Check if given strings are rotations of each other or not

Given a string s1 and a string s2, write a function to check whether s2 is a rotation of s1. 

Examples:

Input: S1 = ABCD, S2 = CDAB
Output: Strings are rotations of each other

Input: S1 = ABCD, S2 = ACBD
Output: Strings are not rotations of each other

Naive Approach: Follow the given steps to solve the problem

• Find all the positions of the first character of the original string in the string to be checked.
• For every position found, consider it to be the starting index of the string to be checked.
• Beginning from the new starting index, compare both strings and check whether they are equal or not.
• (Suppose the original string to is s1, string to be checked to be s2,n is the length of strings and j is the position of the 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.
• Repeat 3rd step for all positions found.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
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 occurrence of firstChar in s2
        for (int idx : indexes) {
            isRotation = checkString(s1, s2, idx, Size);

            if (isRotation)
                break;
        }

        if (isRotation)
            cout << "Strings are rotations of each other"
                 << endl;
        else
            cout
                << "Strings are not rotations of each other"
                << endl;
    }
    return 0;
}

Java

// Java program for the above approach

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 occurrence of firstChar in s2
            for (int idx : indexes) {
                isRotation = checkString(s1, s2, idx, Size);

                if (isRotation)
                    break;
            }

            if (isRotation)
                System.out.println(
                    "Strings are rotations of each other");
            else
                System.out.println(
                    "Strings are not rotations of each other");
        }
    }
}

// This code is contributed by shinjanpatra

Python3

# Python3 program for the above approach


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 occurrence of firstChar in s2
    for idx in indexes:

        isRotation = checkString(s1, s2, idx, Size)

        if(isRotation):
            break

    if(isRotation):
        print("Strings are rotations of each other")
    else:
        print("Strings are not rotations of each other")

# This code is contributed by shinjanpatra

C#

// C# program for the above approach

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 occurrence 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(
                    "Strings are rotations of each other");
            else
                Console.WriteLine(
                    "Strings are not rotations of each other");
        }
    }
}

// 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 occurrence 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

Strings are rotations of each other

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

Program to check if strings are rotations of each other or not using queue:

Follow the given steps to solve the problem

• If the size of both strings is not equal, then it can never be possible.
• Push the original string into a queue q1.
• Push the string to be checked inside another queue q2.
• Keep popping q2‘s and pushing it back into it till the number of such operations is less than the size of the string.
• If q2 becomes equal to q1 at any point during these operations, it is possible. Else not.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#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;
}

// Driver code
int main()
{
    string str1 = "AACD", str2 = "ACDA";

    if (check_rotation(str1, str2))
        printf("Strings are rotations of each other");
    else
        printf("Strings are not rotations of each other");
    return 0;
}

Java

// Java program for the above approach

import java.util.*;

class GFG {
    static boolean check_rotation(String s, String goal)
    {
        if (s.length() != goal.length())
            return false;

        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;
    }

    // Driver code
    public static void main(String[] args)
    {
        String str1 = "AACD";
        String str2 = "ACDA";

        // Function call
        if (check_rotation(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 gauravrajput1

Python3

# Python3 program for the above approach


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


# Driver code
if __name__ == "__main__":

    string1 = "AACD"
    string2 = "ACDA"

    # Function call
    if check_rotation(string1, string2):
        print("Strings are rotations of each other")
    else:
        print("Strings are not rotations of each other")

        # This code is contributed by ukasp.

C#

// Include namespace system
using System;
using System.Threading;
using System.Collections.Generic;

public class GFG
{
  public static bool check_rotation(String s, String goal)
  {
    if (s.Length != goal.Length)
    {
      return false;
    }
    var q1 = new LinkedList<char>();
    for (int i = 0; i < s.Length; i++)
    {
      q1.AddLast(s[i]);
    }
    var q2 = new LinkedList<char>();
    for (int i = 0; i < goal.Length; i++)
    {
      q2.AddLast(goal[i]);
    }
    var k = goal.Length;
    while (k > 0)
    {
      k--;
      var ch = q2.First.Value;
      q2.RemoveFirst();
      q2.AddLast(ch);
      if (!q2.Equals(q1))
      {
        return true;
      }
    }
    return false;
  }

  // Driver code
  public static void Main(String[] args)
  {
    var str1 = "AACD";
    var str2 = "ACDA";

    // Function call
    if (GFG.check_rotation(str1, str2))
    {
      Console.WriteLine("Strings are rotations of each other");
    }
    else
    {
      Console.Write("Strings are not rotations of each other");
    }
  }
}

// This code is contributed by aadityaburujwale.

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

Strings are rotations of each other

Time Complexity: O(N₁ * N₂), where N₁ and N₂ are the lengths of the strings.
Auxiliary Space: O(N)

Efficient Approach: Follow the given steps to solve the problem

• Create a temp string and store concatenation of str1 to str1 in temp, i.e temp = str1.str1
• 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.

Below is the implementation of the above approach:

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 code */
int main()
{
    string str1 = "AACD", str2 = "ACDA";

    // Function call
    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 <stdlib.h>
#include <string.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 code */
int main()
{
    char* str1 = "AACD";
    char* str2 = "ACDA";

    // Function call
    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 code
    public static void main(String[] args)
    {
        String str1 = "AACD";
        String str2 = "ACDA";

        // Fuinction call
        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 code
if __name__ == "__main__":
    string1 = "AACD"
    string2 = "ACDA"

    # Function call
    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 code
    public static void Main()
    {
        String str1 = "FGABCDE";
        String str2 = "ABCDEFG";

        // Function call
        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.

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>

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(strpos($temp, $str2) != false)
{
        return true;
}
else
{
    return false;
}
}

// Driver code
$str1 = "AACD";
$str2 = "ACDA";

// Function call
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

Time Complexity: O(N), where N is the length of the string.
Auxiliary Space: O(N)

Another Approach: Comparing the prefix and suffix

• Check if sizes of two strings are not same, then return false.
• Check if ith character is equal to the first character of str2
  • Check prefix of str1 with suffix of str2
  • Check suffix of str2 with prefix of str1
• If none of the above cases satisfy then answer must be false so, return false

Below is the Implementation of the Above Approach:

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;

    for (int i = 0; i < str1.length(); i++) {
        // check if ith character is equal to the first
        // character of str2
        if (str1[i] == str2[0]) {
            // check suffix of str1 with prefix of str2
            if (str1.substr(i)
                == str2.substr(0, str1.length() - i)) {
                // check prefix of str1 with suffix of str2
                if (str1.substr(0, i)
                    == str2.substr(str1.length() - i))
                    return true;
            }
        }
    }
    // if none of the above cases satisfy then answer must
    // be false so return false
    return false;
}

/* Driver code */
int main()
{
    string str1 = "AACD", str2 = "ACDA";

    // Function call
    if (areRotations(str1, str2))
        printf("Strings are rotations of each other");
    else
        printf("Strings are not rotations of each other");
    return 0;
}
// This code is contributed by Abhishek Sharma

Java

// Java program to check if two given strings are rotations
// of each other

class StringRotation {
    /* Function checks if passed strings (s1 and s2)
       are rotations of each other */
    static boolean areRotations(String s1, String s2)
    {
        /* Check if sizes of two strings are same */
        if (s1.length() != s2.length()) {
            return false;
        }
        else {
            for (int i = 0; i < s1.length(); i++) {
                // checking character at ith index with
                // first character of s2
                if (s1.charAt(i) == s2.charAt(0)) {
                    // checking prefix of s2 with suffix of
                    // s1
                    if (s1.substring(i).equals(s2.substring(
                            0, s1.length() - i))) {
                        // checking prefix of s1 with suffix
                        // of s2
                        if (s1.substring(0, i).equals(
                                s2.substring(s1.length()
                                             - i)))
                            return true;
                    }
                }
            }
        }
        return false;
    }

    // Driver code
    public static void main(String[] args)
    {
        String str1 = "AACD";
        String str2 = "ACDA";

        // Fuinction call
        if (areRotations(str1, str2))
            System.out.println(
                "Strings are rotations of each other");
        else
            System.out.printf(
                "Strings are not rotations of each other");
    }
}

Python

# Python program to check if strings are rotations of
# each other or not

# Function checks if passed strings (s1 and s2)
# are rotations of each other


def areRotations(s1, s2):
    # check length of both strings are equal or not
    if len(s1) != len(s2):
        return False
    else:
        for i in range(len(s1)):
            if s1[i] == s2[0]:  # compare the ith charcter in s1 with first character of s2
                if s1[i:] == s2[:len(s1)-i]:  # compare prefix of s2 with suffix of s1
                    # compare prefix of s1 with suffix of s2
                    if s1[:i] == s2[len(s1)-i:]:
                        return True
    return False


# Driver code
if __name__ == "__main__":
    string1 = "AACD"
    string2 = "ACDA"
    # Function call
    if areRotations(string1, string2):
        print("Strings are rotations of each other")
    else:
        print("Strings are not rotations of each other")

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
        if (str1.Length != str2.Length)
            return false;
        else {
            for (int i = 0; i < str1.Length; i++) {
                // checking character at ith index with
                // first character of s2
                if (str1[i] == str2[0]) {
                    // checking prefix of s2 with suffix of
                    // s1
                    if (str1.Substring(i)
                        == str2.Substring(0, str1.Length
                                                 - i)) {
                        // checking prefix of s1 with suffix
                        // of s2
                        if (str1.Substring(0, i)
                            == str2.Substring(str1.Length
                                              - i))
                            return true;
                    }
                }
            }
        }
        return false;
    }

    // Driver code
    public static void Main()
    {
        String str1 = "FGABCDE";
        String str2 = "ABCDEFG";

        // Function call
        if (areRotations(str1, str2))
            Console.Write("Strings are"
                          + " rotations of each other");
        else
            Console.Write("Strings are "
                          + "not rotations of each other");
    }
}

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)
    {
        /* Check if sizes of two strings are same */
        if (str1.length !== str2.length) {
        return false;
        } else {
        for (let i = 0; i < str1.length; i++) {
            // checking character at ith index with first character of s2
            if (str1[i] === str2[0]) {
               //checking prefix of s2 with suffix of s1
                if (str1.substr(i) === str2.substr(0, str1.length - i)) {
                     // checking prefix of s1 with suffix of s2
                    if (str1.substr(0, i) === str2.substr(str1.length - i)) {
                        return true;
                    }
                }
            }
        }
    }
    return false;
    }

    // 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");

</script>

Output

Strings are rotations of each other

Time Complexity: O(N*N), where N is the length of the string and another N because we compare two string.
Auxiliary Space: O(N) , Because we are creating substring