Dada una array arr que contiene N elementos, encuentre la longitud de la subsecuencia más larga tal que sea una permutación válida de una longitud particular. Si no existe tal secuencia de permutación, imprima 0.
Ejemplos:
Entrada: arr[] = {3, 2, 1, 6, 5}
Salida: 3
Explicación:
La subsecuencia de permutación más larga será [3, 2, 1].
Entrada: arr[]= {2, 3, 4, 5}
Salida: 0
Explicación:
No hay una subsecuencia de permutación válida presente porque falta el elemento 1.
Enfoque: el problema mencionado anteriormente es sobre la subsecuencia de permutación, por lo que el orden de los elementos de la array es irrelevante, solo importa la frecuencia de cada elemento . Si la array tiene una longitud N, entonces la longitud máxima posible para la secuencia de permutación es N y la longitud mínima posible es 0 . Si la subsecuencia de longitud L es una permutación válida, entonces todos los elementos de 1 a L deben estar presentes .
- Cuente la frecuencia de los elementos en el rango [1, N] de la array
- Iterar a través de todos los elementos de 1 a N en la array y contar las iteraciones hasta que se observe una frecuencia de 0. Si la frecuencia de un elemento es ‘0’, devuelve el recuento actual de iteraciones como la longitud requerida.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ Program to find length of
// Longest Permutation Subsequence
// in a given array
#include <bits/stdc++.h>
using namespace std;
// Function to find the
// longest permutation subsequence
int longestPermutation(int a[], int n)
{
// Map data structure to
// count the frequency of each element
map<int, int> freq;
for (int i = 0; i < n; i++) {
freq[a[i]]++;
}
int len = 0;
for (int i = 1; i <= n; i++) {
// If frequency of element is 0,
// then we can not move forward
// as every element should be present
if (freq[i] == 0) {
break;
}
// Increasing the length by one
len++;
}
return len;
}
// Driver Code
int main()
{
int arr[] = { 3, 2, 1, 6, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << longestPermutation(arr, n)
<< "\n";
return 0;
}
Java
// Java Program to find length of
// Longest Permutation Subsequence
// in a given array
import java.util.*;
class GFG{
// Function to find the
// longest permutation subsequence
static int longestPermutation(int arr[], int n)
{
// Map data structure to
// count the frequency of each element
HashMap<Integer,Integer> freq = new HashMap<Integer,Integer>();
for (int i = 0; i < n; i++) {
if(freq.containsKey(arr[i])){
freq.put(arr[i], freq.get(arr[i])+1);
}else{
freq.put(arr[i], 1);
}
}
int len = 0;
for (int i = 1; i <= n; i++) {
// If frequency of element is 0,
// then we can not move forward
// as every element should be present
if (!freq.containsKey(i)) {
break;
}
// Increasing the length by one
len++;
}
return len;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 3, 2, 1, 6, 5 };
int n = arr.length;
System.out.print(longestPermutation(arr, n));
}
}
// This code is contributed by Rajput-Ji
C#
// C# Program to find length of
// longest Permutation Subsequence
// in a given array
using System;
using System.Collections.Generic;
public class GFG{
// Function to find the
// longest permutation subsequence
static int longestPermutation(int []arr, int n)
{
// Map data structure to
// count the frequency of each element
Dictionary<int,int> freq = new Dictionary<int,int>();
for (int i = 0; i < n; i++) {
if(freq.ContainsKey(arr[i])){
freq[arr[i]] = freq[arr[i]] + 1;
}else{
freq.Add(arr[i], 1);
}
}
int len = 0;
for (int i = 1; i <= n; i++) {
// If frequency of element is 0,
// then we can not move forward
// as every element should be present
if (!freq.ContainsKey(i)) {
break;
}
// Increasing the length by one
len++;
}
return len;
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 3, 2, 1, 6, 5 };
int n = arr.Length;
Console.Write(longestPermutation(arr, n));
}
}
// This code is contributed by 29AjayKumar
Python3
# Program to find length of # Longest Permutation Subsequence # in a given array from collections import defaultdict # Function to find the # longest permutation subsequence def longestPermutation(a, n): # Map data structure to # count the frequency of each element freq = defaultdict(int) for i in range( n ): freq[a[i]] += 1 length = 0 for i in range(1 , n + 1): # If frequency of element is 0, # then we can not move forward # as every element should be present if (freq[i] == 0): break # Increasing the length by one length += 1 return length # Driver Code if __name__ == "__main__": arr = [ 3, 2, 1, 6, 5 ] n = len(arr) print(longestPermutation(arr, n)) # This code is contributed by chitranayal
Javascript
<script>
// Javascript Program to find length of
// Longest Permutation Subsequence
// in a given array
// Function to find the
// longest permutation subsequence
function longestPermutation(arr, n)
{
// Map data structure to
// count the frequency of each element
let freq = new Map();
for (let i = 0; i < n; i++) {
if(freq.has(arr[i])){
freq.set(arr[i], freq.get(arr[i])+1);
}else{
freq.set(arr[i], 1);
}
}
let len = 0;
for (let i = 1; i <= n; i++) {
// If frequency of element is 0,
// then we can not move forward
// as every element should be present
if (!freq.has(i)) {
break;
}
// Increasing the length by one
len++;
}
return len;
}
// Driver code
let arr = [ 3, 2, 1, 6, 5 ];
let n = arr.length;
document.write(longestPermutation(arr, n));
</script>
Producción:
3
Complejidad temporal: O(N)
Complejidad espacial auxiliar: O(N)