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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<bits/stdc++.h>
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[0] is always 0
  lps[0] = 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[0] 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[0] 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[0] is always 0
    lps[0] = 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




<script>
 
// javascript program to check if two strings are rotations
// of each other.
function isRotation(a, b)
{
    var n = a.length;
    var m = b.length;
    if (n != m)
        return false;
 
    // create lps that will hold the longest
    // prefix suffix values for pattern
    var lps = Array.from({length: n}, (_, i) => 0);
 
    // length of the previous longest prefix suffix
    var len = 0;
    var i = 1;
    lps[0] = 0; // lps[0] 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 (k = lps[n - 1]; k < m; ++k) {
        if (b.charAt(k) != a.charAt(i++))
            return false;
    }
    return true;
}
 
// Driver code
var s1 = "ABACD";
var s2 = "CDABA";
document.write(isRotation(s1, s2) ? "1" : "0");
 
// This code is contributed by shikhasingrajput.
</script>

Output: 
 

1

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

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