Check if a given string is a rotation of a palindrome

Last Updated : 07 Jul, 2022

Given a string, check if it is a rotation of a palindrome. For example your function should return true for "aab" as it is a rotation of "aba".

Examples:

Input:  str = "aaaad" Output: 1   // "aaaad" is a rotation of a palindrome "aadaa"  Input:  str = "abcd" Output: 0 // "abcd" is not a rotation of any palindrome.

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A Simple Solution is to take the input string, try every possible rotation of it and return true if a rotation is a palindrome. If no rotation is palindrome, then return false.
Following is the implementation of this approach.

Implementation:

C++

#include <iostream> #include <string> using namespace std;  // A utility function to check if a string str is palindrome bool isPalindrome(string str) {     // Start from leftmost and rightmost corners of str     int l = 0;     int h = str.length() - 1;      // Keep comparing characters while they are same     while (h > l)         if (str[l++] != str[h--])             return false;      // If we reach here, then all characters were matching     return true; }  // Function to check if a given string is a rotation of a // palindrome. bool isRotationOfPalindrome(string str) {     // If string itself is palindrome     if (isPalindrome(str))         return true;      // Now try all rotations one by one     int n = str.length();     for (int i = 0; i < n - 1; i++) {         string str1 = str.substr(i + 1, n - i - 1);         string str2 = str.substr(0, i + 1);          // Check if this rotation is palindrome         if (isPalindrome(str1.append(str2)))             return true;     }     return false; }  // Driver program to test above function int main() {     cout << isRotationOfPalindrome("aab") << endl;     cout << isRotationOfPalindrome("abcde") << endl;     cout << isRotationOfPalindrome("aaaad") << endl;     return 0; }

Java

// Java program to check if a given string // is a rotation of a palindrome import java.io.*;  class Palindrome {     // A utility function to check if a string str is palindrome     static boolean isPalindrome(String str)     {         // Start from leftmost and rightmost corners of str         int l = 0;         int h = str.length() - 1;          // Keep comparing characters while they are same         while (h > l)             if (str.charAt(l++) != str.charAt(h--))                 return false;          // If we reach here, then all characters were matching         return true;     }      // Function to check if a given string is a rotation of a     // palindrome     static boolean isRotationOfPalindrome(String str)     {         // If string itself is palindrome         if (isPalindrome(str))             return true;          // Now try all rotations one by one         int n = str.length();         for (int i = 0; i < n - 1; i++) {             String str1 = str.substring(i + 1);             String str2 = str.substring(0, i + 1);              // Check if this rotation is palindrome             if (isPalindrome(str1 + str2))                 return true;         }         return false;     }      // driver program     public static void main(String[] args)     {         System.out.println((isRotationOfPalindrome("aab")) ? 1 : 0);         System.out.println((isRotationOfPalindrome("abcde")) ? 1 : 0);         System.out.println((isRotationOfPalindrome("aaaad")) ? 1 : 0);     } }  // Contributed by Pramod Kumar

Python3

# Python program to check if a given string is a rotation # of a palindrome  # A utility function to check if a string str is palindrome def isPalindrome(string):      # Start from leftmost and rightmost corners of str     l = 0     h = len(string) - 1      # Keep comparing characters while they are same     while h > l:         l+= 1         h-= 1         if string[l-1] != string[h + 1]:             return False      # If we reach here, then all characters were matching         return True  # Function to check if a given string is a rotation of a # palindrome. def isRotationOfPalindrome(string):      # If string itself is palindrome     if isPalindrome(string):         return True      # Now try all rotations one by one     n = len(string)     for i in range(n-1):         string1 = string[i + 1:n]         string2 = string[0:i + 1]          # Check if this rotation is palindrome         string1+=(string2)         if isPalindrome(string1):             return True      return False  # Driver program print ("1" if isRotationOfPalindrome("aab") == True else "0") print ("1" if isRotationOfPalindrome("abcde") == True else "0") print ("1" if isRotationOfPalindrome("aaaad") == True else "0")  # This code is contributed by BHAVYA JAIN

// C# program to check if a given string // is a rotation of a palindrome using System;  class GFG {     // A utility function to check if     // a string str is palindrome     public static bool isPalindrome(string str)     {         // Start from leftmost and         // rightmost corners of str         int l = 0;         int h = str.Length - 1;          // Keep comparing characters         // while they are same         while (h > l) {             if (str[l++] != str[h--]) {                 return false;             }         }          // If we reach here, then all         // characters were matching         return true;     }      // Function to check if a given string     // is a rotation of a palindrome     public static bool isRotationOfPalindrome(string str)     {         // If string itself is palindrome         if (isPalindrome(str)) {             return true;         }          // Now try all rotations one by one         int n = str.Length;         for (int i = 0; i < n - 1; i++) {             string str1 = str.Substring(i + 1);             string str2 = str.Substring(0, i + 1);              // Check if this rotation is palindrome             if (isPalindrome(str1 + str2)) {                 return true;             }         }         return false;     }      // Driver Code     public static void Main(string[] args)     {         Console.WriteLine((isRotationOfPalindrome("aab")) ? 1 : 0);         Console.WriteLine((isRotationOfPalindrome("abcde")) ? 1 : 0);         Console.WriteLine((isRotationOfPalindrome("aaaad")) ? 1 : 0);     } }  // This code is contributed by Shrikant13

JavaScript

<script>  // javascript program to check if a given string // is a rotation of a palindrome      // A utility function to check if     // a string str is palindrome     function  isPalindrome( str)     {              // Start from leftmost and         // rightmost corners of str         var l = 0;         var h = str.length - 1;            // Keep comparing characters         // while they are same         while (h > l) {             if (str[l++] != str[h--]) {                 return false;             }         }            // If we reach here, then all         // characters were matching         return true;     }        // Function to check if a given string     // is a rotation of a palindrome     function isRotationOfPalindrome( str)     {         // If string itself is palindrome         if (isPalindrome(str)) {             return true;         }            // Now try all rotations one by one         var n = str.length;         for (var i = 0; i < n - 1; i++) {             var str1 = str.substring(i + 1);             var str2 = str.substring(0, i + 1);                // Check if this rotation is palindrome             if (isPalindrome(str1 + str2)) {                 return true;             }         }         return false;     }        // Driver Code         document.write((isRotationOfPalindrome("aab")) ? 1 : 0 );         document.write("<br>");         document.write((isRotationOfPalindrome("abcde")) ? 1 : 0 );         document.write("<br>");         document.write((isRotationOfPalindrome("aaaad")) ? 1 : 0);  // This code is contributed by bunnyram19. </script>

Output

1 0 1

Time Complexity: O(n²)
Auxiliary Space: O(n) for storing rotations.

Note that the above algorithm can be optimized to work in O(1) extra space as we can rotate a string in O(n) time and O(1) extra space.

An Optimized Solution can work in O(n) time. The idea here is to use Manacher's algorithm to solve the above problem.

Let the given string be 's' and length of string be 'n'.
Preprocess/Prepare the string to use Manachers Algorithm, to help find even length palindrome, for which we concatenate the given string with itself (to find if rotation will give a palindrome) and add '$' symbol at the beginning and '#' characters between letters. We end the string using '@'. So for 'aaad' input the reformed string will be - '$#a#a#a#a#d#a#a#a#a#d#@'
Now the problem reduces to finding Longest Palindromic Substring using Manacher's algorithm of length n or greater in the string.
If there is palindromic substring of length n, then return true, else return false. If we find a palindrome of greater length then we check if the size of our input is even or odd, correspondingly our palindrome length found should also be even or odd.

For eg. if our input size is 3 and while performing Manacher's Algorithm we get a palindrome size of 5 it obviously would contain a substring of size of 3 which is a palindrome but the same cannot be said for a palindrome of length of 4. Hence we check if both the size of the input and the size of palindrome found at any instance is both even or both odd.

Boundary case would be a word with same letters that would defy the above property but for that case our algorithm will find both even length and odd length palindrome one of them being a substring, hence it wont be a problem.

Below is the implementation of the above algorithm:

C++

// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;  // Function to check if we have found // a palindrome of same length as the input // which is a rotation of the input string bool checkPal(int x, int len) {     if (x == len)         return true;     else if (x > len) {         if ((x % 2 == 0 && len % 2 == 0)             || (x % 2 != 0 && len % 2 != 0))             return true;     }     return false; }  // Function to preprocess the string // for Manacher's Algorithm string reform(string s) {     string s1 = "$#";      // Adding # between the characters     for (int i = 0; i < s.size(); i++) {         s1 += s[i];         s1 += '#';     }      s1 += '@';     return s1; }  // Function to find the longest palindromic // substring using Manacher's Algorithm bool longestPal(string s, int len) {      // Current Left Position     int mirror = 0;      // Center Right Position     int R = 0;      // Center Position     int C = 0;      // LPS Length Array     int P[s.size()] = { 0 };     int x = 0;      // Get currentLeftPosition Mirror     // for currentRightPosition i     for (int i = 1; i < s.size() - 1; i++) {         mirror = 2 * C - i;          // If currentRightPosition i is         // within centerRightPosition R         if (i < R)             P[i] = min((R - i), P[mirror]);          // Attempt to expand palindrome centered         // at currentRightPosition i         while (s[i + (1 + P[i])] == s[i - (1 + P[i])]) {             P[i]++;         }          // Check for palindrome         bool ans = checkPal(P[i], len);         if (ans)             return true;          // If palindrome centered at currentRightPosition i         // expand beyond centerRightPosition R,         // adjust centerPosition C based on expanded palindrome         if (i + P[i] > R) {             C = i;             R = i + P[i];         }     }      return false; }  // Driver code int main() {     string s = "aaaad";     int len = s.size();     s += s;     s = reform(s);     cout << longestPal(s, len);      return 0; }  // This code is contributed by Vindusha Pankajakshan

Java

// Java implementation of the approach  import java.util.*;  class GFG  {  // Function to check if we have found  // a palindrome of same length as the input  // which is a rotation of the input string  static boolean checkPal(int x, int len)  {     if (x == len)     {         return true;     }      else if (x > len)      {         if ((x % 2 == 0 && len % 2 == 0) ||             (x % 2 != 0 && len % 2 != 0))          {             return true;         }     }     return false; }  // Function to preprocess the string  // for Manacher's Algorithm  static String reform(String s) {     String s1 = "$#";      // Adding # between the characters      for (int i = 0; i < s.length(); i++)      {         s1 += s.charAt(i);         s1 += '#';     }      s1 += '@';     return s1; }  // Function to find the longest palindromic  // substring using Manacher's Algorithm  static boolean longestPal(String s, int len)  {      // Current Left Position      int mirror = 0;      // Center Right Position      int R = 0;      // Center Position      int C = 0;      // LPS Length Array      int[] P = new int[s.length()];     int x = 0;      // Get currentLeftPosition Mirror      // for currentRightPosition i      for (int i = 1; i < s.length() - 1; i++)     {         mirror = 2 * C - i;          // If currentRightPosition i is          // within centerRightPosition R          if (i < R)          {             P[i] = Math.min((R - i), P[mirror]);         }          // Attempt to expand palindrome centered          // at currentRightPosition i          while (s.charAt(i + (1 + P[i])) ==                 s.charAt(i - (1 + P[i])))         {             P[i]++;         }          // Check for palindrome          boolean ans = checkPal(P[i], len);         if (ans)          {             return true;         }          // If palindrome centered at currentRightPosition i          // expand beyond centerRightPosition R,          // adjust centerPosition C based on expanded palindrome          if (i + P[i] > R)          {             C = i;             R = i + P[i];         }     }     return false; }  // Driver code  public static void main(String[] args)  {     String s = "aaaad";     int len = s.length();     s += s;     s = reform(s);     System.out.println(longestPal(s, len) ? 1 : 0); } }   // This code is contributed by PrinciRaj1992

Python3

# Python implementation of the approach   # Function to check if we have found  # a palindrome of same length as the input  # which is a rotation of the input string  def checkPal (x, Len):      if (x == Len):         return True     elif (x > Len):         if ((x % 2 == 0 and Len % 2 == 0) or (x % 2 != 0 and Len % 2 != 0)):             return True      return False  # Function to preprocess the string  # for Manacher's Algorithm  def reform (s):      s1 = "$#"      # Adding '#' between the characters     for i in range(len(s)):         s1 += s[i]         s1 += "#"      s1 += "@"     return s1  # Function to find the longest palindromic  # substring using Manacher's Algorithm  def longestPal (s, Len):      # Current Left Position      mirror = 0         # Center Right Position      R = 0        # Center Position      C = 0         # LPS Length Array      P = [0] * len(s)     x = 0         # Get currentLeftPosition Mirror      # for currentRightPosition i      for i in range(1, len(s) - 1):         mirror = 2 * C - i           # If currentRightPosition i is          # within centerRightPosition R          if (i < R):             P[i] = min((R-i), P[mirror])          # Attempt to expand palindrome centered          # at currentRightPosition i         while (s[i + (1 + P[i])] == s[i - (1 + P[i])]):             P[i] += 1          # Check for palindrome         ans = checkPal(P[i], Len)         if (ans):             return True                  # If palindrome centered at current         # RightPosition i expand beyond          # centerRightPosition R, adjust centerPosition         # C based on expanded palindrome         if (i + P[i] > R):             C = i              R = i + P[i]      return False  # Driver Code if __name__ == '__main__':          s = "aaaad"     Len = len(s)     s += s     s = reform(s)     print(longestPal(s, Len))  # This code is contributed by himanshu77

// C# implementation of the approach  using System; using System.Collections.Generic;   class GFG  {  // Function to check if we have found  // a palindrome of same length as the input  // which is a rotation of the input string  static bool checkPal(int x, int len)  {     if (x == len)     {         return true;     }      else if (x > len)      {         if ((x % 2 == 0 && len % 2 == 0) ||             (x % 2 != 0 && len % 2 != 0))          {             return true;         }     }     return false; }  // Function to preprocess the string  // for Manacher's Algorithm  static String reform(String s) {     String s1 = "$#";      // Adding # between the characters      for (int i = 0; i < s.Length; i++)      {         s1 += s[i];         s1 += '#';     }      s1 += '@';     return s1; }  // Function to find the longest palindromic  // substring using Manacher's Algorithm  static bool longestPal(String s, int len)  {      // Current Left Position      int mirror = 0;      // Center Right Position      int R = 0;      // Center Position      int C = 0;      // LPS Length Array      int[] P = new int[s.Length];     int x = 0;      // Get currentLeftPosition Mirror      // for currentRightPosition i      for (int i = 1; i < s.Length - 1; i++)     {         mirror = 2 * C - i;          // If currentRightPosition i is          // within centerRightPosition R          if (i < R)          {             P[i] = Math.Min((R - i), P[mirror]);         }          // Attempt to expand palindrome centered          // at currentRightPosition i          while (s[i + (1 + P[i])] == s[i - (1 + P[i])])         {             P[i]++;         }          // Check for palindrome          bool ans = checkPal(P[i], len);         if (ans)          {             return true;         }          // If palindrome centered at currentRightPosition i          // expand beyond centerRightPosition R,          // adjust centerPosition C based on expanded palindrome          if (i + P[i] > R)          {             C = i;             R = i + P[i];         }     }     return false; }  // Driver code  public static void Main(String[] args)  {     String s = "aaaad";     int len = s.Length;     s += s;     s = reform(s);     Console.WriteLine(longestPal(s, len) ? 1 : 0); } }   // This code is contributed by Rajput-Ji

JavaScript

<script> // javascript implementation of the approach       // Function to check if we have found     // a palindrome of same length as the input     // which is a rotation of the input string     function checkPal(x , len) {         if (x == len) {             return true;         } else if (x > len) {             if ((x % 2 == 0 && len % 2 == 0) ||              (x % 2 != 0 && len % 2 != 0))              {                 return true;             }         }         return false;     }      // Function to preprocess the string     // for Manacher's Algorithm      function reform( s) {         var s1 = "$#";          // Adding # between the characters         for (i = 0; i < s.length; i++) {             s1 += s.charAt(i);             s1 += '#';         }          s1 += '@';         return s1;     }      // Function to find the longest palindromic     // substring using Manacher's Algorithm     function longestPal( s , len) {          // Current Left Position         var mirror = 0;          // Center Right Position         var R = 0;          // Center Position         var C = 0;          // LPS Length Array         var P = Array(s.length).fill(0);         var x = 0;          // Get currentLeftPosition Mirror         // for currentRightPosition i         for (i = 1; i < s.length - 1; i++) {             mirror = 2 * C - i;              // If currentRightPosition i is             // within centerRightPosition R             if (i < R) {                 P[i] = Math.min((R - i), P[mirror]);             }              // Attempt to expand palindrome centered             // at currentRightPosition i             while (s.charAt(i + (1 + P[i])) == s.charAt(i - (1 + P[i]))) {                 P[i]++;             }              // Check for palindrome             var ans = checkPal(P[i], len);             if (ans) {                 return true;             }              // If palindrome centered at currentRightPosition i             // expand beyond centerRightPosition R,             // adjust centerPosition C based on expanded palindrome             if (i + P[i] > R) {                 C = i;                 R = i + P[i];             }         }         return false;     }      // Driver code         var s = "aaaad";         var len = s.length;         s += s;         s = reform(s);         document.write(longestPal(s, len) ? 1 : 0);  // This code is contributed by umadevi9616  </script>

Output

Time Complexity : O(n²)

Auxiliary Space: O(n)

Check if a given string is a rotation of a palindrome

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