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’
- Concatenate ‘str’ with itself and store in a temporary string say ‘concat’.
- Create an array of strings to store all rotations of ‘str’. Let the array be ‘arr’.
- Find all rotations of ‘str’ by taking substrings of ‘concat’ at index 0, 1, 2..n-1. Store these rotations in arr[]
- Sort arr[] and return arr[0].
Following is the implementation of above solution.
C++
#include <iostream>
#include <algorithm>
using namespace std;
string minLexRotation(string str)
{
int n = str.length();
string arr[n];
string concat = str + str;
for ( int i = 0; i < n; i++)
arr[i] = concat.substr(i, n);
sort(arr, arr+n);
return arr[0];
}
int main()
{
cout << minLexRotation( "GEEKSFORGEEKS" ) << endl;
cout << minLexRotation( "GEEKSQUIZ" ) << endl;
cout << minLexRotation( "BCABDADAB" ) << endl;
}
|
Java
import java.util.*;
class GFG
{
static String minLexRotation(String str)
{
int n = str.length();
String arr[] = new String[n];
String concat = str + str;
for ( int i = 0 ; i < n; i++)
{
arr[i] = concat.substring(i, i + n);
}
Arrays.sort(arr);
return arr[ 0 ];
}
public static void main(String[] args)
{
System.out.println(minLexRotation( "GEEKSFORGEEKS" ));
System.out.println(minLexRotation( "GEEKSQUIZ" ));
System.out.println(minLexRotation( "BCABDADAB" ));
}
}
|
Python3
def minLexRotation(str_) :
n = len (str_)
arr = [ 0 ] * n
concat = str_ + str_
for i in range (n) :
arr[i] = concat[i : n + i]
arr.sort()
return arr[ 0 ]
print (minLexRotation( "GEEKSFORGEEKS" ))
print (minLexRotation( "GEEKSQUIZ" ))
print (minLexRotation( "BCABDADAB" ))
|
C#
using System;
class GFG
{
static String minLexRotation(String str)
{
int n = str.Length;
String []arr = new String[n];
String concat = str + str;
for ( int i = 0; i < n; i++)
{
arr[i] = concat.Substring(i, n);
}
Array.Sort(arr);
return arr[0];
}
public static void Main(String[] args)
{
Console.WriteLine(minLexRotation( "GEEKSFORGEEKS" ));
Console.WriteLine(minLexRotation( "GEEKSQUIZ" ));
Console.WriteLine(minLexRotation( "BCABDADAB" ));
}
}
|
Javascript
<script>
function minLexRotation(str)
{
let n = str.length;
let arr = new Array(n);
let concat = str + str;
for (let i = 0; i < n; i++)
{
arr[i] = concat.substring(i, i + n);
}
arr.sort();
return arr[0];
}
document.write(minLexRotation( "GEEKSFORGEEKS" ) + "</br>" );
document.write(minLexRotation( "GEEKSQUIZ" ) + "</br>" );
document.write(minLexRotation( "BCABDADAB" ) + "</br>" );
</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.
Last Updated :
07 Jul, 2022
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