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Count number of primes in an array

  • Difficulty Level : Easy
  • Last Updated : 21 May, 2021
Geek Week

Given an array arr[] of N positive integers. The task is to write a program to count the number of prime elements in the given array.
Examples
 

Input: arr[] = {1, 3, 4, 5, 7}
Output: 3
There are three primes, 3, 5 and 7

Input: arr[] = {1, 2, 3, 4, 5, 6, 7}
Output: 4

 

Naive Approach: A simple solution is to traverse the array and keep checking for every element if it is prime or not and keep the count of the prime elements at the same time.
Efficient Approach: Generate all primes upto maximum element of the array using sieve of Eratosthenes and store them in a hash. Now traverse the array and find the count of those elements which are prime using the hash table. 
Below is the implementation of above approach: 
 

C++




// CPP program to find count of
// primes in given array.
#include <bits/stdc++.h>
using namespace std;
 
// Function to find count of prime
int primeCount(int arr[], int n)
{
    // Find maximum value in the array
    int max_val = *max_element(arr, arr+n);
 
    // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
    // THAN OR EQUAL TO max_val
    // Create a boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    vector<bool> prime(max_val + 1, true);
 
    // Remaining part of SIEVE
    prime[0] = false;
    prime[1] = false;
    for (int p = 2; p * p <= max_val; p++) {
 
        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p] == true) {
 
            // Update all multiples of p
            for (int i = p * 2; i <= max_val; i += p)
                prime[i] = false;
        }
    }
 
    // Find all primes in arr[]
    int count = 0;
    for (int i = 0; i < n; i++)
        if (prime[arr[i]])
            count++;   
 
    return count;
}
 
// Driver code
int main()
{
 
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << primeCount(arr, n);
 
    return 0;
}

Java




import java.util.Arrays;
import java.util.Vector;
 
// Java program to find count of
// primes in given array.
class GFG
{
 
    // Function to find count of prime
    static int primeCount(int arr[], int n)
    {
        // Find maximum value in the array
        //.*max_element(arr, arr+n);
        int max_val = Arrays.stream(arr).max().getAsInt();
 
        // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
        // THAN OR EQUAL TO max_val
        // Create a boolean array "prime[0..n]". A
        // value in prime[i] will finally be false
        // if i is Not a prime, else true.
        Boolean[] prime = new Boolean[max_val + 1];
        for (int i = 0; i < max_val + 1; i++)
        {
            prime[i] = true;
        }
 
        // Remaining part of SIEVE
        prime[0] = false;
        prime[1] = false;
        for (int p = 2; p * p <= max_val; p++)
        {
 
            // If prime[p] is not changed, then
            // it is a prime
            if (prime[p] == true)
            {
 
                // Update all multiples of p
                for (int i = p * 2; i <= max_val; i += p)
                {
                    prime[i] = false;
                }
            }
        }
 
        // Find all primes in arr[]
        int count = 0;
        for (int i = 0; i < n; i++)
        {
            if (prime[arr[i]])
            {
                count++;
            }
        }
 
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {1, 2, 3, 4, 5, 6, 7};
        int n = arr.length;
        System.out.println(primeCount(arr, n));
    }
}
 
// This code is contributed by
// PrinciRaj1992

Python3




# Python 3 program to find count of
# primes in given array.
from math import sqrt
 
# Function to find count of prime
def primeCount(arr, n):
     
    # Find maximum value in the array
    max_val = arr[0];
    for i in range(len(arr)):
        if(arr[i] > max_val):
            max_val = arr[i]
 
    # USE SIEVE TO FIND ALL PRIME NUMBERS
    # LESS THAN OR EQUAL TO max_val
    # Create a boolean array "prime[0..n]".
    # A value in prime[i] will finally be
    # false if i is Not a prime, else true.
    prime =[ True for i in range(max_val + 1)]
 
    # Remaining part of SIEVE
    prime[0] = False
    prime[1] = False
    k = int(sqrt(max_val)) + 1
    for p in range(2, k, 1):
         
        # If prime[p] is not changed,
        # then it is a prime
        if (prime[p] == True):
             
            # Update all multiples of p
            for i in range(p * 2, max_val + 1, p):
                prime[i] = False
 
    # Find all primes in arr[]
    count = 0
    for i in range(0, n, 1):
        if (prime[arr[i]]):
            count += 1
 
    return count
 
# Driver code
if __name__ == '__main__':
    arr = [1, 2, 3, 4, 5, 6, 7]
    n = len(arr)
 
    print(primeCount(arr, n))
 
# This code is contributed by
# Shashank_Sharma

C#




// C# program to find count of
// primes in given array.
using System;
using System.Linq;
 
class GFG
{
 
    // Function to find count of prime
    static int primeCount(int []arr, int n)
    {
         
        // Find maximum value in the array
        //.*max_element(arr, arr+n);
        int max_val = arr.Max();
 
        // USE SIEVE TO FIND ALL PRIME NUMBERS LESS
        // THAN OR EQUAL TO max_val
        // Create a boolean array "prime[0..n]". A
        // value in prime[i] will finally be false
        // if i is Not a prime, else true.
        Boolean[] prime = new Boolean[max_val + 1];
        for (int i = 0; i < max_val + 1; i++)
        {
            prime[i] = true;
        }
 
        // Remaining part of SIEVE
        prime[0] = false;
        prime[1] = false;
        for (int p = 2; p * p <= max_val; p++)
        {
 
            // If prime[p] is not changed, then
            // it is a prime
            if (prime[p] == true)
            {
 
                // Update all multiples of p
                for (int i = p * 2; i <= max_val; i += p)
                {
                    prime[i] = false;
                }
            }
        }
 
        // Find all primes in arr[]
        int count = 0;
        for (int i = 0; i < n; i++)
        {
            if (prime[arr[i]])
            {
                count++;
            }
        }
        return count;
    }
 
    // Driver code
    public static void Main()
    {
        int []arr = {1, 2, 3, 4, 5, 6, 7};
        int n = arr.Length;
        Console.WriteLine(primeCount(arr, n));
    }
}
 
//This code is contributed by 29AjayKumar

PHP




<?php
// PHP program to find count
// of primes in given array.
 
// Function to find count of prime
function primeCount($arr, $n)
{
    // Find maximum value in the array
    $max_val = max($arr);
 
    // Use Sieve to find all Prime Numbers
    // less than or equal to max_val
    // Create a boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    $prime = array_fill(0, $max_val + 1, true);
 
    // Remaining part of SIEVE
    $prime[0] = false;
    $prime[1] = false;
    for ($p = 2; $p * $p <= $max_val; $p++)
    {
 
        // If prime[p] is not changed,
        // then it is a prime
        if ($prime[$p] == true)
        {
 
            // Update all multiples of p
            for ($i = $p * 2;
                 $i <= $max_val; $i += $p)
                $prime[$i] = false;
        }
    }
 
    // Find all primes in arr[]
    $count = 0;
    for ($i = 0; $i < $n; $i++)
        if ($prime[$arr[$i]])
            $count++;
 
    return $count;
}
 
// Driver code
$arr = array(1, 2, 3, 4, 5, 6, 7 );
$n = sizeof($arr);
 
echo primeCount($arr, $n);
 
// This code is contributed by mits
?>

Javascript




<script>
// Javascript program to find count
// of primes in given array.
 
// Function to find count of prime
function primeCount(arr, n)
{
    // Find maximum value in the array
    let max_val = arr.sort((a, b) => b - a)[0];
 
    // Use Sieve to find all Prime Numbers
    // less than or equal to max_val
    // Create a boolean array "prime[0..n]". A
    // value in prime[i] will finally be false
    // if i is Not a prime, else true.
    let prime = new Array(max_val + 1).fill(true);
 
    // Remaining part of SIEVE
    prime[0] = false;
    prime[1] = false;
    for (let p = 2; p * p <= max_val; p++)
    {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true)
        {
 
            // Update all multiples of p
            for (let i = p * 2;
                i <= max_val; i += p)
                prime[i] = false;
        }
    }
 
    // Find all primes in arr[]
    let count = 0;
    for (let i = 0; i < n; i++)
        if (prime[arr[i]])
            count++;
 
    return count;
}
 
// Driver code
let arr = new Array(1, 2, 3, 4, 5, 6, 7 );
let n = arr.length;
 
document.write(primeCount(arr, n));
 
// This code is contributed by _saurabh_jaiswal
</script>
Output: 
4

 




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