Sum 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 sum of all prime elements in the given array.

Examples: 

Input: arr[] = {1, 3, 4, 5, 7} 
Output: 15 
There are three primes, 3, 5 and 7 whose sum =15.

Input: arr[] = {1, 2, 3, 4, 5, 6, 7} 
Output: 17 

Naive Approach: A simple solution is to traverse the array and keep checking for every element if it is prime or not and add 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 sum of those elements which are prime using the sieve.

Below is the implementation of the efficient approach: 

C++

// CPP program to find sum of
// primes in given array.
#include <bits/stdc++.h>
using namespace std;

// Function to find count of prime
int primeSum(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;
        }
    }

    // Sum all primes in arr[]
    int sum = 0;
    for (int i = 0; i < n; i++)
        if (prime[arr[i]])
            sum += arr[i];

    return sum;
}

// Driver code
int main()
{

    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << primeSum(arr, n);

    return 0;
}

Java

// Java program to find sum of
// primes in given array.
import java.util.*;

class GFG
{

// Function to find count of prime
static int primeSum(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<>(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);
        }
    }

    // Sum all primes in arr[]
    int sum = 0;
    for (int i = 0; i < n; i++)
        if (prime.get(arr[i]))
            sum += arr[i];

    return sum;
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int n = arr.length;
    System.out.print(primeSum(arr, n));
}
}

/* This code contributed by PrinciRaj1992 */

Python

# Python3 program to find sum of
# primes in given array.

# Function to find count of prime
def primeSum( 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=[True for i in range(max_val + 1)]

    # Remaining part of SIEVE
    prime[0] = False
    prime[1] = False
    for p in range(2, max_val + 1):
        if(p * p > max_val):
            break

        # 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

    # Sum all primes in arr[]
    sum = 0

    for i in range(n):
        if (prime[arr[i]]):
            sum += arr[i]

    return sum

# Driver code
arr =[1, 2, 3, 4, 5, 6, 7]

n = len(arr)

print(primeSum(arr, n))

# This code is contributed by mohit kumar 29

C#

// C# program to find sum of
// primes in given array.
using System;
using System.Linq;
using System.Collections.Generic;

class GFG
{

// Function to find count of prime
static int primeSum(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);
        }
    }

    // Sum all primes in arr[]
    int sum = 0;
    for (int i = 0; i < n; i++)
        if (prime[arr[i]])
            sum += arr[i];

    return sum;
}

// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 5, 6, 7 };
    int n = arr.Length;
    Console.WriteLine(primeSum(arr, n));
}
}

// This code contributed by Rajput-Ji

Javascript

<script>
// Javascript program to find sum of
// primes in given array.

// Function to find count of prime
function primeSum(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;
        }
    }

    // Sum all primes in arr[]
    let sum = 0;
    for (let i = 0; i < n; i++)
        if (prime[arr[i]])
            sum += arr[i];

    return sum;
}

// Driver code

let arr = [1, 2, 3, 4, 5, 6, 7];
let n = arr.length;

document.write(primeSum(arr, n));

// This code is contributed by _saurabh_jaiswal.
</script>

Output

17

Time complexity: O(M*loglogM) (where, M = maximum element present in arr)
Auxiliary Space: O(M), for the extra vector used. (where, M = maximum element present in arr)