Lexicographically minimum string rotation | Set 1

Write code to find lexicographic minimum in a circular array, e.g. for the array BCABDADAB, the lexicographic minimum is ABBCABDAD.
Source: Google Written Test

More Examples: 

Input:  GEEKSQUIZ
Output: EEKSQUIZG

Input:  GFG
Output: FGG

Input:  GEEKSFORGEEKS
Output: EEKSFORGEEKSG

Following is a simple solution. Let the given string be ‘str’ 

1. Concatenate ‘str’ with itself and store in a temporary string say ‘concat’. 
2. Create an array of strings to store all rotations of ‘str’. Let the array be ‘arr’. 
3. Find all rotations of ‘str’ by taking substrings of ‘concat’ at index 0, 1, 2..n-1. Store these rotations in arr[] 
4.  Sort arr[] and return arr[0].

Following is the implementation of above solution. 

C++

// A simple C++ program to find lexicographically minimum rotation
// of a given string
#include <iostream>
#include <algorithm>
using namespace std;

// This functionr return lexicographically minimum
// rotation of str
string minLexRotation(string str)
{
    // Find length of given string
    int n = str.length();

    // Create an array of strings to store all rotations
    string arr[n];

    // Create a concatenation of string with itself
    string concat = str + str;

    // One by one store all rotations of str in array.
    // A rotation is obtained by getting a substring of concat
    for (int i = 0; i < n; i++)
        arr[i] = concat.substr(i, n);

    // Sort all rotations
    sort(arr, arr+n);

    // Return the first rotation from the sorted array
    return arr[0];
}

// Driver program to test above function
int main()
{
    cout << minLexRotation("GEEKSFORGEEKS") << endl;
    cout << minLexRotation("GEEKSQUIZ") << endl;
    cout << minLexRotation("BCABDADAB") << endl;
}

Java

// A simple Java program to find
// lexicographically minimum rotation
// of a given String
import java.util.*;

class GFG
{

    // This functionr return lexicographically
    // minimum rotation of str
    static String minLexRotation(String str)
    {
        // Find length of given String
        int n = str.length();

        // Create an array of strings
        // to store all rotations
        String arr[] = new String[n];

        // Create a concatenation of
        // String with itself
        String concat = str + str;

        // One by one store all rotations
        // of str in array. A rotation is
        // obtained by getting a substring of concat
        for (int i = 0; i < n; i++)
        {
            arr[i] = concat.substring(i, i + n);
        }

        // Sort all rotations
        Arrays.sort(arr);

        // Return the first rotation
        // from the sorted array
        return arr[0];
    }

    // Driver code
    public static void main(String[] args)
    {
        System.out.println(minLexRotation("GEEKSFORGEEKS"));
        System.out.println(minLexRotation("GEEKSQUIZ"));
        System.out.println(minLexRotation("BCABDADAB"));
    }
}

// This code is contributed by 29AjayKumar

Python3

# A simple Python3 program to find lexicographically
# minimum rotation of a given string

# This function return lexicographically minimum
# rotation of str
def minLexRotation(str_) :

    # Find length of given string
    n = len(str_)

    # Create an array of strings to store all rotations
    arr = [0] * n

    # Create a concatenation of string with itself
    concat = str_ + str_

    # One by one store all rotations of str in array.
    # A rotation is obtained by getting a substring of concat
    for i in range(n) :
        arr[i] = concat[i : n + i]

    # Sort all rotations
    arr.sort()

    # Return the first rotation from the sorted array
    return arr[0]

# Driver Code
print(minLexRotation("GEEKSFORGEEKS"))
print(minLexRotation("GEEKSQUIZ"))
print(minLexRotation("BCABDADAB"))

# This code is contributed by divyamohan123

C#

// A simple C# program to find
// lexicographically minimum rotation
// of a given String
using System;

class GFG
{

    // This functionr return lexicographically
    // minimum rotation of str
    static String minLexRotation(String str)
    {
        // Find length of given String
        int n = str.Length;

        // Create an array of strings
        // to store all rotations
        String []arr = new String[n];

        // Create a concatenation of
        // String with itself
        String concat = str + str;

        // One by one store all rotations
        // of str in array. A rotation is
        // obtained by getting a substring of concat
        for (int i = 0; i < n; i++)
        {
            arr[i] = concat.Substring(i, n);
        }

        // Sort all rotations
        Array.Sort(arr);

        // Return the first rotation
        // from the sorted array
        return arr[0];
    }

    // Driver code
    public static void Main(String[] args)
    {
        Console.WriteLine(minLexRotation("GEEKSFORGEEKS"));
        Console.WriteLine(minLexRotation("GEEKSQUIZ"));
        Console.WriteLine(minLexRotation("BCABDADAB"));
    }
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// A simple Javascript program to find
// lexicographically minimum rotation
// of a given String

// This functionr return lexicographically
// minimum rotation of str
function minLexRotation(str)
{

    // Find length of given String
    let n = str.length;

    // Create an array of strings
    // to store all rotations
    let arr = new Array(n);

    // Create a concatenation of
    // String with itself
    let concat = str + str;

    // One by one store all rotations
    // of str in array. A rotation is
    // obtained by getting a substring of concat
    for(let i = 0; i < n; i++)
    {
        arr[i] = concat.substring(i, i + n);
    }

    // Sort all rotations
    arr.sort();

    // Return the first rotation
    // from the sorted array
    return arr[0];
}

// Driver code
document.write(minLexRotation("GEEKSFORGEEKS") + "</br>");
document.write(minLexRotation("GEEKSQUIZ") + "</br>");
document.write(minLexRotation("BCABDADAB") + "</br>");

// This code is contributed by divyeshrabadiya07

</script>

Output

EEKSFORGEEKSG
EEKSQUIZG
ABBCABDAD

Lexicographically smallest rotated sequence | Set 2

Time complexity of the above solution is O(n²Logn) under the assumption that we have used a O(nLogn) sorting algorithm. 

Auxiliary Space: O(n)
This problem can be solved using more efficient methods like Booth’s Algorithm which solves the problem in O(n) time. We will soon be covering these methods as separate posts.