Check if a string can be obtained by rotating another string 2 places
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
// rotated anti-clockwise
Input: string1 = “amazon”, string2 = “onamaz”
Output: Yes
// rotated clockwise
Asked in: Amazon Interview
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> |
Yes
Time Complexity : O(n)
Auxiliary Space: O(n)
Exercise : Check if string2 is obtained by rotating string1 by k places
Method 2 – Without using any extra space:
We could check directly if the string is rotated or not by comparing the two strings.
Steps –
- Check if the string is rotated in clockwise manner.
- Check if the string is rotated in anticlockwise manner.
- 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!
See the code for better understanding.
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 |
Yes
Time Complexity – O(n)
Space Complexity – O(1)
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