Minimum rotations required to get the same string
Given a string, we need to find the minimum number of rotations required to get the same string.
Examples:
Input : s = "geeks" Output : 5 Input : s = "aaaa" Output : 1
The idea is based on below post.
A Program to check if strings are rotations of each other or not
Step 1 : Initialize result = 0 (Here result is count of rotations)
Step 2 : Take a temporary string equals to original string concatenated with itself.
Step 3 : Now take the substring of temporary string of size same as original string starting from second character (or index 1).
Step 4 : Increase the count.
Step 5 : Check whether the substring becomes equal to original string. If yes, then break the loop. Else go to step 2 and repeat it from the next index.
C/C++
// C++ program to determine minimum number // of rotations required to yield same // string. #include <iostream> using namespace std; // Returns count of rotations to get the // same string back. int findRotations(string str) { // tmp is the concatenated string. string tmp = str + str; int n = str.length(); for ( int i = 1; i <= n; i++) { // substring from i index of original // string size. string substring = tmp.substr(i, str.size()); // if substring matches with original string // then we will come out of the loop. if (str == substring) return i; } return n; } // Driver code int main() { string str = "abc" ; cout << findRotations(str) << endl; return 0; } |
Java
// Java program to determine minimum number // of rotations required to yield same // string. import java.util.*; class GFG { // Returns count of rotations to get the // same string back. static int findRotations(String str) { // tmp is the concatenated string. String tmp = str + str; int n = str.length(); for ( int i = 1 ; i <= n; i++) { // substring from i index of original // string size. String substring = tmp.substring(i, str.length()); // if substring matches with original string // then we will come out of the loop. if (str == substring) return i; } return n; } // Driver Method public static void main(String[] args) { String str = "abc" ; System.out.println(findRotations(str)); } } /* This code is contributed by Mr. Somesh Awasthi */ |
Python3
# Python 3 program to determine minimum # number of rotations required to yield # same string. # Returns count of rotations to get the # same string back. def findRotations( str ): # tmp is the concatenated string. tmp = str + str n = len ( str ) for i in range ( 1 , n + 1 ): # substring from i index of # original string size. substring = tmp[i: n] # if substring matches with # original string then we will # come out of the loop. if ( str = = substring): return i return n # Driver code if __name__ = = '__main__' : str = "abc" print (findRotations( str )) # This code is contributed # by 29AjayKumar. |
C#
// C# program to determine minimum number // of rotations required to yield same // string. using System; class GFG { // Returns count of rotations to get // the same string back. static int findRotations(String str) { // tmp is the concatenated string. String tmp = str + str; int n = str.Length; for ( int i = 1; i <= n; i++) { // substring from i index of // original string size. String substring = tmp.Substring(i, str.Length); // if substring matches with // original string then we will // come out of the loop. if (str == substring) return i; } return n; } // Driver Method public static void Main() { String str = "abc" ; Console.Write(findRotations(str)); } } // This code is contributed by nitin mittal. |
PHP
<?php // PHP program to determine minimum // number of rotations required to // yield same string. // Returns count of rotations // to get the same string back. function findRotations( $str ) { // tmp is the concatenated string. $tmp = ( $str + $str ); $n = strlen ( $str ); for ( $i = 1; $i <= $n ; $i ++) { // substring from i index // of original string size. $substring = $tmp . substr ( $i , strlen ( $str )); // if substring matches with // original string then we will // come out of the loop. if ( $str == $substring ) return $i ; } return $n ; } // Driver code $str = "abc" ; echo findRotations( $str ), "\n" ; // This code is contributed // by Sachin ?> |
Output:
3
Time Complexity: O(n2)
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