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++

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// 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;
}

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Java

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// 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 */

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Python

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# 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

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C#

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// 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

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Output:

17

Time complexity : O(n*loglogn)



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