Sum of all prime numbers in an Array
Last Updated :
04 Aug, 2023
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++
#include <bits/stdc++.h>
using namespace std;
int primeSum( int arr[], int n)
{
int max_val = *max_element(arr, arr + n);
vector< bool > prime(max_val + 1, true );
prime[0] = false ;
prime[1] = false ;
for ( int p = 2; p * p <= max_val; p++) {
if (prime[p] == true ) {
for ( int i = p * 2; i <= max_val; i += p)
prime[i] = false ;
}
}
int sum = 0;
for ( int i = 0; i < n; i++)
if (prime[arr[i]])
sum += arr[i];
return sum;
}
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
import java.util.*;
class GFG
{
static int primeSum( int arr[], int n)
{
int max_val = Arrays.stream(arr).max().getAsInt();
Vector<Boolean> prime = new Vector<>(max_val + 1 );
for ( int i = 0 ; i < max_val + 1 ; i++)
prime.add(i,Boolean.TRUE);
prime.add( 0 ,Boolean.FALSE);
prime.add( 1 ,Boolean.FALSE);
for ( int p = 2 ; p * p <= max_val; p++)
{
if (prime.get(p) == true )
{
for ( int i = p * 2 ; i <= max_val; i += p)
prime.add(i,Boolean.FALSE);
}
}
int sum = 0 ;
for ( int i = 0 ; i < n; i++)
if (prime.get(arr[i]))
sum += arr[i];
return sum;
}
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));
}
}
|
Python
def primeSum( arr, n):
max_val = max (arr)
prime = [ True for i in range (max_val + 1 )]
prime[ 0 ] = False
prime[ 1 ] = False
for p in range ( 2 , max_val + 1 ):
if (p * p > max_val):
break
if (prime[p] = = True ):
for i in range (p * 2 , max_val + 1 , p):
prime[i] = False
sum = 0
for i in range (n):
if (prime[arr[i]]):
sum + = arr[i]
return sum
arr = [ 1 , 2 , 3 , 4 , 5 , 6 , 7 ]
n = len (arr)
print (primeSum(arr, n))
|
C#
using System;
using System.Linq;
using System.Collections.Generic;
class GFG
{
static int primeSum( int []arr, int n)
{
int max_val = arr.Max();
List< bool > prime = new List< bool >(max_val + 1);
for ( int i = 0; i < max_val + 1; i++)
prime.Insert(i, true );
prime.Insert(0, false );
prime.Insert(1, false );
for ( int p = 2; p * p <= max_val; p++)
{
if (prime[p] == true )
{
for ( int i = p * 2; i <= max_val; i += p)
prime.Insert(i, false );
}
}
int sum = 0;
for ( int i = 0; i < n; i++)
if (prime[arr[i]])
sum += arr[i];
return sum;
}
public static void Main(String[] args)
{
int []arr = { 1, 2, 3, 4, 5, 6, 7 };
int n = arr.Length;
Console.WriteLine(primeSum(arr, n));
}
}
|
Javascript
<script>
function primeSum(arr, n) {
let max_val = arr.sort((a, b) => b - a)[0];
let prime = new Array(max_val + 1).fill( true );
prime[0] = false ;
prime[1] = false ;
for (let p = 2; p * p <= max_val; p++) {
if (prime[p] == true ) {
for (let i = p * 2; i <= max_val; i += p)
prime[i] = false ;
}
}
let sum = 0;
for (let i = 0; i < n; i++)
if (prime[arr[i]])
sum += arr[i];
return sum;
}
let arr = [1, 2, 3, 4, 5, 6, 7];
let n = arr.length;
document.write(primeSum(arr, n));
</script>
|
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)
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