Check whether the given string is Palindrome using Stack

Given string str, the task is to find whether the given string is a palindrome or not using a stack.

Examples:

Input: str = “geeksforgeeks” 
Output: No
Input: str = “madam” 
Output: Yes  

Approach:

• Find the length of the string say len. Now, find the mid as mid = len / 2.
• Push all the elements till mid into the stack i.e. str[0…mid-1].
• If the length of the string is odd then neglect the middle character.
• Till the end of the string, keep popping elements from the stack and compare them with the current character i.e. string[i].
• If there is a mismatch then the string is not a palindrome. If all the elements match then the string is a palindrome.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function that returns true
// if string is a palindrome
bool isPalindrome(string s)
{
    int length = s.size();

    // Creating a Stack
    stack<char> st;

    // Finding the mid
    int i, mid = length / 2;

    for (i = 0; i < mid; i++) {
        st.push(s[i]);
    }

    // Checking if the length of the string
    // is odd, if odd then neglect the
    // middle character
    if (length % 2 != 0) {
        i++;
    }

    char ele;
    // While not the end of the string
    while (s[i] != '\0')
    {
         ele = st.top();
         st.pop();

    // If the characters differ then the
    // given string is not a palindrome
    if (ele != s[i])
        return false;
        i++;
    }

return true;
}

// Driver code
int main()
{
    string s = "madam";

    if (isPalindrome(s)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }

    return 0;
}

// This Code is Contributed by Harshit Srivastava

C

// C implementation of the approach
#include <malloc.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* stack;
int top = -1;

// push function
void push(char ele)
{
    stack[++top] = ele;
}

// pop function
char pop()
{
    return stack[top--];
}

// Function that returns 1
// if str is a palindrome
int isPalindrome(char str[])
{
    int length = strlen(str);

    // Allocating the memory for the stack
    stack = (char*)malloc(length * sizeof(char));

    // Finding the mid
    int i, mid = length / 2;

    for (i = 0; i < mid; i++) {
        push(str[i]);
    }

    // Checking if the length of the string
    // is odd, if odd then neglect the
    // middle character
    if (length % 2 != 0) {
        i++;
    }

    // While not the end of the string
    while (str[i] != '\0') {
        char ele = pop();

        // If the characters differ then the
        // given string is not a palindrome
        if (ele != str[i])
            return 0;
        i++;
    }

    return 1;
}

// Driver code
int main()
{
    char str[] = "madam";

    if (isPalindrome(str)) {
        printf("Yes");
    }
    else {
        printf("No");
    }

    return 0;
}

Java

// Java implementation of the approach
class GFG
{
static char []stack;
static int top = -1;

// push function
static void push(char ele)
{
    stack[++top] = ele;
}

// pop function
static char pop()
{
    return stack[top--];
}

// Function that returns 1
// if str is a palindrome
static int isPalindrome(char str[])
{
    int length = str.length;

    // Allocating the memory for the stack
    stack = new char[length * 4];

    // Finding the mid
    int i, mid = length / 2;

    for (i = 0; i < mid; i++)
    {
        push(str[i]);
    }

    // Checking if the length of the String
    // is odd, if odd then neglect the
    // middle character
    if (length % 2 != 0)
    {
        i++;
    }

    // While not the end of the String
    while (i < length)
    {
        char ele = pop();

        // If the characters differ then the
        // given String is not a palindrome
        if (ele != str[i])
            return 0;
        i++;
    }

    return 1;
}

// Driver code
public static void main(String[] args)
{
    char str[] = "madam".toCharArray();

    if (isPalindrome(str) == 1)
    {
        System.out.printf("Yes");
    }
    else
    {
        System.out.printf("No");
    }
}
}

// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
stack = []
top = -1

# push function
def push(ele: str):
    global top
    top += 1
    stack[top] = ele

# pop function
def pop():
    global top
    ele = stack[top]
    top -= 1
    return ele

# Function that returns 1
# if str is a palindrome
def isPalindrome(string: str) -> bool:
    global stack
    length = len(string)

    # Allocating the memory for the stack
    stack = ['0'] * (length + 1)

    # Finding the mid
    mid = length // 2
    i = 0
    while i < mid:
        push(string[i])
        i += 1

    # Checking if the length of the string
    # is odd, if odd then neglect the
    # middle character
    if length % 2 != 0:
        i += 1

    # While not the end of the string
    while i < length:
        ele = pop()

        # If the characters differ then the
        # given string is not a palindrome
        if ele != string[i]:
            return False
        i += 1
    return True

# Driver Code
if __name__ == "__main__":

    string = "madam"

    if isPalindrome(string):
        print("Yes")
    else:
        print("No")

# This code is contributed by
# sanjeev2552

C#

// C# implementation of the approach
using System;

class GFG
{

static char []stack;
static int top = -1;

// push function
static void push(char ele)
{
    stack[++top] = ele;
}

// pop function
static char pop()
{
    return stack[top--];
}

// Function that returns 1
// if str is a palindrome
static int isPalindrome(char []str)
{
    int length = str.Length;

    // Allocating the memory for the stack
    stack = new char[length * 4];

    // Finding the mid
    int i, mid = length / 2;

    for (i = 0; i < mid; i++)
    {
        push(str[i]);
    }

    // Checking if the length of the String
    // is odd, if odd then neglect the
    // middle character
    if (length % 2 != 0)
    {
        i++;
    }

    // While not the end of the String
    while (i < length)
    {
        char ele = pop();

        // If the characters differ then the
        // given String is not a palindrome
        if (ele != str[i])
            return 0;
        i++;
    }
    return 1;
}

// Driver code
public static void Main(String[] args)
{
    char []str = "madam".ToCharArray();

    if (isPalindrome(str) == 1)
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>
// Javascript implementation of the approach

// Function that returns true
// if string is a palindrome
function isPalindrome(s)
{
    var length = s.length;

    // Creating a Stack
    var st = [];

    // Finding the mid
    var i, mid = parseInt(length / 2);

    for (i = 0; i < mid; i++) {
        st.push(s[i]);
    }

    // Checking if the length of the string
    // is odd, if odd then neglect the
    // middle character
    if (length % 2 != 0) {
        i++;
    }

    var ele;
    // While not the end of the string
    while (i != s.length)
    {
         ele = st[st.length-1];
         st.pop();

    // If the characters differ then the
    // given string is not a palindrome
    if (ele != s[i])
        return false;
        i++;
    }

return true;
}

// Driver code
var s = "madam";
if (isPalindrome(s)) {
    document.write( "Yes");
}
else {
    document.write( "No");
}

</script>

Output:

Yes

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