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:
// 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 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 */ |
# 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# 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 |
<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> |
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)