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++ 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, i+str.length()); // if substring matches with original string // then we will come out of the loop. if (str.equals(substring)) return i; } return n; } // Driver Method public static void main(String[] args) { String str = "aaaa" ; 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: 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 ?> |
Javascript
<script> // javascript 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. var tmp = str + str; var n = str.length; for ( var i = 1; i <= n; i++) { // substring from i index of original // string size. var 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 var str = "abc" ; document.write(findRotations(str)); // This code contributed by gauravrajput1 </script> |
Output:
3
Time Complexity: O(n2)
Space Complexity : O(2n) ~ O(n)
We can solve this problem without using any temporary variable as extra space . We will traverse the original string and at each position we partition it and concatenate the right substring and left substring and check weather it is equal to original string
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) { int ans = 0; //to store the answer int n = str.length(); //length of the string //All the length where we can partition for ( int i=1;i<str.length()-1;i++) { //right part + left part = rotated string // we are checking weather the rotated string is equal to //original string if (str.substr(i,n-i) + str.substr(0,i) == str) { ans = i; break ; } } if (ans == 0) return n; return ans; } // Driver code int main() { string str = "abc" ; cout << findRotations(str) << endl; return 0; } |
Python3
# Python 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 ): ans = 0 # to store the answer n = len ( Str ) # length of the String # All the length where we can partition for i in range ( 1 , len ( Str ) - 1 ): # right part + left part = rotated String # we are checking weather the rotated String is equal to # original String if ( Str [i: n] + Str [ 0 : i] = = Str ): ans = i break if (ans = = 0 ): return n return ans # Driver code Str = "abc" print (findRotations( Str )) # This code is contributed by shinjanpatra |
Javascript
<script> // JavaScript 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) { let ans = 0; // to store the answer let n = str.length; // length of the string // All the length where we can partition for (let i = 1; i < str.length - 1; i++) { // right part + left part = rotated string // we are checking weather the rotated string is equal to // original string if (str.substr(i, n - i) + str.substr(0, i) == str) { ans = i; break ; } } if (ans == 0) return n; return ans; } // Driver code let str = "abc" ; document.write(findRotations(str), "</br>" ); // This code is contributed by shinjanpatra </script> |
3
Time Complexity : O(n2)
Space Complexity : O(1)
Alternate Implementation in Python :
C++
// C++ program to determine minimum // number of rotations required to yield // same string. #include <iostream> using namespace std; // Driver program int main() { string String = "aaaa" ; string check = "" ; for ( int r = 1; r < String.length() + 1; r++) { // checking the input after each rotation check = String.substr(0, r) + String.substr(r, String.length()-r); // following if statement checks if input is // equals to check , if yes it will print r and // break out of the loop if (check == String){ cout<<r; break ; } } return 0; } // This code is contributed by shinjanpatra |
Python3
# Python 3 program to determine minimum # number of rotations required to yield # same string. # input string = 'aaaa' check = '' for r in range ( 1 , len (string) + 1 ): # checking the input after each rotation check = string[r:] + string[:r] # following if statement checks if input is # equals to check , if yes it will print r and # break out of the loop if check = = string: print (r) break # This code is contributed # by nagasowmyanarayanan. |
Javascript
<script> // JavaScript program to determine minimum // number of rotations required to yield // same string. // input let string = 'aaaa' let check = '' for (let r=1;r<string.length+1;r++){ // checking the input after each rotation check = string.substring(r) + string.substring(0,r) // following if statement checks if input is // equals to check , if yes it will print r and // break out of the loop if (check == string){ document.write(r) break } } // This code is contributed // by shinjanpatra </script> |
Java
/*package whatever //do not write package name here */ import java.io.*; class GFG { public static void main (String[] args) { String string = "aaaa" ; String check = "" ; for ( int r = 1 ; r < string.length() + 1 ; r++) { // checking the input after each rotation check = string.substring( 0 , r) + string.substring(r, string.length()); // following if statement checks if input is // equals to check , if yes it will print r and // break out of the loop if (check.equals(string)){ System.out.println(r); break ; } } } } |
Output:
1
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