# 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.

More Examples:

```Input:  GEEKSQUIZ
Output: EEKSQUIZG

Input:  GFG
Output: FGG

Input:  GEEKSFORGEEKS
Output: EEKSFORGEEKSG```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 C++ implementation of above solution.

```// 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;
}
```

Output:

```EEKSFORGEEKSG
EEKSQUIZG

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.
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.

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