A Program to check if strings are rotations of each other or not

Difficulty Level : Easy
Last Updated : 09 Sep, 2021

Given a string s1 and a string s2, write a snippet to say whether s2 is a rotation of s1?
(eg given s1 = ABCD and s2 = CDAB, return true, given s1 = ABCD, and s2 = ACBD , return false)

Algorithm: areRotations(str1, str2)

    1. Create a temp string and store concatenation of str1 to
       str1 in temp.
                          temp = str1.str1
    2. If str2 is a substring of temp then str1 and str2 are 
       rotations of each other.

    Example:                 
                     str1 = "ABACD"
                     str2 = "CDABA"

     temp = str1.str1 = "ABACDABACD"
     Since str2 is a substring of temp, str1 and str2 are 
     rotations of each other.

C++

// C++ program to check if two given strings
// are rotations of  each other
# include <bits/stdc++.h>
using namespace std;
 
/* Function checks if passed strings (str1
   and str2) are rotations of each other */
bool areRotations(string str1, string str2)
{
   /* Check if sizes of two strings are same */
   if (str1.length() != str2.length())
        return false;
 
   string temp = str1 + str1;
  return (temp.find(str2) != string::npos);
}
 
/* Driver program to test areRotations */
int main()
{
   string str1 = "AACD", str2 = "ACDA";
   if (areRotations(str1, str2))
     printf("Strings are rotations of each other");
   else
      printf("Strings are not rotations of each other");
   return 0;
}

C

// C program to check if two given strings are rotations of
// each other
# include <stdio.h>
# include <string.h>
# include <stdlib.h>
 
/* Function checks if passed strings (str1 and str2)
   are rotations of each other */
int areRotations(char *str1, char *str2)
{
  int size1   = strlen(str1);
  int size2   = strlen(str2);
  char *temp;
  void *ptr;
 
  /* Check if sizes of two strings are same */
  if (size1 != size2)
     return 0;
 
  /* Create a temp string with value str1.str1 */
  temp   = (char *)malloc(sizeof(char)*(size1*2 + 1));
  temp[0] = '';
  strcat(temp, str1);
  strcat(temp, str1);
 
  /* Now check if str2 is a substring of temp */
  ptr = strstr(temp, str2);
 
  free(temp); // Free dynamically allocated memory
 
  /* strstr returns NULL if the second string is NOT a
    substring of first string */
  if (ptr != NULL)
    return 1;
  else
    return 0;
}
 
/* Driver program to test areRotations */
int main()
{
    char *str1 = "AACD";
    char *str2 = "ACDA";
 
    if (areRotations(str1, str2))
       printf("Strings are rotations of each other");
    else
       printf("Strings are not rotations of each other");
 
    getchar();
    return 0;
}

Java

// Java program to check if two given strings are rotations of
// each other
 
class StringRotation
{
    /* Function checks if passed strings (str1 and str2)
       are rotations of each other */
    static boolean areRotations(String str1, String str2)
    {
        // There lengths must be same and str2 must be
        // a substring of str1 concatenated with str1. 
        return (str1.length() == str2.length()) &&
               ((str1 + str1).indexOf(str2) != -1);
    }
     
    // Driver method
    public static void main (String[] args)
    {
        String str1 = "AACD";
        String str2 = "ACDA";
 
        if (areRotations(str1, str2))
            System.out.println("Strings are rotations of each other");
        else
            System.out.printf("Strings are not rotations of each other");
    }
}
// This code is contributed by  munjal

Python

# Python program to check if strings are rotations of
# each other or not
 
# Function checks if passed strings (str1 and str2)
# are rotations of each other
def areRotations(string1, string2):
    size1 = len(string1)
    size2 = len(string2)
    temp = ''
 
    # Check if sizes of two strings are same
    if size1 != size2:
        return 0
 
    # Create a temp string with value str1.str1
    temp = string1 + string1
 
    # Now check if str2 is a substring of temp
    # string.count returns the number of occurrences of
    # the second string in temp
    if (temp.count(string2)> 0):
        return 1
    else:
        return 0
 
# Driver program to test the above function
string1 = "AACD"
string2 = "ACDA"
 
if areRotations(string1, string2):
    print "Strings are rotations of each other"
else:
    print "Strings are not rotations of each other"
 
# This code is contributed by Bhavya Jain

C#

// C# program to check if two given strings
// are rotations of each other
using System;
 
class GFG {
     
    /* Function checks if passed strings
    (str1 and str2) are rotations of
    each other */
    static bool areRotations(String str1,
                                 String str2)
    {
         
        // There lengths must be same and
        // str2 must be a substring of
        // str1 concatenated with str1.
        return (str1.Length == str2.Length )
             && ((str1 + str1).IndexOf(str2)
                                     != -1);
    }
     
    // Driver method
    public static void Main ()
    {
        String str1 = "FGABCDE";
        String str2 = "ABCDEFG";
 
        if (areRotations(str1, str2))
            Console.Write("Strings are"
            + " rotation s of each other");
        else
            Console.Write("Strings are "
           + "not rotations of each other");
    }
}
 
// This code is contributed by nitin mittal.

PHP

<?php
// Php program to check if
// two given strings are
// rotations of each other
 
/* Function checks if passed
strings (str1 and str2) are
rotations of each other */
function areRotations($str1, $str2)
{
/* Check if sizes of two
   strings are same */
if (strlen($str1) != strlen($str2))
{
        return false;
}
 
$temp = $str1 + $str1;
if ($temp.count($str2)> 0)
{
        return true;
}
else
{
    return false;
}
}
 
// Driver code
$str1 = "AACD";
$str2 = "ACDA";
if (areRotations($str1, $str2))
{
    echo "Strings are rotations ".
                  "of each other";
}
else
{
    echo "Strings are not " .
         "rotations of each other" ;
}
 
// This code is contributed
// by Shivi_Aggarwal.
?>

Output

Strings are rotations of each other

Method 2(Using STL):

Algorithm :

1. If the size of both the strings is not equal, then it can never be possible.

2. Push the original string into a queue q1.

3. Push the string to be checked inside another queue q2.

4. Keep popping q2‘s and pushing it back into it till the number of such operations are less than the size of the string.

5. If q2 becomes equal to q1 at any point during these operations, it is possible. Else not.

C++

#include <bits/stdc++.h>
using namespace std;
bool check_rotation(string s, string goal)
{
    if (s.size() != goal.size())
        ;
    queue<char> q1;
    for (int i = 0; i < s.size(); i++) {
        q1.push(s[i]);
    }
    queue<char> q2;
    for (int i = 0; i < goal.size(); i++) {
        q2.push(goal[i]);
    }
    int k = goal.size();
    while (k--) {
        char ch = q2.front();
        q2.pop();
        q2.push(ch);
        if (q2 == q1)
            return true;
    }
    return false;
}
int main()
{
    string s1 = "ABCD";
    string s2 = "CDAB";
    if (check_rotation(s1, s2))
        cout << s2 << " is a rotated form of " << s1
             << endl;
    else
        cout << s2 << " is not a rotated form of " << s1
             << endl;
    string s3 = "ACBD";
    if (check_rotation(s1, s3))
        cout << s3 << " is a rotated form of " << s1
             << endl;
    else
        cout << s3 << " is not a rotated form of " << s1
             << endl;
    return 0;
}

Output

CDAB is a rotated form of ABCD
ACBD is not a rotated form of ABCD

Library Functions Used:
strstr:
strstr finds a sub-string within a string.
Prototype: char * strstr(const char *s1, const char *s2);
See http://www.lix.polytechnique.fr/Labo/Leo.Liberti/public/computing/prog/c/C/MAN/strstr.htm for more details
strcat:
strncat concatenate two strings
Prototype: char *strcat(char *dest, const char *src);
See http://www.lix.polytechnique.fr/Labo/Leo.Liberti/public/computing/prog/c/C/MAN/strcat.htm for more details

Time Complexity: Time complexity of this problem depends on the implementation of strstr function.
If the implementation of strstr is done using KMP matcher then complexity of the above program is (-)(n1 + n2) where n1 and n2 are lengths of strings. KMP matcher takes (-)(n) time to find a substring in a string of length n where length of substring is assumed to be smaller than the string.

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