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

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  • Difficulty Level : Easy
  • Last Updated : 29 Jul, 2022

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.

Implementation:

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


Python3




# 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.
?>


Javascript




<script>
// javascript 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)
    {
        // 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
 
        var str1 = "AACD";
        var str2 = "ACDA";
 
        if (areRotations(str1, str2))
            document.write("Strings are rotations of each other");
        else
            document.write("Strings are not rotations of each other");
 
// This code is contributed by umadevi9616
</script>


Output

Strings are rotations of each other

Time Complexity will be O(n*n), where n is the length of the string.
Auxiliary Space: O(n)

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.

Implementation:

C++




#include <bits/stdc++.h>
using namespace std;
bool check_rotation(string s, string goal)
{
    if (s.size() != goal.size())
        return false;
    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;
}
//CDAB
//ACBD


Java




import java.util.*;
 
class GFG{
static boolean check_rotation(String s, String goal)
{
    if (s.length() != goal.length())
        ;
    Queue<Character> q1 = new LinkedList<>();
    for (int i = 0; i < s.length(); i++) {
        q1.add(s.charAt(i));
    }
    Queue<Character> q2 = new LinkedList<>();
    for (int i = 0; i < goal.length(); i++) {
        q2.add(goal.charAt(i));
    }
    int k = goal.length();
    while (k>0) {
        k--;
        char ch = q2.peek();
        q2.remove();
        q2.add(ch);
        if (q2.equals(q1))
            return true;
    }
    return false;
}
public static void main(String[] args)
{
    String s1 = "ABCD";
    String s2 = "CDAB";
    if (check_rotation(s1, s2))
        System.out.print(s2+ " is a rotated form of " +  s1
             +"\n");
    else
        System.out.print(s2+ " is not a rotated form of " +  s1
             +"\n");
    String s3 = "ACBD";
    if (check_rotation(s1, s3))
        System.out.print(s3+ " is a rotated form of " +  s1
             +"\n");
    else
        System.out.print(s3+ " is not a rotated form of " +  s1
             +"\n");
}
}
 
// This code is contributed by gauravrajput1


Python3




def check_rotation(s, goal):
 
    if (len(s) != len(goal)):
        skip
 
    q1 = []
    for i in range(len(s)):
        q1.insert(0, s[i])
 
    q2 = []
    for i in range(len(goal)):
        q2.insert(0, goal[i])
 
    k = len(goal)
    while (k > 0):
        ch = q2[0]
        q2.pop(0)
        q2.append(ch)
        if (q2 == q1):
            return True
 
        k -= 1
 
    return False
 
 
if __name__ == "__main__":
 
    s1 = "ABCD"
    s2 = "CDAB"
    if (check_rotation(s1, s2)):
        print(s2, " is a rotated form of ", s1)
 
    else:
        print(s2, " is not a rotated form of ", s1)
 
    s3 = "ACBD"
    if (check_rotation(s1, s3)):
        print(s3, " is a rotated form of ", s1)
 
    else:
        print(s3, " is not a rotated form of ", s1)
 
        # This code is contributed by ukasp.


C#




using System;
using System.Collections;
public class GFG{
   
  public static bool check_rotation(string s, string goal)
{
    if (s.Length != goal.Length)
           return false;
    Queue q1 = new Queue();
    for (int i = 0; i < s.Length; i++) {
        q1.Enqueue(s[i]);
    }
    Queue q2 = new Queue();
    for (int i = 0; i < goal.Length; i++) {
        q2.Enqueue(goal[i]);
    }
    int k = goal.Length;
    for(int i=k-1;i>=0;i--) {
        char ch = (char)(q2.Peek());
        q2.Dequeue();
        q2.Enqueue(ch);
        if (q2.Equals(q1))
           return true;
    }
    return false;
}
 
    static public void Main (){
 
      string s1 = "ABCD";
    string s2 = "CDAB";
    if (!check_rotation(s1, s2))
    {  Console.Write(s2);
     Console.Write(" is a rotated form of ");
     Console.WriteLine(s1);
    }
    else
    {
    Console.Write(s2);
     Console.Write(" is not a rotated form of ");
     Console.WriteLine(s1);
    }
    string s3 = "ACBD";
    if (check_rotation(s1, s3))
    { Console.Write(s3);
     Console.Write(" is a rotated form of ");
     Console.WriteLine(s1);
    }
    else
    {
    Console.Write(s3);
     Console.Write(" is not a rotated form of ");
     Console.WriteLine(s1);
    }
    }
}
 
// This code is contributed by akashish_.


Javascript




<script>
 
function check_rotation(s, goal){
 
    if (s.length != goal.length){
        return false;
    }
 
    let q1 = []
    for(let i=0;i<s.length;i++)
        q1.push(s[i])
 
    let q2 = []
    for(let i=0;i<goal.length;i++)
        q2.push(goal[i])
 
    let k = goal.length
    while (k--){
        let ch = q2[0]
        q2.shift()
        q2.push(ch)
        if (JSON.stringify(q2) == JSON.stringify(q1))
            return true
    }
 
    return false
}
 
 
// driver code
 
let s1 = "ABCD"
let s2 = "CDAB"
if (check_rotation(s1, s2))
    document.write(s2, " is a rotated form of ", s1,"</br>")
 
else
    document.write(s2, " is not a rotated form of ", s1,"</br>")
 
let s3 = "ACBD"
if (check_rotation(s1, s3))
    document.write(s3, " is a rotated form of ", s1,"</br>")
 
else
    document.write(s3, " is not a rotated form of ", s1,"</br>")
 
// This code is contributed by shinjanpatra.
 
</script>


Output

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

Time Complexity will be O(n1 + n2) , where n is the length of the strings.
Auxiliary Space: O(n)

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 O(n1 + n2) where n1 and n2 are lengths of strings. KMP matcher takes O(n) time to find a substring in a string of length n where length of substring is assumed to be smaller than the string. 

Method 3:

Algorithm:

  1. Find all the positions of first character of original string in the string to be checked.
  2. For every position found, consider it to be the starting index of the string to be checked.
  3. Beginning from the new starting index, compare both strings and check whether they are equal or not.
  4. (Suppose original string to be s1,string to be checked be s2,n is length of strings and j is the position of first character of s1 in s2, then for i < (length of original string) ,check if s1[i]==s2[(j+1)%n). Return false if any character mismatch is found, else return true.
  5. Repeat 3rd step for all positions found.

Implementation:

C++




#include <iostream>
#include <vector>
using namespace std;
 
bool checkString(string &s1, string &s2, int indexFound, int Size)
{
    for(int i=0;i<Size;i++){
       //check whether the character is equal or not
        if(s1[i]!=s2[(indexFound+i)%Size])return false;
      // %Size keeps (indexFound+i) in bounds, since it
      // ensures it's value is always less than Size
    }
   
     return true;
}
   
int main() {
     
  string s1="abcd";
  string s2="cdab";
   
  if(s1.length()!=s2.length()){
      cout<<"s2 is not a rotation on s1"<<endl;
  }
  else{
      // store occurrences of the first character of s1
      vector<int> indexes;
       
      int Size = s1.length();
       
      char firstChar = s1[0];
     
     
      for(int i=0;i<Size;i++)
      {
          if(s2[i]==firstChar)
          {
            indexes.push_back(i);
          }
      }
     
 
      bool isRotation=false;
 
     // check if the strings are rotation of each other for
     // every occurence of firstChar in s2 
      for(int idx: indexes)
      {
          isRotation = checkString( s1, s2, idx, Size);
 
          if(isRotation)
            break;
      }
 
      if(isRotation)cout<<"s2 is rotation of s1"<<endl;
      else cout<<"s2 is not a rotation of s1"<<endl;
     
    }
    return 0;
}


Java




/*package whatever //do not write package name here */
 
import java.io.*;
import java.util.*;
 
class GFG
{
   
    // java program to check if two strings are rotation of each other or not
    static boolean checkString(String s1, String s2, int indexFound, int Size)
{
    for(int i=0;i<Size;i++)
    {
       
       //check whether the character is equal or not
        if(s1.charAt(i) != s2.charAt((indexFound+i)%Size))return false;
       
      // %Size keeps (indexFound+i) in bounds,
      // since it ensures it's value is always less than Size
    }
   
     return true;
}
   
 
    // Driver code
    public static void main(String args[])
    {
        String s1="abcd";
        String s2="cdab";
   
        if(s1.length() != s2.length()){
            System.out.println("s2 is not a rotation on s1");
        }
        else{
     
            ArrayList<Integer>indexes = new ArrayList<Integer>(); //store occurrences of the first character of s1
       
            int Size = s1.length();
             
            char firstChar = s1.charAt(0);
     
     
            for(int i=0;i<Size;i++)
            {
                if(s2.charAt(i)==firstChar)
                {
                  indexes.add(i);
                }
            }
     
 
            boolean isRotation=false;
 
             // check if the strings are rotation of each other for every occurence of firstChar in s2 
            for(int idx: indexes)
            {
                isRotation = checkString(s1, s2, idx, Size);
 
                if(isRotation)
                    break;
            }
 
            if(isRotation)System.out.println("s2 is rotation of s1");
            else System.out.println("s2 is not a rotation of s1");
     
        }
    }
}
 
// This code is contributed by shinjanpatra


Python3




def checkString(s1, s2, indexFound, Size):
    for i in range(Size):
     
        # check whether the character is equal or not
        if(s1[i] != s2[(indexFound + i) % Size]):
            return False
         
        # %Size keeps (indexFound+i) in bounds,
        # since it ensures it's value is always less than Size
    return True
 
# driver code
s1 = "abcd"
s2 = "cdab"
 
if(len(s1) != len(s2)):
    print("s2 is not a rotation on s1")
 
else:
     
    indexes = [] #store occurrences of the first character of s1
    Size = len(s1)
    firstChar = s1[0]
    for i in range(Size):
        if(s2[i] == firstChar):
            indexes.append(i)
 
    isRotation = False
 
    # check if the strings are rotation of each other
    # for every occurence of firstChar in s2
    for idx in indexes:
 
        isRotation = checkString(s1, s2, idx, Size)
 
        if(isRotation):
            break
 
    if(isRotation):
        print("s2 is rotation of s1")
    else:
        print("s2 is not a rotation of s1")
 
# This code is contributed by shinjanpatra


C#




using System;
 
public class GFG{
 
  public static  bool checkString(string s1, string s2, int indexFound, int Size)
  {
    for(int i = 0; i < Size; i++)
    {
       
      //check whether the character is equal or not
      if(s1[i] != s2[(indexFound+i)%Size])
        return false;
       
      // %Size keeps (indexFound+i) in bounds, since it
      // ensures it's value is always less than Size
    }
 
    return true;
  }
 
  static public void Main (){
 
    string s1="abcd";
    string s2="cdab";
 
    if(s1.Length != s2.Length){
      Console.WriteLine("s2 is not a rotation on s1");
    }
    else
    {
       
      // store occurrences of the first character of s1
      int[] indexes = new int[1000];
      int j = 0;
 
      int Size = s1.Length;
 
      char firstChar = s1[0];
 
 
      for(int i = 0; i < Size; i++)
      {
        if(s2[i] == firstChar)
        {
          indexes[j] = i;
          j++;
        }
      }
 
 
      bool isRotation=false;
 
      // check if the strings are rotation of each other for
      // every occurence of firstChar in s2 
      for(int idx = 0; idx < indexes.Length; idx++)
      {
        isRotation = checkString( s1, s2, idx, Size);
 
        if(isRotation)
          break;
      }
 
      if(isRotation)
        Console.WriteLine("s2 is rotation of s1");
      else
        Console.WriteLine("s2 is not a rotation of s1");
 
    }
  }
}
 
// This code is contributed by akashish__


Javascript




<script>
 
 
function checkString(s1, s2, indexFound, Size)
{
    for(let i = 0; i < Size; i++)
    {
     
        //check whether the character is equal or not
        if(s1[i] != s2[(indexFound + i) % Size])return false;
         
        // %Size keeps (indexFound+i) in bounds, since it ensures it's value is always less than Size
    }
 
    return true;
}
 
// driver code
let s1 = "abcd";
let s2 = "cdab";
 
if(s1.length != s2.length)
{
    document.write("s2 is not a rotation on s1");
}
else
{
     
    let indexes = []; //store occurrences of the first character of s1
    let Size = s1.length;
    let firstChar = s1[0];
    for(let i = 0; i < Size; i++)
    {
        if(s2[i] == firstChar)
        {
            indexes.push(i);
        }
    }
 
    let isRotation = false;
 
    // check if the strings are rotation of each other for every occurence of firstChar in s2
    for(let idx of indexes)
    {
        isRotation = checkString(s1, s2, idx, Size);
 
        if(isRotation)
            break;
    }
 
    if(isRotation)document.write("s2 is rotation of s1")
    else document.write("s2 is not a rotation of s1")
}
 
// This code is contributed by shinjanpatra
 
</script>


Output

s2 is rotation of s1

Time Complexity will be O(n*n) in the worst case, where n is the length of the string.
Auxiliary Space: O(1)


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