Given a string, write a recursive function that checks if the given string is a palindrome, else, not a palindrome.
Examples:
Input : malayalam Output : Yes Reverse of malayalam is also malayalam. Input : max Output : No Reverse of max is not max.
We have discussed an iterative function here.
The idea of a recursive function is simple:
1) If there is only one character in string return true. 2) Else compare first and last characters and recur for remaining substring.
Below is the implementation of the above idea:
C++
// A recursive C++ program to // check whether a given number// is palindrome or not#include <bits/stdc++.h>using namespace std;// A recursive function that// check a str[s..e] is// palindrome or not.bool isPalRec(char str[], int s, int e){ // If there is only one character if (s == e) return true; // If first and last // characters do not match if (str[s] != str[e]) return false; // If there are more than // two characters, check if // middle substring is also // palindrome or not. if (s < e + 1) return isPalRec(str, s + 1, e - 1); return true;}bool isPalindrome(char str[]){ int n = strlen(str); // An empty string is // considered as palindrome if (n == 0) return true; return isPalRec(str, 0, n - 1);}// Driver Codeint main(){ char str[] = "geeg"; if (isPalindrome(str)) cout << "Yes"; else cout << "No"; return 0;}// This code is contributed by shivanisinghss2110 |
C
// A recursive C program to // check whether a given number// is palindrome or not#include <stdio.h>#include <string.h>#include <stdbool.h>// A recursive function that// check a str[s..e] is// palindrome or not.bool isPalRec(char str[], int s, int e){ // If there is only one character if (s == e) return true; // If first and last // characters do not match if (str[s] != str[e]) return false; // If there are more than // two characters, check if // middle substring is also // palindrome or not. if (s < e + 1) return isPalRec(str, s + 1, e - 1); return true;}bool isPalindrome(char str[]){int n = strlen(str);// An empty string is // considered as palindromeif (n == 0) return true;return isPalRec(str, 0, n - 1);}// Driver Codeint main(){ char str[] = "geeg"; if (isPalindrome(str)) printf("Yes"); else printf("No"); return 0;} |
Java
// A recursive JAVA program to // check whether a given String // is palindrome or notimport java.io.*;class GFG{ // A recursive function that // check a str(s..e) is // palindrome or not. static boolean isPalRec(String str, int s, int e) { // If there is only one character if (s == e) return true; // If first and last // characters do not match if ((str.charAt(s)) != (str.charAt(e))) return false; // If there are more than // two characters, check if // middle substring is also // palindrome or not. if (s < e + 1) return isPalRec(str, s + 1, e - 1); return true; } static boolean isPalindrome(String str) { int n = str.length(); // An empty string is // considered as palindrome if (n == 0) return true; return isPalRec(str, 0, n - 1); } // Driver Code public static void main(String args[]) { String str = "geeg"; if (isPalindrome(str)) System.out.println("Yes"); else System.out.println("No"); }}// This code is contributed// by Nikita Tiwari |
Python
# A recursive Python program # to check whether a given # number is palindrome or not# A recursive function that # check a str[s..e] is # palindrome or not.def isPalRec(st, s, e) : # If there is only one character if (s == e): return True # If first and last # characters do not match if (st[s] != st[e]) : return False # If there are more than # two characters, check if # middle substring is also # palindrome or not. if (s < e + 1) : return isPalRec(st, s + 1, e - 1); return Truedef isPalindrome(st) : n = len(st) # An empty string is # considered as palindrome if (n == 0) : return True return isPalRec(st, 0, n - 1);# Driver Codest = "geeg"if (isPalindrome(st)) : print "Yes"else : print "No" # This code is contributed# by Nikita Tiwari. |
C#
// A recursive C# program to // check whether a given number// is palindrome or notusing System;class GFG{ // A recursive function that // check a str(s..e) // is palindrome or not. static bool isPalRec(String str, int s, int e) { // If there is only one character if (s == e) return true; // If first and last character // do not match if ((str[s]) != (str[e])) return false; // If there are more than two // characters, check if middle // substring is also // palindrome or not. if (s < e + 1) return isPalRec(str, s + 1, e - 1); return true; } static bool isPalindrome(String str) { int n = str.Length; // An empty string is considered // as palindrome if (n == 0) return true; return isPalRec(str, 0, n - 1); } // Driver Code public static void Main() { String str = "geeg"; if (isPalindrome(str)) Console.Write("Yes"); else Console.Write("No"); }}// This code is contributed by Nitin Mittal. |
PHP
<?php// A recursive php program to // check whether a given number// is palindrome or not// A recursive function that // check a str[s..e] is // palindrome or not.function isPalRec($str, $s,$e){ // If there is only one character if ($s == $e) return true; // If first and last // characters do not match if ($str[$s] != $str[$e]) return false; // If there are more than two // characters, check if middle // substring is also palindrome or not. if ($s < $e + 1) return isPalRec($str, $s + 1, $e - 1); return true;}function isPalindrome($str){$n = strlen($str);// An empty string is // considered as palindromeif ($n == 0) return true;return isPalRec($str, 0, $n - 1);}// Driver Code{ $str = "geeg"; if (isPalindrome($str)) echo("Yes"); else echo("No"); return 0;}// This code is contributed // by nitin mittal.?> |
Javascript
<script> // A recursive javascript program to // check whether a given String // is palindrome or not // A recursive function that // check a str(s..e) is // palindrome or not. function isPalRec( str , s , e) { // If there is only one character if (s == e) return true; // If first and last // characters do not match if ((str.charAt(s)) != (str.charAt(e))) return false; // If there are more than // two characters, check if // middle substring is also // palindrome or not. if (s < e + 1) return isPalRec(str, s + 1, e - 1); return true; } function isPalindrome( str) { var n = str.length; // An empty string is // considered as palindrome if (n == 0) return true; return isPalRec(str, 0, n - 1); } // Driver Code var str = "geeg"; if (isPalindrome(str)) document.write("Yes"); else document.write("No");// This code contributed by gauravrajput1 </script> |
Output
Yes
Time Complexity: O(n)
Auxiliary Space: O(n)
Another Approach :
Basically while traversing check whether ith and n-i-1th index are equal or not.
If there are not equal return false and if they are equal then continue with the recursion calls.
C++
#include <iostream>using namespace std;bool isPalindrome(string s, int i){ if(i > s.size()/2){ return true ; } return s[i] == s[s.size()-i-1] && isPalindrome(s, i+1) ; } int main(){ string str = "geeg" ; if (isPalindrome(str, 0)) cout << "Yes"; else cout << "No"; return 0;} |
Java
/*package whatever //do not write package name here */import java.io.*;class GFG { public static boolean isPalindrome(String s, int i){ if(i > s.length()/2) { return true ; } return s.charAt(i) == s.charAt(s.length()-i-1) && isPalindrome(s, i+1) ; } public static void main (String[] args) { String str = "geeg" ; if (isPalindrome(str, 0)) { System.out.println("Yes"); } else { System.out.println("No"); } }}// This code is contributed by akashish. |
Python3
def isPalindrome(s, i): if(i > len(s)/2): return True ans = False if((s[i] is s[len(s) - i - 1]) and isPalindrome(s, i + 1)): ans = True return ansstr = "geeg"if (isPalindrome(str, 0)): print("Yes")else: print("No") # This code is contributed by akashish__ |
C#
using System;public class GFG{ public static bool isPalindrome(string s, int i){ if(i > s.Length/2){ return true ; } return s[i] == s[s.Length-i-1] && isPalindrome(s, i+1) ; } public static void Main (){ // Code string str = "geeg" ; if (isPalindrome(str, 0)) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } }}// This code is contributed by akashish_. |
Javascript
<script>function isPalindrome(s,i){if(i > s.length/2) {return true;} return s[i] == s[s.length-i-1] && isPalindrome(s, i+1)} let str = "geeg";let ans = isPalindrome(str, 0);if (ans == true) { console.log("Yes");}else { console.log("No");} // This code is contributed by akashish__</script> |
Output
Yes
Time Complexity: O(n)
Auxiliary Space: O(n)
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