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 */


Output:

3

Time Complexity : O(n)

This article is contributed by Jatin Goyal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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