Minimum rotations required to get the same string - GeeksforGeeks

21 Apr, 2022
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 notStep 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++
Java
Python3
C#
PHP
Javascript
// 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 codeint main(){    string str = "abc";    cout << findRotations(str) << endl;    return 0;}
// 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 */
# 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 codeif __name__ == '__main__':     str = "abc"    print(findRotations(str)) # This code is contributed# by 29AjayKumar.
// 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 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?>
<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++
Python3
Javascript
// 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 codeint main(){    string str = "abc";    cout << findRotations(str) << endl;    return 0;}
# 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 codeStr = "abc"print(findRotations(Str)) # This code is contributed by shinjanpatra
<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 :  
Python3
# Python 3 program to determine minimum# number of rotations required to yield# same string.  # inputstring = '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.
Output:  
1
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nitin mittal
Sach_Code
29AjayKumar
gp6
nagasowmyanarayanan
GauravRajput1
kaivalyamanit
rotation
Strings
Strings
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