Check whether the given string is Palindrome using Stack

Namachila 3 weeks ago

0 0 5 minutes read

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

Examples:

Input: str = “zambiatek”
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).

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