# Check if strings are rotations of each other or not | Set 2

• Difficulty Level : Medium
• Last Updated : 19 Aug, 2022

Given two strings s1 and s2, check whether s2 is a rotation of s1.

Examples:

```Input : ABACD, CDABA
Output : True

Input : GEEKS, EKSGE
Output : True```

We have discussed an approach in earlier post which handles substring match as a pattern. In this post, we will be going to use KMP algorithm’s lps (longest proper prefix which is also suffix) construction, which will help in finding the longest match of the prefix of string b and suffix of string a. By which we will know the rotating point, from this point match the characters. If all the characters are matched, then it is a rotation, else not.

Below is the basic implementation of the above approach.

## C++

 `// C++ program to check if``// two strings are rotations``// of each other``#include``using` `namespace` `std;``bool` `isRotation(string a,``                ``string b)``{``  ``int` `n = a.length();``  ``int` `m = b.length();``  ``if` `(n != m)``    ``return` `false``;` `  ``// create lps[] that``  ``// will hold the longest``  ``// prefix suffix values``  ``// for pattern``  ``int` `lps[n];` `  ``// length of the previous``  ``// longest prefix suffix``  ``int` `len = 0;``  ``int` `i = 1;``  ` `  ``// lps is always 0``  ``lps = 0;` `  ``// the loop calculates``  ``// lps[i] for i = 1 to n-1``  ``while` `(i < n)``  ``{``    ``if` `(a[i] == b[len])``    ``{``      ``lps[i] = ++len;``      ``++i;``    ``}``    ``else``    ``{``      ``if` `(len == 0)``      ``{``        ``lps[i] = 0;``        ``++i;``      ``}``      ``else``      ``{``        ``len = lps[len - 1];``      ``}``    ``}``  ``}` `  ``i = 0;` `  ``// Match from that rotating``  ``// point``  ``for` `(``int` `k = lps[n - 1];``           ``k < m; ++k)``  ``{``    ``if` `(b[k] != a[i++])``      ``return` `false``;``  ``}``  ``return` `true``;``}` `// Driver code``int` `main()``{``  ``string s1 = ``"ABACD"``;``  ``string s2 = ``"CDABA"``;``  ``cout << (isRotation(s1, s2) ?``           ``"1"` `: ``"0"``);``}` `// This code is contributed by Chitranayal`

## Java

 `// Java program to check if two strings are rotations``// of each other.``import` `java.util.*;``import` `java.lang.*;``import` `java.io.*;``class` `stringMatching {``    ``public` `static` `boolean` `isRotation(String a, String b)``    ``{``        ``int` `n = a.length();``        ``int` `m = b.length();``        ``if` `(n != m)``            ``return` `false``;` `        ``// create lps[] that will hold the longest``        ``// prefix suffix values for pattern``        ``int` `lps[] = ``new` `int``[n];` `        ``// length of the previous longest prefix suffix``        ``int` `len = ``0``;``        ``int` `i = ``1``;``        ``lps[``0``] = ``0``; ``// lps is always 0` `        ``// the loop calculates lps[i] for i = 1 to n-1``        ``while` `(i < n) {``            ``if` `(a.charAt(i) == b.charAt(len)) {``                ``lps[i] = ++len;``                ``++i;``            ``}``            ``else` `{``                ``if` `(len == ``0``) {``                    ``lps[i] = ``0``;``                    ``++i;``                ``}``                ``else` `{``                    ``len = lps[len - ``1``];``                ``}``            ``}``        ``}` `        ``i = ``0``;` `        ``// match from that rotating point``        ``for` `(``int` `k = lps[n - ``1``]; k < m; ++k) {``            ``if` `(b.charAt(k) != a.charAt(i++))``                ``return` `false``;``        ``}``        ``return` `true``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String s1 = ``"ABACD"``;``        ``String s2 = ``"CDABA"``;` `        ``System.out.println(isRotation(s1, s2) ? ``"1"` `: ``"0"``);``    ``}``}`

## Python3

 `# Python program to check if``# two strings are rotations``# of each other``def` `isRotation(a: ``str``, b: ``str``) ``-``> ``bool``:``    ``n ``=` `len``(a)``    ``m ``=` `len``(b)``    ``if` `(n !``=` `m):``        ``return` `False` `    ``# create lps[] that``    ``# will hold the longest``    ``# prefix suffix values``    ``# for pattern``    ``lps ``=` `[``0` `for` `_ ``in` `range``(n)]` `    ``# length of the previous``    ``# longest prefix suffix``    ``length ``=` `0``    ``i ``=` `1` `    ``# lps is always 0``    ``lps[``0``] ``=` `0` `    ``# the loop calculates``    ``# lps[i] for i = 1 to n-1``    ``while` `(i < n):``        ``if` `(a[i] ``=``=` `b[length]):``            ``length ``+``=` `1``            ``lps[i] ``=` `length``            ``i ``+``=` `1``        ``else``:``            ``if` `(length ``=``=` `0``):``                ``lps[i] ``=` `0``                ``i ``+``=` `1``            ``else``:``                ``length ``=` `lps[length ``-` `1``]``    ``i ``=` `0` `    ``# Match from that rotating``    ``# point``    ``for` `k ``in` `range``(lps[n ``-` `1``], m):``        ``if` `(b[k] !``=` `a[i]):``            ``return` `False``        ``i ``+``=` `1``    ``return` `True` `# Driver code``if` `__name__ ``=``=` `"__main__"``:` `    ``s1 ``=` `"ABACD"``    ``s2 ``=` `"CDABA"``    ``print``(``"1"` `if` `isRotation(s1, s2) ``else` `"0"``)` `# This code is contributed by sanjeev2552`

## C#

 `// C# program to check if``// two strings are rotations``// of each other.``using` `System;` `class` `GFG``{``public` `static` `bool` `isRotation(``string` `a,``                              ``string` `b)``{``    ``int` `n = a.Length;``    ``int` `m = b.Length;``    ``if` `(n != m)``        ``return` `false``;` `    ``// create lps[] that will``    ``// hold the longest prefix``    ``// suffix values for pattern``    ``int` `[]lps = ``new` `int``[n];` `    ``// length of the previous``    ``// longest prefix suffix``    ``int` `len = 0;``    ``int` `i = 1;``    ` `    ``// lps is always 0``    ``lps = 0;` `    ``// the loop calculates``    ``// lps[i] for i = 1 to n-1``    ``while` `(i < n)``    ``{``        ``if` `(a[i] == b[len])``        ``{``            ``lps[i] = ++len;``            ``++i;``        ``}``        ``else``        ``{``            ``if` `(len == 0)``            ``{``                ``lps[i] = 0;``                ``++i;``            ``}``            ``else``            ``{``                ``len = lps[len - 1];``            ``}``        ``}``    ``}` `    ``i = 0;` `    ``// match from that``    ``// rotating point``    ``for` `(``int` `k = lps[n - 1]; k < m; ++k)``    ``{``        ``if` `(b[k] != a[i++])``            ``return` `false``;``    ``}``    ``return` `true``;``}` `// Driver code``public` `static` `void` `Main()``{``    ``string` `s1 = ``"ABACD"``;``    ``string` `s2 = ``"CDABA"``;` `    ``Console.WriteLine(isRotation(s1, s2) ?``                                     ``"1"` `: ``"0"``);``}``}` `// This code is contributed``// by anuj_67.`

## Javascript

 ``

Output

`1`

Complexity Analysis:

• Time Complexity: O(n)
• Auxiliary Space: O(n)

My Personal Notes arrow_drop_up