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# Check if given strings are rotations of each other or not

• Difficulty Level : Easy
• Last Updated : 03 Feb, 2023

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 ` `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 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

 ``

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(1)

## 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 ` `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 q1 = ``new` `LinkedList<>();` `        ``for` `(``int` `i = ``0``; i < s.length(); i++) {` `            ``q1.add(s.charAt(i));` `        ``}`   `        ``Queue 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

 ``

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 ` `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 ` `#include ` `#include `   `/* 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.`

## PHP

 ``

## Javascript

 ``

Output

`Strings are rotations of each other`

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

No extra space approach:

Instead of creating a new string as str1+str1, we use two pointers.

One pointer per string, also a count of the length of string that has been found the same since the last mismatch.

Every time there is a mismatch, we reset one of the pointer to 0 and start comparing again, once one pointer has run over any one string twice, we know we have checked for all possible rotations. Therefore we can return 0.

We keep track of a reset variable that helps us implement this, since i should only be incremented only when current character at i has been compared with current at j, when j is reset, the value of j changes to 0, now if we increment i without comparing s1[i]==s2[0] then we miss this comparison (we skip it). Therefore a reset variable keeps track of if there has been a reset given there was a mismatch, if yes then it increments i else i is only incremented when a match is found.

## C++

 `// Code to check if two strings are rotation of one another` `#include ` `using` `namespace` `std;`   `int` `isCyclicRotation(string& p, string& q)` `{` `    ``int` `i = 0, j = 0, k = 0, n = p.length();` `    ``bool` `reset = ``false``;` `    ``while` `(i < 2 * n) {` `        ``if` `(k == n)` `            ``return` `1;` `        ``if` `(p[i % n] == q[j % n]) {` `            ``i++;` `            ``j++;` `            ``k++;` `        ``}` `        ``else` `if` `(reset) {` `            ``reset = ``false``;` `            ``i++;` `        ``}` `        ``else` `{` `            ``reset = ``true``;` `            ``j = 0;` `            ``k = 0;` `        ``}` `    ``}` `    ``return` `0;` `}`   `int` `main()` `{` `    ``string str1 = ``"AACD"``, str2 = ``"ACDA"``;` `    ``if` `(isCyclicRotation(str1, str2)) {` `        ``cout << ``"YES"``;` `    ``}` `    ``else` `{` `        ``cout << ``"NO"``;` `    ``}` `    ``return` `0;` `}`

## Java

 `// Java Code to check if two strings are rotation of one another` `import` `java.util.Scanner;`   `public` `class` `GFG {` `    ``public` `static` `boolean` `isCyclicRotation(String p,` `                                           ``String q)` `    ``{` `        ``int` `i = ``0``, j = ``0``, k = ``0``, n = p.length();` `        ``boolean` `reset = ``false``;` `        ``while` `(i < ``2` `* n) {` `            ``if` `(k == n) {` `                ``return` `true``;` `            ``}` `            ``if` `(p.charAt(i % n) == q.charAt(j % n)) {` `                ``i++;` `                ``j++;` `                ``k++;` `            ``}` `            ``else` `if` `(reset) {` `                ``reset = ``false``;` `                ``i++;` `            ``}` `            ``else` `{` `                ``reset = ``true``;` `                ``j = ``0``;` `                ``k = ``0``;` `            ``}` `        ``}` `        ``return` `false``;` `    ``}`   `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``Scanner sc = ``new` `Scanner(System.in);` `        ``String str1 = ``"AACD"``, str2 = ``"ACDA"``;` `        ``if` `(isCyclicRotation(str1, str2)) {` `            ``System.out.println(``"YES"``);` `        ``}` `        ``else` `{` `            ``System.out.println(``"NO"``);` `        ``}` `    ``}` `}` `// This code is contributed by Prasad Kandekar(prasad264)`

## Python3

 `# python code to check if two strings are rotation of one another` `def` `isCyclicRotation(p, q):` `    ``i, j, k, n ``=` `0``, ``0``, ``0``, ``len``(p)` `    ``reset ``=` `False` `    ``while` `i < ``2` `*` `n:` `        ``if` `k ``=``=` `n:` `            ``return` `1` `        ``if` `p[i ``%` `n] ``=``=` `q[j ``%` `n]:` `            ``i ``+``=` `1` `            ``j ``+``=` `1` `            ``k ``+``=` `1` `        ``elif` `reset:` `            ``reset ``=` `False` `            ``i ``+``=` `1` `        ``else``:` `            ``reset ``=` `True` `            ``j ``=` `0` `            ``k ``=` `0` `    ``return` `0`     `str1 ``=` `"AACD"` `str2 ``=` `"ACDA"` `if` `isCyclicRotation(str1, str2):` `    ``print``(``"YES"``)` `else``:` `    ``print``(``"NO"``)` `# This code is contributed by Prasad Kandekar(prasad264)`

## C#

 `// C# Code to check if two strings are rotation of one another` `using` `System;`   `public` `class` `GFG {` `    ``static` `bool` `IsCyclicRotation(``string` `p, ``string` `q)` `    ``{` `        ``int` `i = 0, j = 0, k = 0, n = p.Length;` `        ``bool` `reset = ``false``;` `        ``while` `(i < 2 * n) {` `            ``if` `(k == n) {` `                ``return` `true``;` `            ``}` `            ``if` `(p[i % n] == q[j % n]) {` `                ``i++;` `                ``j++;` `                ``k++;` `            ``}` `            ``else` `if` `(reset) {` `                ``reset = ``false``;` `                ``i++;` `            ``}` `            ``else` `{` `                ``reset = ``true``;` `                ``j = 0;` `                ``k = 0;` `            ``}` `        ``}` `        ``return` `false``;` `    ``}`   `    ``static` `void` `Main(``string``[] args)` `    ``{` `        ``string` `str1 = ``"AACD"``, str2 = ``"ACDA"``;` `        ``if` `(IsCyclicRotation(str1, str2)) {` `            ``Console.WriteLine(``"YES"``);` `        ``}` `        ``else` `{` `            ``Console.WriteLine(``"NO"``);` `        ``}` `    ``}` `}` `// This code is contributed by Prasad Kandekar(prasad264)`

## Javascript

 `// Javascript Code to check if two strings are rotation of one another` `function` `isCyclicRotation(p, q) {` `    ``let i = 0, j = 0, k = 0, n = p.length;` `    ``let reset = ``false``;` `    ``while` `(i < 2 * n) {` `        ``if` `(k === n)` `            ``return` `true``;` `        ``if` `(p[i % n] === q[j % n]) {` `            ``i++;` `            ``j++;` `            ``k++;` `        ``}` `        ``else` `if` `(reset) {` `            ``reset = ``false``;` `            ``i++;` `        ``}` `        ``else` `{` `            ``reset = ``true``;` `            ``j = 0;` `            ``k = 0;` `        ``}` `    ``}` `    ``return` `false``;` `}`   `let str1 = ``"AACD"``, str2 = ``"ACDA"``;` `if` `(isCyclicRotation(str1, str2)) {` `    ``console.log(``"YES"``);` `}` `else` `{` `    ``console.log(``"NO"``);` `}` `// This code is contributed by Prasad Kandekar(prasad264)`

Output

`YES`

Time Complexity: O(2*n) where n is the length of the string, this can be approximated to O(n)
Auxiliary Space: O(1) i.e. constant extra space.

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