Given an array arr containing non-negative integers, the task is to print the length of the longest subsequence of prime numbers in the array.
Examples:
Input: arr[] = { 3, 4, 11, 2, 9, 21 }
Output: 3
Longest Prime Subsequence is {3, 2, 11} and hence the answer is 3.
Input: arr[] = { 6, 4, 10, 13, 9, 25 }
Output: 1
Approach:
- Traverse the given array.
- For each element in the array, check if it prime or not.
- If the element is prime, it will be in Longest Prime Subsequence. Hence, increment the required length of Longest Prime Subsequence by 1
Below is the implementation of the above approach:
C++
// 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 primesvoid 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 numbersint 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 codeint 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
// Java program to find the length of// Longest Prime Subsequence in an Arrayimport java.util.*;class GFG{static final int N = 100005; // Function to create Sieve// to check primesstatic 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 numbersstatic 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 codepublic 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 |
Python3
# Python 3 program to find the length of# Longest Prime Subsequence in an Array N = 100005 # Function to create Sieve# to check primesdef 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 numbersdef 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 codeif __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#
// C# program to find the length of// longest Prime Subsequence in an Arrayusing System;class GFG{static readonly int N = 100005; // Function to create Sieve// to check primesstatic 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 numbersstatic 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 codepublic 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 |
Javascript
<script>// JavaScript program to find the length of// Longest Prime Subsequence in an Arraylet N = 100005// Function to create Sieve// to check primesfunction 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 numbersfunction 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 codelet arr = [ 3, 4, 11, 2, 9, 21 ];let n = arr.length// Function calldocument.write(longestPrimeSubsequence(arr, n) + "<br>");// This code is contributed by gfgking</script> |
Output
3
Time Complexity: O(N log (log N))
Auxiliary Space: O(N)
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