Longitud de la subsecuencia principal más larga en una array

Dada una array arr que contiene enteros no negativos, la tarea es imprimir la longitud de la subsecuencia más larga de números primos en la array.
Ejemplos: 
 

Entrada: arr[] = { 3, 4, 11, 2, 9, 21 } 
Salida: 3 
La subsecuencia principal más larga es {3, 2, 11} y, por lo tanto, la respuesta es 3.
Entrada: arr[] = { 6, 4 , 10, 13, 9, 25 } 
Salida: 1 
 

Acercarse:

• Recorre la array dada.
• Para cada elemento de la array, compruebe si es primo o no .
• Si el elemento es primo, estará en la subsecuencia principal más larga. Por lo tanto, incremente la longitud requerida de la subsecuencia principal más larga en 1

A continuación se muestra la implementación del enfoque anterior: 
 

// C++ program to find the length of
// Longest Prime Subsequence in an Array
 
#include <bits/stdc++.h>
using namespace std;
#define N 100005
 
// Function to create Sieve
// to check primes
void SieveOfEratosthenes(
    bool prime[], int p_size)
{
 
    // False here indicates
    // that it is not prime
    prime[0] = false;
    prime[1] = false;
 
    for (int p = 2; p * p <= p_size; p++) {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p]) {
 
            // Update all multiples of p,
            // set them to non-prime
            for (int i = p * 2;
                 i <= p_size;
                 i += p)
                prime[i] = false;
        }
    }
}
 
// Function to find the longest subsequence
// which contain all prime numbers
int longestPrimeSubsequence(int arr[], int n)
{
    bool prime[N + 1];
    memset(prime, true, sizeof(prime));
 
    // Precompute N primes
    SieveOfEratosthenes(prime, N);
 
    int answer = 0;
 
    // Find the length of
    // longest prime subsequence
    for (int i = 0; i < n; i++) {
        if (prime[arr[i]]) {
            answer++;
        }
    }
 
    return answer;
}
 
// Driver code
int main()
{
    int arr[] = { 3, 4, 11, 2, 9, 21 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    cout << longestPrimeSubsequence(arr, n)
         << endl;
 
    return 0;
}

// Java program to find the length of
// Longest Prime Subsequence in an Array
import java.util.*;
 
class GFG
{
static final int N = 100005;
  
// Function to create Sieve
// to check primes
static void SieveOfEratosthenes(
    boolean prime[], int p_size)
{
  
    // False here indicates
    // that it is not prime
    prime[0] = false;
    prime[1] = false;
  
    for (int p = 2; p * p <= p_size; p++) {
  
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p]) {
  
            // Update all multiples of p,
            // set them to non-prime
            for (int i = p * 2;
                 i <= p_size;
                 i += p)
                prime[i] = false;
        }
    }
}
  
// Function to find the longest subsequence
// which contain all prime numbers
static int longestPrimeSubsequence(int arr[], int n)
{
    boolean []prime = new boolean[N + 1];
    Arrays.fill(prime, true);
  
    // Precompute N primes
    SieveOfEratosthenes(prime, N);
  
    int answer = 0;
  
    // Find the length of
    // longest prime subsequence
    for (int i = 0; i < n; i++) {
        if (prime[arr[i]]) {
            answer++;
        }
    }
  
    return answer;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 3, 4, 11, 2, 9, 21 };
    int n = arr.length;
  
    // Function call
    System.out.print(longestPrimeSubsequence(arr, n)
         +"\n");
}
}
 
// This code is contributed by 29AjayKumar

# Python 3 program to find the length of
# Longest Prime Subsequence in an Array
N = 100005
  
# Function to create Sieve
# to check primes
def SieveOfEratosthenes(prime,  p_size):
  
    # False here indicates
    # that it is not prime
    prime[0] = False
    prime[1] = False
  
    p = 2
    while  p * p <= p_size:
  
        # If prime[p] is not changed,
        # then it is a prime
        if (prime[p]):
  
            # Update all multiples of p,
            # set them to non-prime
            for i in range( p * 2, p_size + 1, p):
                prime[i] = False
 
        p += 1
       
# Function to find the longest subsequence
# which contain all prime numbers
def longestPrimeSubsequence( arr, n):
    prime = [True]*(N + 1)
  
    # Precompute N primes
    SieveOfEratosthenes(prime, N)
  
    answer = 0
  
    # Find the length of
    # longest prime subsequence
    for i in range (n):
        if (prime[arr[i]]):
            answer += 1
  
    return answer
  
# Driver code
if __name__ == "__main__":
    arr = [ 3, 4, 11, 2, 9, 21 ]
    n = len(arr)
  
    # Function call
    print (longestPrimeSubsequence(arr, n))
 
# This code is contributed by chitranayal

// C# program to find the length of
// longest Prime Subsequence in an Array
using System;
 
class GFG
{
static readonly int N = 100005;
   
// Function to create Sieve
// to check primes
static void SieveOfEratosthenes(
    bool []prime, int p_size)
{
   
    // False here indicates
    // that it is not prime
    prime[0] = false;
    prime[1] = false;
   
    for (int p = 2; p * p <= p_size; p++) {
   
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p]) {
   
            // Update all multiples of p,
            // set them to non-prime
            for (int i = p * 2;
                 i <= p_size;
                 i += p)
                prime[i] = false;
        }
    }
}
   
// Function to find the longest subsequence
// which contain all prime numbers
static int longestPrimeSubsequence(int []arr, int n)
{
    bool []prime = new bool[N + 1];
    for (int i = 0; i < N+1; i++)
        prime[i] = true;
   
    // Precompute N primes
    SieveOfEratosthenes(prime, N);
   
    int answer = 0;
   
    // Find the length of
    // longest prime subsequence
    for (int i = 0; i < n; i++) {
        if (prime[arr[i]]) {
            answer++;
        }
    }
   
    return answer;
}
   
// Driver code
public static void Main(String[] args)
{
    int []arr = { 3, 4, 11, 2, 9, 21 };
    int n = arr.Length;
   
    // Function call
    Console.Write(longestPrimeSubsequence(arr, n)
         +"\n");
}
}
  
// This code is contributed by 29AjayKumar

<script>
 
// JavaScript program to find the length of
// Longest Prime Subsequence in an Array
 
let N = 100005
 
// Function to create Sieve
// to check primes
function SieveOfEratosthenes(prime, p_size)
{
 
    // False here indicates
    // that it is not prime
    prime[0] = false;
    prime[1] = false;
 
    for (let p = 2; p * p <= p_size; p++) {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p]) {
 
            // Update all multiples of p,
            // set them to non-prime
            for (let i = p * 2;
                i <= p_size;
                i += p)
                prime[i] = false;
        }
    }
}
 
// Function to find the longest subsequence
// which contain all prime numbers
function longestPrimeSubsequence(arr, n)
{
    let prime = new Array(N + 1);
    prime.fill(true)
 
    // Precompute N primes
    SieveOfEratosthenes(prime, N);
 
    let answer = 0;
 
    // Find the length of
    // longest prime subsequence
    for (let i = 0; i < n; i++) {
        if (prime[arr[i]]) {
            answer++;
        }
    }
 
    return answer;
}
 
// Driver code
 
let arr = [ 3, 4, 11, 2, 9, 21 ];
let n = arr.length
 
// Function call
document.write(longestPrimeSubsequence(arr, n) + "<br>");
 
// This code is contributed by gfgking
 
</script>

3

Complejidad de tiempo: O(N log (log N)) 

Espacio Auxiliar: O(N)