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

• Difficulty Level : Medium
• Last Updated : 14 Apr, 2021

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; // 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       # 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

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

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