# 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
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 ` `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 ` `#include ` `#include ` `#include `   `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

 ``

Output:

`Yes`

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

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