents of the temporary stack
        // to the original stack in order to get
        // the original order of the elements
        while (!temp.isEmpty())
            stack.push(temp.pop());
 
        // Print the modified stack content
        printStack(stack);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        Stack<Integer> stack = new Stack<>();
        stack.push(16);
        stack.push(15);
        stack.push(29);
        stack.push(24);
        stack.push(19);
 
        deleteEven(stack);
    }
}

Python3




# Python implementation of the approach
  
# Utility function to print
# the contents of a stack
def printStack(s):
    while (len(s)!=0):
        print(s.pop(),end=" ")
 
# Function to delete all even
# elements from the stack
def deleteEven(s):
    temp = []
     
    # While stack is not empty
    while (len(s)!=0):
        val=s.pop()
         
        # If value is odd then push
        # it to the temporary stack
        if (val % 2 == 1):
            temp.append(val)
     
    # Transfer the contents of the temporary stack
    # to the original stack in order to get
    # the original order of the elements
    while (len(temp)!=0):
        s.append(temp.pop())
     
    # Print the modified stack content
    printStack(s);
 
# Driver Code
s = []
s.append(16)
s.append(15)
s.append(29)
s.append(24)
s.append(19)
deleteEven(s)
 
# This code is contributed by rag2127.

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Utility function to print
    // the contents of a stack
    static void printStack(Stack<int> stack)
    {
        while (stack.Count != 0)
            Console.Write(stack.Pop() + " ");
    }
 
    // Function to delete all even
    // elements from the stack
    static void deleteEven(Stack<int> stack)
    {
        Stack<int> temp = new Stack<int>();
 
        // While stack is not empty
        while (stack.Count != 0)
        {
            int val = stack.Pop();
 
            // If value is odd then push
            // it to the temporary stack
            if (val % 2 == 1)
                temp.Push(val);
        }
 
        // Transfer the contents of the temporary stack
        // to the original stack in order to get
        // the original order of the elements
        while (temp.Count != 0)
            stack.Push(temp.Pop());
 
        // Print the modified stack content
        printStack(stack);
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        Stack<int> stack = new Stack<int>();
        stack.Push(16);
        stack.Push(15);
        stack.Push(29);
        stack.Push(24);
        stack.Push(19);
 
        deleteEven(stack);
    }
}
 
// This code has been contributed by 29AjayKumar

Javascript




<script>
 
// JavaScript implementation of the approach
 
// Utility function to print
// the contents of a stack
function printStack(s)
{
    while (s.length!=0)
    {
        document.write( s[s.length-1] + " ");
        s.pop();
    }
}
 
// Function to delete all even
// elements from the stack
function deleteEven(s)
{
    var temp = [];
    // While stack is not empty
    while (s.length!=0)
    {
        var val = s[s.length-1];
        s.pop();
 
        // If value is odd then push
        // it to the temporary stack
        if (val % 2 == 1)
            temp.push(val);
    }
 
    // Transfer the contents of the temporary stack
    // to the original stack in order to get
    // the original order of the elements
    while (temp.length!=0)
    {
        s.push(temp[temp.length-1]);
        temp.pop();
    }
 
    // Print the modified stack content
    printStack(s);
}
 
// Driver Code
var s = [];
s.push(16);
s.push(15);
s.push(29);
s.push(24);
s.push(19);
deleteEven(s);
 
 
</script>

Output: 

19 29 15

 

Time Complexity: O(N)
Auxiliary Space: O(N) 


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