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:

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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 */
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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
chevron_right

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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
chevron_right

Output:
17

Time complexity : O(n*loglogn)




Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Article Tags :
Practice Tags :