Dada una array arr[] de tamaño N que es una permutación de 1 a N , la tarea es encontrar la permutación lexicográficamente más pequeña que se puede formar invirtiendo como máximo un subarreglo.
Ejemplos:
Entrada: arr[] = {1, 3, 4, 2, 5}
Salida: 1 2 4 3 5
Explicación: El subarreglo del índice 1 al índice 3 se puede invertir para obtener la permutación lexicográficamente más pequeña.Entrada: arr[] = {4, 3, 1, 2}
Salida: 1 3 4 2
Enfoque: La idea para resolver el problema se basa en el recorrido de la array .
- En el problema dado, la permutación lexicográficamente más pequeña se puede obtener colocando el menor número en su lugar correcto mediante una inversión recorriendo desde la izquierda y comprobando que (i+1) es igual a arr[i]
- Si no es igual, busque el índice de ese i+1 en el arreglo e invierta el subarreglo desde el i-ésimo índice hasta la posición (i+1) .
A continuación se muestra la implementación del enfoque anterior.
C++
// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the
// Lexicographically smallest
// Permutation by one subarray reversal
void lexsmallest(vector<int>& arr, int n)
{
// Initialize the variables
// To store the first and last
// Position of the subarray
int first = -1, flag = 0, find = -1, last = -1;
// Traverse the array
// And check if arr[i]!=i+1
for (int i = 0; i < n; i++) {
if (arr[i] != i + 1) {
flag = 1;
// Mark the first position
// Of the Subarray to be reversed
first = i;
find = i + 1;
break;
}
}
// If flag == 0, it is the
// Smallest permutation,
// So print the array
if (flag == 0) {
for (int i = 0; i < n; i++) {
cout << arr[i] << " ";
}
}
// Check where the minimum element is present
else {
for (int i = 0; i < n; i++) {
// It is the last position
// Of the subarray to be
// Reversed
if (arr[i] == find) {
last = i;
break;
}
}
// Reverse the subarray
// And print the array
reverse(arr.begin() + first,
arr.begin() + last + 1);
for (int i = 0; i < n; i++) {
cout << arr[i] << " ";
}
}
}
// Driver Code
int main()
{
// Initialize the array arr[]
vector<int> arr = { 1, 3, 4, 2, 5 };
int N = arr.size();
// Function call
lexsmallest(arr, N);
return 0;
}
Java
// Java code for the above approach
import java.util.*;
class GFG{
// Function to find the
// Lexicographically smallest
// Permutation by one subarray reversal
static void lexsmallest(int []arr, int n)
{
// Initialize the variables
// To store the first and last
// Position of the subarray
int first = -1, flag = 0, find = -1, last = -1;
// Traverse the array
// And check if arr[i]!=i+1
for (int i = 0; i < n; i++) {
if (arr[i] != i + 1) {
flag = 1;
// Mark the first position
// Of the Subarray to be reversed
first = i;
find = i + 1;
break;
}
}
// If flag == 0, it is the
// Smallest permutation,
// So print the array
if (flag == 0) {
for (int i = 0; i < n; i++) {
System.out.print(arr[i]+ " ");
}
}
// Check where the minimum element is present
else {
for (int i = 0; i < n; i++) {
// It is the last position
// Of the subarray to be
// Reversed
if (arr[i] == find) {
last = i;
break;
}
}
// Reverse the subarray
// And print the array
arr = reverse(arr,first,last);
for (int i = 0; i < n; i++) {
System.out.print(arr[i]+ " ");
}
}
}
static int[] reverse(int str[], int start, int end) {
// Temporary variable to store character
int temp;
while (start <= end) {
// Swapping the first and last character
temp = str[start];
str[start] = str[end];
str[end] = temp;
start++;
end--;
}
return str;
}
// Driver Code
public static void main(String[] args)
{
// Initialize the array arr[]
int []arr = { 1, 3, 4, 2, 5 };
int N = arr.length;
// Function call
lexsmallest(arr, N);
}
}
// This code contributed by shikhasingrajput
Python3
# Python code for the above approach # Function to find the # Lexicographically smallest # Permutation by one subarray reversal def lexsmallest(arr, n): # Initialize the variables # To store the first and last # Position of the subarray first = -1 flag = 0 find = -1 last = -1 # Traverse the array # And check if arr[i]!=i+1 for i in range(0, n): if (arr[i] != i + 1): flag = 1 # Mark the first position # Of the Subarray to be reversed first = i find = i + 1 break # If flag == 0, it is the # Smallest permutation, # So print the array if (flag == 0): for i in range(0, n): print(arr[i], end=" ") # Check where the minimum element is present else: for i in range(0, n): # It is the last position # Of the subarray to be # Reversed if (arr[i] == find): last = i break # Reverse the subarray # And print the array arr[first: last + 1] = arr[first: last + 1][::-1] print(*arr) # Driver Code # Initialize the array arr[] arr = [1, 3, 4, 2, 5] N = len(arr) # Function call lexsmallest(arr, N) # This code is contributed by Samim Hossain Mondal.
C#
// C# code for the above approach
using System;
class GFG{
// Function to find the
// Lexicographically smallest
// Permutation by one subarray reversal
static void lexsmallest(int []arr, int n)
{
// Initialize the variables
// To store the first and last
// Position of the subarray
int first = -1, flag = 0, find = -1, last = -1;
// Traverse the array
// And check if arr[i]!=i+1
for (int i = 0; i < n; i++) {
if (arr[i] != i + 1) {
flag = 1;
// Mark the first position
// Of the Subarray to be reversed
first = i;
find = i + 1;
break;
}
}
// If flag == 0, it is the
// Smallest permutation,
// So print the array
if (flag == 0) {
for (int i = 0; i < n; i++) {
Console.Write(arr[i]+ " ");
}
}
// Check where the minimum element is present
else {
for (int i = 0; i < n; i++) {
// It is the last position
// Of the subarray to be
// Reversed
if (arr[i] == find) {
last = i;
break;
}
}
// Reverse the subarray
// And print the array
arr = reverse(arr,first,last);
for (int i = 0; i < n; i++) {
Console.Write(arr[i]+ " ");
}
}
}
static int[] reverse(int[] str, int start, int end) {
// Temporary variable to store character
int temp;
while (start <= end)
{
// Swapping the first and last character
temp = str[start];
str[start] = str[end];
str[end] = temp;
start++;
end--;
}
return str;
}
// Driver Code
static public void Main (){
// Initialize the array arr[]
int [] arr = { 1, 3, 4, 2, 5 };
int N = arr.Length;
// Function call
lexsmallest(arr, N);
}
}
// This code is contributed by hrithikgarg03188.
Javascript
<script>
// JavaScript code for the above approach
// Function to find the
// Lexicographically smallest
// Permutation by one subarray reversal
const lexsmallest = (arr, n) => {
// Initialize the variables
// To store the first and last
// Position of the subarray
let first = -1, flag = 0, find = -1, last = -1;
// Traverse the array
// And check if arr[i]!=i+1
for (let i = 0; i < n; i++) {
if (arr[i] != i + 1) {
flag = 1;
// Mark the first position
// Of the Subarray to be reversed
first = i;
find = i + 1;
break;
}
}
// If flag == 0, it is the
// Smallest permutation,
// So print the array
if (flag == 0) {
for (let i = 0; i < n; i++) {
document.write(`${arr[i]} `);
}
}
// Check where the minimum element is present
else {
for (let i = 0; i < n; i++) {
// It is the last position
// Of the subarray to be
// Reversed
if (arr[i] == find) {
last = i;
break;
}
}
// Reverse the subarray
// And print the array
arr.splice(first, last, ...arr.slice(first, last + 1).reverse());
for (let i = 0; i < n; i++) {
document.write(`${arr[i]} `);
}
}
}
// Driver Code
// Initialize the array arr[]
let arr = [1, 3, 4, 2, 5];
let N = arr.length;
// Function call
lexsmallest(arr, N);
// This code is contributed by rakeshsahni
</script>
Producción
1 2 4 3 5
Complejidad temporal: O(N)
Espacio auxiliar: O(1)
Tema relacionado: Subarrays, subsecuencias y subconjuntos en array
Publicación traducida automáticamente
Artículo escrito por lokeshpotta20 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA