Check whether the sum of prime elements of the array is prime or not

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

Complexity Analysis:

• Time Complexity: O(n * log(log n))
• Auxiliary Space: O(MAX)

Approach 2:Without Sieve Array(No Extra Space)

The previous code used a sieve to pre-compute all primes up to a certain limit, and then used this pre-computed information to check if the sum of primes in the array is also prime. This approach requires extra memory to store the sieve array and runs in O(MAX*log(log(MAX))) time complexity, where MAX is the limit up to which primes are computed.

The new code optimizes the previous approach by checking if each number in the array is prime as we go through it. Instead of pre-computing all primes up to a certain limit, we only need to check if each number is divisible by any prime less than or equal to its square root. This approach does not require extra memory to store the sieve array.

C++

#include <bits/stdc++.h>
#define ll long long int
using namespace std;

// Function to check if a number is prime
bool isPrime(int n)
{
    if (n <= 1)
        return false;

    for (int i = 2; i*i <= n; i++)
        if (n % i == 0)
            return false;

    return true;
}

// 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 (isPrime(arr[i]))
            sum += arr[i];

    // if sum is prime
    // then return yes
    if (isPrime(sum))
        return true;

    return false;
}

// Driver code
int main()
{
    // array of elements
    int arr[] = { 2,3,2,2 };
    int n = sizeof(arr) / sizeof(arr[0]);

    if (checkArray(arr, n))
        cout << "Yes";
    else
        cout << "No";

    return 0;
}

Java

/*package whatever //do not write package name here */

import java.lang.Math;

public class Main {
    // Function to check if a number is prime
    static boolean isPrime(int n) {
        if (n <= 1) {
            return false;
        }

        for (int i = 2; i <= Math.sqrt(n); i++) {
            if (n % i == 0) {
                return false;
            }
        }

        return true;
    }

    // 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 (isPrime(arr[i])) {
                sum += arr[i];
            }
        }

        // if sum is prime then return true
        if (isPrime(sum)) {
            return true;
        }

        return false;
    }

    // Driver code
    public static void main(String[] args) {
        int[] arr = {1, 2, 3};
        int n = arr.length;

        if (checkArray(arr, n)) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }
}

Python3

# Function to check if a number is prime
def isPrime(n):
    if n <= 1:
        return False

    for i in range(2, int(n**0.5) + 1):
        if n % i == 0:
            return False

    return True

# 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 isPrime(arr[i]):
            sum += arr[i]

    # if sum is prime
    # then return yes
    if isPrime(sum):
        return True

    return False

# Driver code
arr = [1, 2, 3]
n = len(arr)

if checkArray(arr, n):
    print("Yes")
else:
    print("No")

C#

using System;

public class MainClass {
    // Function to check if a number is prime
    static bool IsPrime(int n) {
        if (n <= 1) {
            return false;
        }

        for (int i = 2; i <= Math.Sqrt(n); i++) {
            if (n % i == 0) {
                return false;
            }
        }

        return true;
    }

    // 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 (IsPrime(arr[i])) {
                sum += arr[i];
            }
        }

        // if sum is prime then return true
        if (IsPrime(sum)) {
            return true;
        }

        return false;
    }

    // Driver code
    public static void Main() {
        int[] arr = {1, 2, 3};
        int n = arr.Length;

        if (CheckArray(arr, n)) {
            Console.WriteLine("Yes");
        } else {
            Console.WriteLine("No");
        }
    }
}

Javascript

// Function to check if a number is prime
function isPrime(n) {
    if (n <= 1) {
        return false;
    }

    for (let i = 2; i <= Math.sqrt(n); i++) {
        if (n % i === 0) {
            return false;
        }
    }

    return true;
}

// Function to check if the sum of
// prime is prime or not
function checkArray(arr, n) {
    // find sum of all prime number
    let sum = 0;
    for (let i = 0; i < n; i++) {
        if (isPrime(arr[i])) {
            sum += arr[i];
        }
    }

    // if sum is prime
    // then return yes
    if (isPrime(sum)) {
        return true;
    }

    return false;
}

// Driver code
const arr = [1, 2, 3];
const n = arr.length;

if (checkArray(arr, n)) {
    console.log("Yes");
} else {
    console.log("No");
}

// Contributed by adityasha4x71

Output

Yes

Complexity Analysis:

Time Complexity: O(n * sqrt(max(arr))).
Auxiliary Space: O(1)