# Product of all prime numbers in an Array

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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:

 `// CPP program to find product of ` `// primes in given array. ` `#include ` `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 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 prime = ``new` `Vector(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 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# 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 */`

 ` `

Output:
```210
```

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