Given string str, the task is to find whether the given string is a palindrome or not using a stack.
Examples:
Input: str = “neveropen”
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 palindromebool 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 codeint 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 functionvoid push(char ele){ stack[++top] = ele;}// pop functionchar pop(){ return stack[top--];}// Function that returns 1// if str is a palindromeint 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 codeint main(){ char str[] = "madam"; if (isPalindrome(str)) { printf("Yes"); } else { printf("No"); } return 0;} |
Java
// Java implementation of the approachclass GFG{static char []stack;static int top = -1;// push functionstatic void push(char ele){ stack[++top] = ele;}// pop functionstatic char pop(){ return stack[top--];}// Function that returns 1// if str is a palindromestatic 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 codepublic 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 approachstack = []top = -1# push functiondef push(ele: str): global top top += 1 stack[top] = ele# pop functiondef pop(): global top ele = stack[top] top -= 1 return ele# Function that returns 1# if str is a palindromedef 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 Codeif __name__ == "__main__": string = "madam" if isPalindrome(string): print("Yes") else: print("No")# This code is contributed by# sanjeev2552 |
C#
// C# implementation of the approachusing System;class GFG{static char []stack;static int top = -1;// push functionstatic void push(char ele){ stack[++top] = ele;}// pop functionstatic char pop(){ return stack[top--];}// Function that returns 1// if str is a palindromestatic 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 codepublic 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 palindromefunction 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 codevar s = "madam";if (isPalindrome(s)) { document.write( "Yes");}else { document.write( "No");}</script> |
Output:
Yes
Time Complexity: O(N).
Auxiliary Space: O(N).
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
