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Product of all prime numbers in an Array

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

Given an array arr[] of N positive integers. The task is to write a program to find the product of all the prime numbers of the given array.
Examples
 

Input: arr[] = {1, 3, 4, 5, 7} 
Output: 105 
There are three primes, 3, 5 and 7 whose product = 105.
Input: arr[] = {1, 2, 3, 4, 5, 6, 7} 
Output: 210 
 

 

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

C++




// CPP program to find product of
// primes in given array.
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the product of prime numbers
// in the given array
int primeProduct(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;
        }
    }
 
    // Product all primes in arr[]
    int prod = 1;
    for (int i = 0; i < n; i++)
        if (prime[arr[i]])
            prod *= arr[i];
 
    return prod;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << primeProduct(arr, n);
 
    return 0;
}

Java




// Java program to find product of
// primes in given array.
import java.util.*;
 
class GFG
{
 
// Function to find the product of prime numbers
// in the given array
static int primeProduct(int arr[], int n)
{
    // Find maximum value in the array
    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.
    Vector<Boolean> prime = new Vector<Boolean>(max_val + 1);
    for(int i = 0; i < max_val + 1; i++)
        prime.add(i, Boolean.TRUE);
 
    // Remaining part of SIEVE
    prime.add(0, Boolean.FALSE);
    prime.add(1, Boolean.FALSE);
    for (int p = 2; p * p <= max_val; p++)
    {
 
        // If prime[p] is not changed, then
        // it is a prime
        if (prime.get(p) == true)
        {
 
            // Update all multiples of p
            for (int i = p * 2; i <= max_val; i += p)
                prime.add(i, Boolean.FALSE);
        }
    }
 
    // Product all primes in arr[]
    int prod = 1;
    for (int i = 0; i < n; i++)
        if (prime.get(arr[i]))
            prod *= arr[i];
 
    return prod;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int n = arr.length;
 
    System.out.print(primeProduct(arr, n));
}
}
 
// This code has been contributed by 29AjayKumar

Python3




# Python3 program to find product of
# primes in given array
import math as mt
 
# function to find the product of prime
# numbers in the given array
def primeProduct(arr, n):
     
    # find the 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 = [True for i in range(max_val + 1)]
     
    # remaining part of SIEVE
    prime[0] = False
    prime[1] = False
     
    for p in range(mt.ceil(mt.sqrt(max_val))):
         
        # Remaining part of SIEVE
         
        # if prime[p] is not changed,
        # than it is prime
        if prime[p]:
             
            # update all multiples of p
            for i in range(p * 2, max_val + 1, p):
                prime[i] = False
     
    # product all primes in arr[]
    prod = 1
     
    for i in range(n):
        if prime[arr[i]]:
            prod *= arr[i]
     
    return prod
 
# Driver code
arr = [1, 2, 3, 4, 5, 6, 7]
 
n = len(arr)
 
print(primeProduct(arr, n))
 
# This code is contributed
# by Mohit kumar 29
                

C#




// C# program to find product of
// primes in given array.
using System;
using System.Linq;
using System.Collections.Generic;
 
class GFG
{
 
// Function to find the product of prime numbers
// in the given array
static int primeProduct(int []arr, int n)
{
    // Find maximum value in the array
    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.
    List<bool> prime = new List<bool>(max_val + 1);
    for(int i = 0; i < max_val + 1; i++)
        prime.Insert(i, true);
 
    // Remaining part of SIEVE
    prime.Insert(0, false);
    prime.Insert(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.Insert(i, false);
        }
    }
 
    // Product all primes in arr[]
    int prod = 1;
    for (int i = 0; i < n; i++)
        if (prime[arr[i]])
            prod *= arr[i];
 
    return prod;
}
 
// Driver code
public static void Main()
{
    int []arr = { 1, 2, 3, 4, 5, 6, 7 };
    int n = arr.Length;
 
    Console.Write(primeProduct(arr, n));
}
}
 
/* This code contributed by PrinciRaj1992 */

PHP




<?php
// PHP program to find product of
// primes in given array.
 
// Function to find the product of
// prime numbers in the given array
function primeProduct($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;
        }
    }
 
    // Product all primes in arr[]
    $prod = 1;
    for ($i = 0; $i < $n; $i++)
        if ($prime[$arr[$i]])
            $prod *= $arr[$i];
 
    return $prod;
}
 
// Driver code
$arr = array(1, 2, 3, 4, 5, 6, 7);
$n = sizeof($arr);
 
echo(primeProduct($arr, $n));
 
// This code contributed by Code_Mech
?>

Javascript




<script>
// Javascript program to find product of
// primes in given array.
 
// Function to find the product of
// prime numbers in the given array
function primeProduct(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;
        }
    }
 
    // Product all primes in arr[]
    let prod = 1;
    for (let i = 0; i < n; i++)
        if (prime[arr[i]])
            prod *= arr[i];
 
    return prod;
}
 
// Driver code
let arr = new Array(1, 2, 3, 4, 5, 6, 7);
let n = arr.length;
 
document.write(primeProduct(arr, n));
 
// This code contributed by _Saurabh_Jaiswal.
</script>
Output: 
210

 

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