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Check whether the sum of prime elements of the array is prime or not
  • Last Updated : 23 Apr, 2021

Given an array having N elements. The task is to check if the sum of prime elements of the array is prime or not.
Examples: 

Input: arr[] = {1, 2, 3}
Output: Yes
As there are two primes in the array i.e. 2 and 3. 
So, the sum of prime is 2 + 3 = 5 and 5 is also prime. 

Input: arr[] = {2, 3, 2, 2}
Output: No

Approach: First find prime number up to 10^5 using Sieve. Then iterate over all elements of the array. If the number is prime then add it to sum. And finally, check whether the sum is prime or not. If prime then prints Yes otherwise No.
Below is the implementation of the above approach:  

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
#define ll long long int
#define MAX 100000
using namespace std;
bool prime[MAX];
 
// Sieve to find prime
void sieve()
{
    memset(prime, true, sizeof(prime));
    prime[0] = prime[1] = false;
    for (int i = 2; i < MAX; i++)
        if (prime[i])
            for (int j = 2 * i; j < MAX; j += i)
                prime[j] = false;
         
     
}
 
// Function to check if the sum of
// prime is prime or not
bool checkArray(int arr[], int n)
{
    // find sum of all prime number
    ll sum = 0;
    for (int i = 0; i < n; i++)
        if (prime[arr[i]])
            sum += arr[i];
 
    // if sum is prime
    // then return yes
    if (prime[sum])
        return true;
 
    return false;
}
 
// Driver code
int main()
{
    // array of elements
    int arr[] = { 1, 2, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    sieve();
 
    if (checkArray(arr, n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java




// Java implementation of the above approach
import java.io.*;
 
class GFG {
     
static int MAX =100000;
 
static boolean prime[] = new boolean[MAX];
 
// Sieve to find prime
static void sieve()
{
    for(int i=0;i<MAX;i++)
    {
        prime[i] =true;
    }
    prime[0] = prime[1] = false;
    for (int i = 2; i < MAX; i++)
        if (prime[i])
            for (int j = 2 * i; j < MAX; j += i)
                prime[j] = false;
         
     
}
 
// Function to check if the sum of
// prime is prime or not
static boolean checkArray(int arr[], int n)
{
    // find sum of all prime number
    int sum = 0;
    for (int i = 0; i < n; i++)
        if (prime[arr[i]])
            sum += arr[i];
 
    // if sum is prime
    // then return yes
    if (prime[sum])
        return true;
 
    return false;
}
 
// Driver code
 
    public static void main (String[] args) {
    // array of elements
    int arr[] = { 1, 2, 3 };
    int n = arr.length;
 
    sieve();
 
    if (checkArray(arr, n))
        System.out.println("Yes");
    else
         System.out.println("No");
 
    }
}
// This code is contributed by shs..

Python3




# Python3 implementation of above approach
from math import gcd, sqrt
 
MAX = 100000
 
prime = [True] * MAX
 
# Sieve to find prime
def sieve() :
     
    # 0 and 1 are not prime numbers
    prime[0] = False
    prime[1] = False
     
    for i in range(2, MAX) :
 
        if prime[i] :
            for j in range(2**i, MAX, i) :
                prime[j] = False
     
# Function to check if the sum of
# prime is prime or not
def checkArray(arr, n) :
 
    # find sum of all prime number
    sum = 0
    for i in range(n) :
 
        if prime[arr[i]] :
            sum += arr[i]
 
    # if sum is prime
    # then return yes
    if prime[sum] :
        return True
 
    return False
 
# Driver code
if __name__ == "__main__" :
 
    # list of elements
    arr = [1, 2, 3]
    n = len(arr)
 
    sieve()
 
    if checkArray(arr, n) :
        print("Yes")
    else :
        print("No")
         
# This code is contributed by ANKITRAI1

C#




// C# implementation of the above approach
using System;
 
class GFG
{
static int MAX = 100000;
 
static bool[] prime = new bool[MAX];
 
// Sieve to find prime
static void sieve()
{
    for(int i = 0; i < MAX; i++)
    {
        prime[i] = true;
    }
    prime[0] = prime[1] = false;
    for (int i = 2; i < MAX; i++)
        if (prime[i])
            for (int j = 2 * i;
                     j < MAX; j += i)
                prime[j] = false;
}
 
// Function to check if the sum of
// prime is prime or not
static bool checkArray(int[] arr, int n)
{
    // find sum of all prime number
    int sum = 0;
    for (int i = 0; i < n; i++)
        if (prime[arr[i]])
            sum += arr[i];
 
    // if sum is prime
    // then return yes
    if (prime[sum])
        return true;
 
    return false;
}
 
// Driver code
public static void Main ()
{
    // array of elements
    int[] arr = new int[] { 1, 2, 3 };
    int n = arr.Length;
     
    sieve();
     
    if (checkArray(arr, n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by mits

PHP




<?php
// PHP implementation of the
// above approach
 
// Sieve to find prime
function sieve()
{
    $MAX = 100000;
    $prime = array($MAX);
    for($i = 0; $i < $MAX; $i++)
    {
        $prime[$i] = true;
    }
    $prime[0] = $prime[1] = false;
    for ($i = 2; $i < $MAX; $i++)
        if ($prime[$i])
            for ($j = 2 * $i;
                 $j < $MAX; $j += $i)
                $prime[$j] = false;
}
 
// Function to check if the sum of
// prime is prime or not
function checkArray($arr, $n)
{
    $prime = array(100000);
     
    // find sum of all prime number
    $sum = 0;
    for ($i = 0; $i < $n; $i++)
        if ($prime[$arr[$i]])
            $sum += $arr[$i];
 
    // if sum is prime
    // then return yes
    if ($prime[$sum])
        return true;
 
    return false;
}
 
// Driver code
$arr= array(1, 2, 3);
$n = sizeof($arr);
 
sieve();
 
if (checkArray($arr, $n))
    echo "Yes";
else
    echo "No";
 
// This code is contributed
// by Akanksha Rai
?>

Javascript




<script>
 
// JavaScript implementation of the
// above approach
 
// function check whether a number
// is prime or not
function isPrime(n)
{
    // Corner case
    if (n <= 1)
        return 0;
   
    // Check from 2 to n-1
    for (let i = 2; i < n; i++)
        if (n % i == 0)
            return 0;
   
    return 1;
}
 
var prime = new Array(5);
   
// Sieve to find prime
function sieve()
{
    for(i = 0; i <=5; i++)
    {
        prime[i] = isPrime(i);
    }
}
 
// Function to check if the sum of
// prime is prime or not
function checkArray(arr, n)
{
     
     
    // find sum of all prime number
    sum = 0;
    for (i = 0; i <= n; i++)
        if (prime[arr[i]])
            sum += arr[i];
 
    // if sum is prime
    // then return yes
    if (sum)
        return 1;
 
    return 0;
}
 
var arr= [1, 2, 3];
n = 3;
 
sieve();
 
if (checkArray(arr, n))
    document.write("Yes");
else
    document.write("No");
 
 
</script>
Output: 
Yes

 

Time Complexity: O(n * log(log n))

Auxiliary Space: O(MAX)

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