Check if a given string is a rotation of a palindrome (original) (raw)

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.

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 #include 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#

// 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

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#

// 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

Time Complexity : O(n2)

Auxiliary Space: O(n)

Check if a given string is a rotation of a palindrome (original) (raw)

We strongly recommend that you click here and practice it, before moving on to the solution.

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

Function to check if a given string is a rotation of a

palindrome.

Driver program

This code is contributed by BHAVYA JAIN

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

Function to preprocess the string

for Manacher's Algorithm

Function to find the longest palindromic

substring using Manacher's Algorithm

Driver Code

This code is contributed by himanshu77