Minimum rotations required to get the same string

Difficulty Level : Medium
Last Updated : 09 May, 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 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:

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>

Output

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:

This article is contributed by Jatin Goyal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

My Personal Notes

Related Articles

Table of Contents

Minimum rotations required to get the same string

C++

Java

Python3

C#

PHP

Javascript

C++

Python3

Javascript

C++

Python3

Javascript

Java

Start Your Coding Journey Now!