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Check if a string can be obtained by rotating another string 2 places

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  • Difficulty Level : Easy
  • Last Updated : 24 Aug, 2022
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Given two strings, the task is to find if a string can be obtained by rotating another string two places. 

Examples: 

Input: string1 = “amazon”, string2 = “azonam” 
Output: Yes 
Explanation: Rotating string1 by 2 places in anti-clockwise gives the string2.

Input: string1 = “amazon”, string2 = “onamaz” 
Output: Yes 
Explanation: Rotating string1 by 2 places in clockwise gives the string2.

Asked in: Amazon Interview

Approach 1 (by Rotating String Clockwise and Anti-clockwise):

The idea is to Rotate the String1 in both clockwise and ant-clockwise directions. Then if this rotated string is equal to String2.

Illustration:

str1 = “amazon”
str2 = “azonam”

Initialise: clock_rot = “”, anticlock_rot = “”

str1 after 2 places clockwise rotation:
clock_rot = “onamaz”

str1 after 2 places anti-clockwise rotation:
anticlock_rot = “azonam”

Therefore, anticlock_rot and str2 are same.

Hence, str2 can be achieved from str1

Follow the steps below to solve the problem:

  • Initialize two empty strings which keep the clockwise and anticlockwise strings respectively.
  • After rotating the str1 compare both clockwise and anticlockwise strings with str2.
  • If any of them matches the str2 return true, otherwise false.

Below is the implementation of the above approach.

C++




// C++ program to check if a string is two time
// rotation of another string.
#include<bits/stdc++.h>
using namespace std;
 
// Function to check if string2 is obtained by
// string 1
bool isRotated(string str1, string str2)
{
    if (str1.length() != str2.length())
        return false;
    if(str1.length()<2){
      return str1.compare(str2) == 0;
    }
    string clock_rot = "";
    string anticlock_rot = "";
    int len = str2.length();
 
    // Initialize string as anti-clockwise rotation
    anticlock_rot = anticlock_rot +
                    str2.substr(len-2, 2) +
                    str2.substr(0, len-2) ;
 
    // Initialize string as clock wise rotation
    clock_rot = clock_rot +
                str2.substr(2) +
                str2.substr(0, 2) ;
 
    // check if any of them is equal to string1
    return (str1.compare(clock_rot) == 0 ||
            str1.compare(anticlock_rot) == 0);
}
 
// Driver code
int main()
{
    string str1 = "geeks";
    string str2 = "eksge";
 
    isRotated(str1, str2) ? cout << "Yes"
                          : cout << "No";
    return 0;
}


Java




// Java program to check if a string is two time
// rotation of another string.
 
class Test
{
    // Method to check if string2 is obtained by
    // string 1
    static boolean isRotated(String str1, String str2)
    {
        if (str1.length() != str2.length())
            return false;
        if(str1.length() < 2)
        {
            return str1.equals(str2);
        }
      
        String clock_rot = "";
        String anticlock_rot = "";
        int len = str2.length();
      
        // Initialize string as anti-clockwise rotation
        anticlock_rot = anticlock_rot +
                        str2.substring(len-2, len) +
                        str2.substring(0, len-2) ;
      
        // Initialize string as clock wise rotation
        clock_rot = clock_rot +
                    str2.substring(2) +
                    str2.substring(0, 2) ;
      
        // check if any of them is equal to string1
        return (str1.equals(clock_rot) ||
                str1.equals(anticlock_rot));
    }
     
    // Driver method
    public static void main(String[] args)
    {
        String str1 = "geeks";
        String str2 = "eksge";
      
        System.out.println(isRotated(str1, str2) ?  "Yes"
                              : "No");
    }
}


Python3




# Python 3 program to check if a string
# is two time rotation of another string.
 
# Function to check if string2 is
# obtained by string 1
def isRotated(str1, str2):
 
    if (len(str1) != len(str2)):
        return False
     
    if(len(str1) < 2):
        return str1 == str2
    clock_rot = ""
    anticlock_rot = ""
    l = len(str2)
 
    # Initialize string as anti-clockwise rotation
    anticlock_rot = (anticlock_rot + str2[l - 2:] +
                                     str2[0: l - 2])
     
    # Initialize string as clock wise rotation
    clock_rot = clock_rot + str2[2:] + str2[0:2]
 
    # check if any of them is equal to string1
    return (str1 == clock_rot or
            str1 == anticlock_rot)
 
# Driver code
if __name__ == "__main__":
     
    str1 = "geeks"
    str2 = "eksge"
if isRotated(str1, str2):
    print("Yes"
else:
    print("No")
 
# This code is contributed by ita_c


C#




using System;
 
// C# program to check if a string is two time
// rotation of another string.
 
public class Test {
    // Method to check if string2 is obtained by
    // string 1
    public static bool isRotated(string str1, string str2)
    {
        if (str1.Length != str2.Length) {
            return false;
        }
 
        if (str1.Length < 2) {
            return str1.Equals(str2);
        }
 
        string clock_rot = "";
        string anticlock_rot = "";
        int len = str2.Length;
 
        // Initialize string as anti-clockwise rotation
        anticlock_rot
            = anticlock_rot
              + str2.Substring(len - 2, len - (len - 2))
              + str2.Substring(0, len - 2);
 
        // Initialize string as clock wise rotation
        clock_rot = clock_rot + str2.Substring(2)
                    + str2.Substring(0, 2);
 
        // check if any of them is equal to string1
        return (str1.Equals(clock_rot)
                || str1.Equals(anticlock_rot));
    }
 
    // Driver code
    public static void Main(string[] args)
    {
        string str1 = "geeks";
        string str2 = "eksge";
 
        Console.WriteLine(isRotated(str1, str2) ? "Yes"
                                                : "No");
    }
}
 
// This code is contributed by Shrikant13


Javascript




<script>
 
// Javascript program to check if a
// string is two time rotation of
// another string.
     
// Method to check if string2 is
// obtained by string 1
function isRotated(str1, str2)
{
    if (str1.length != str2.length)
        return false;
         
    if (str1.length < 2)
    {
        return str1.localeCompare(str2);
    }
   
    let clock_rot = "";
    let anticlock_rot = "";
    let len = str2.length;
   
    // Initialize string as anti-clockwise rotation
    anticlock_rot = anticlock_rot +
                    str2.substring(len - 2, len + 1) +
                    str2.substring(0, len - 1) ;
   
    // Initialize string as clock wise rotation
    clock_rot = clock_rot +
                str2.substring(2, str2.length - 2 + 1) +
                str2.substring(0, 2 + 1);
   
    // Check if any of them is equal to string1
    return (str1.localeCompare(clock_rot) ||
            str1.localeCompare(anticlock_rot));
}
 
// Driver code
let str1 = "geeks";
let str2 = "eksge";
 
document.write(isRotated(str1, str2) ? 
               "Yes" : "No");
 
// This code is contributed by rag2127
 
</script>


Output

Yes

Time Complexity: O(n), Time is taken to rotate the string and then compare the string.
Auxiliary Space: O(n), Space for storing clockwise and anticlockwise strings.

Approach 2(Without Using auxiliary space):

We could check directly if the string is rotated or not by comparing the two strings. 

Illustration:

Steps –

  1. Check if the string is rotated in clockwise manner.
  2. Check if the string is rotated in anticlockwise manner. 
  3. Return true if any one of the above is true

We compare for clockwise and anticlockwise by using for loops and the modulo operator-

Note that – 

For clockwise – str1[i] == str2[(i + 2) % n]

For anticlockwise – str1[(i + 2) % n] == str2[i]

Here n is length of string 

Check using the above two conditions and the problem will be solved!

Below is the implementation of the above approach:

C++




// C++ code to find if string is rotated by 2 positions
 
#include <iostream>
using namespace std;
 
bool isRotated(string str1, string str2)
{
    // Your code here
    // clockwise direction check
    int n = str1.length();
    bool clockwise = true, anticlockwise = true;
    for (int i = 0; i < n; i++)
    {
        if (str1[i] != str2[(i + 2) % n])
        {
            clockwise = false; // not rotated clockwise
            break;
        }
    }
 
    for (int i = 0; i < n; i++)
    {
        if (str1[(i + 2) % n] != str2[i])
        {
            anticlockwise = false; // not rotated anticlockwise
            break;
        }
    }
 
    return clockwise or anticlockwise; // if any of both is true, return true
}
int main()
{
    string str1 = "geeks";
    string str2 = "eksge";
 
    isRotated(str1, str2) ? cout << "Yes"
                          : cout << "No";
    return 0;
}
 
//code contributed by Anshit Bansal


Java




/*package whatever //do not write package name here */
 
import java.io.*;
 
class GFG {
  //Method to check if string2 is obtained by string1
    static boolean isRotated(String str1, String str2)
    {
        int n = str1.length();
        int m = str2.length();
        if (n != m) //check both the string length equal or not
            return false;
        boolean clockwise = true;
        boolean anticlockwise = true;
      //check clockwise rotation is possible or not
        for (int i = 0; i < n; i++) {
            if (str1.charAt(i)!= str2.charAt((i + 2) % n)) {
                clockwise = false;
                break;
            }
        }
      // check anticlockwise rotation is possible or not
        for (int i = 0; i < m; i++) {
            if (str1.charAt((i + 2) % n)!= str2.charAt(i)) {
                anticlockwise = false;
                break;
            }
        }
        return (clockwise || anticlockwise);
    }
    public static void main (String[] args) {
        String str1 = "geeks";
    String str2 = "eksge";
 
    System.out.println(isRotated(str1, str2) ? "Yes"
                                             : "No");
}
}


Output

Yes

Time Complexity – O(n), Iterating over the string 2 times for comparing both the strings.
Space Complexity – O(1)


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