Escriba código para encontrar el mínimo lexicográfico en una array circular, por ejemplo, para la array BCABDADAB, el mínimo lexicográfico es ABBCABDAD.
Fuente: prueba escrita de Google
Más ejemplos:
Input: GEEKSQUIZ Output: EEKSQUIZG Input: GFG Output: FGG Input: GEEKSFORGEEKS Output: EEKSFORGEEKSG
La siguiente es una solución simple. Deje que la string dada sea ‘str’
- Concatene ‘str’ consigo mismo y guárdelo en una string temporal, diga ‘concat’.
- Cree una array de strings para almacenar todas las rotaciones de ‘str’. Deje que la array sea ‘arr’.
- Encuentre todas las rotaciones de ‘str’ tomando substrings de ‘concat’ en el índice 0, 1, 2..n-1. Guarde estas rotaciones en arr[]
- Ordenar arr[] y devolver arr[0].
A continuación se muestra la implementación de la solución anterior.
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>
Producción
EEKSFORGEEKSG EEKSQUIZG ABBCABDAD
Secuencia rotada lexicográficamente más pequeña | conjunto 2
La complejidad temporal de la solución anterior es O(n 2 Logn) bajo el supuesto de que hemos utilizado un algoritmo de clasificación O(nLogn).
Espacio Auxiliar: O(n)
Este problema se puede resolver usando métodos más eficientes como el Algoritmo de Booth que resuelve el problema en tiempo O(n). Pronto cubriremos estos métodos como publicaciones separadas.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA