Print prime numbers with prime sum of digits in an array

Given an array arr[] and the task is to print the additive primes in an array. 
Additive primes: Primes such that the sum of their digits is also a prime, such as 2, 3, 7, 11, 23 are additive primes but not 13, 19, 31 etc.
Examples:

Input: arr[] = {2, 4, 6, 11, 12, 18, 7}
Output: 2, 11, 7 

Input: arr[] = {2, 3, 19, 13, 25, 7}
Output: 2, 3, 7

A simple approach is to traverse through all array elements. For every element check if it is Additive prime or not. 
This above approach is fine when array is small or when array values are large. For large sized arrays having relatively small values, we use Sieve to store primes up to maximum element of the array. Then check if the current element is prime or not. If yes then check the sum of its digit is also prime or not. If yes then print that number. 
Below is the implementation of the above approach: 

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to store the primes
void sieve(int maxEle, int prime[])
{
    prime[0] = prime[1] = 1;

    for (int i = 2; i * i <= maxEle; i++) {
        if (!prime[i]) {
            for (int j = 2 * i; j <= maxEle; j += i)
                prime[j] = 1;
        }
    }
}

// Function to return the sum of digits
int digitSum(int n)
{
    int sum = 0;
    while (n) {
        sum += n % 10;
        n = n / 10;
    }
    return sum;
}

// Function to print additive primes
void printAdditivePrime(int arr[], int n)
{

    int maxEle = *max_element(arr, arr + n);

    int prime[maxEle + 1];
    memset(prime, 0, sizeof(prime));
    sieve(maxEle, prime);

    for (int i = 0; i < n; i++) {

        // If the number is prime
        if (prime[arr[i]] == 0) {
            int sum = digitSum(arr[i]);

            // Check if it's digit sum is prime
            if (prime[sum] == 0)
                cout << arr[i] << " ";
        }
    }
}

// Driver code
int main()
{

    int a[] = { 2, 4, 6, 11, 12, 18, 7 };
    int n = sizeof(a) / sizeof(a[0]);

    printAdditivePrime(a, n);

    return 0;
}

Java

// Java implementation of the above approach
import java.util.Arrays;

class GFG
{

// Function to store the primes
static void sieve(int maxEle, int prime[])
{
    prime[0] = prime[1] = 1;

    for (int i = 2; i * i <= maxEle; i++)
    {
        if (prime[i]==0)
        {
            for (int j = 2 * i; j <= maxEle; j += i)
                prime[j] = 1;
        }
    }
}

// Function to return the sum of digits
static int digitSum(int n)
{
    int sum = 0;
    while (n > 0)
    {
        sum += n % 10;
        n = n / 10;
    }
    return sum;
}

// Function to print additive primes
static void printAdditivePrime(int arr[], int n)
{

    int maxEle = Arrays.stream(arr).max().getAsInt();

    int prime[] = new int[maxEle + 1];
    sieve(maxEle, prime);

    for (int i = 0; i < n; i++)
    {

        // If the number is prime
        if (prime[arr[i]] == 0)
        {
            int sum = digitSum(arr[i]);

            // Check if it's digit sum is prime
            if (prime[sum] == 0)
                System.out.print(arr[i]+" ");
        }
    }
}

// Driver code
public static void main(String[] args)
{

    int a[] = { 2, 4, 6, 11, 12, 18, 7 };
    int n =a.length;
    printAdditivePrime(a, n);
}
}

// This code is contributed by chandan_jnu

Python3

# Python3 implementation of the
# above approach

# from math lib import sqrt
from math import sqrt

# Function to store the primes
def sieve(maxEle, prime) :

    prime[0], prime[1] = 1 , 1

    for i in range(2, int(sqrt(maxEle)) + 1) :
        if (not prime[i]) :
            for j in range(2 * i , maxEle + 1, i) :
                prime[j] = 1

# Function to return the sum of digits
def digitSum(n) :
    sum = 0
    while (n) :

        sum += n % 10
        n = n // 10
    return sum

# Function to print additive primes
def printAdditivePrime(arr, n):
    maxEle = max(arr)
    prime = [0] * (maxEle + 1)
    sieve(maxEle, prime)
    for i in range(n) :

        # If the number is prime
        if (prime[arr[i]] == 0):
            sum = digitSum(arr[i])

            # Check if it's digit sum is prime
            if (prime[sum] == 0) :
                print(arr[i], end = " ")

# Driver code
if __name__ == "__main__" :
    a = [ 2, 4, 6, 11, 12, 18, 7 ]
    n = len(a)
    printAdditivePrime(a, n)

# This code is contributed by Ryuga

C#

// C# implementation of the above approach
using System.Linq;
using System;

class GFG
{

// Function to store the primes
static void sieve(int maxEle, int[] prime)
{
    prime[0] = prime[1] = 1;

    for (int i = 2; i * i <= maxEle; i++)
    {
        if (prime[i] == 0)
        {
            for (int j = 2 * i; j <= maxEle; j += i)
                prime[j] = 1;
        }
    }
}

// Function to return the sum of digits
static int digitSum(int n)
{
    int sum = 0;
    while (n > 0)
    {
        sum += n % 10;
        n = n / 10;
    }
    return sum;
}

// Function to print additive primes
static void printAdditivePrime(int []arr, int n)
{

    int maxEle = arr.Max();

    int[] prime = new int[maxEle + 1];
    sieve(maxEle, prime);

    for (int i = 0; i < n; i++)
    {

        // If the number is prime
        if (prime[arr[i]] == 0)
        {
            int sum = digitSum(arr[i]);

            // Check if it's digit sum is prime
            if (prime[sum] == 0)
                Console.Write(arr[i] + " ");
        }
    }
}

// Driver code
static void Main()
{
    int[] a = { 2, 4, 6, 11, 12, 18, 7 };
    int n = a.Length;
    printAdditivePrime(a, n);
}
}

// This code is contributed by chandan_jnu

PHP

<?php
// PHP implementation of the above approach

// Function to store the primes
function sieve($maxEle, &$prime)
{
    $prime[0] = $prime[1] = 1;

    for ($i = 2; $i * $i <= $maxEle; $i++)
    {
        if (!$prime[$i])
        {
            for ($j = 2 * $i;
                 $j <= $maxEle; $j += $i)
                $prime[$j] = 1;
        }
    }
}

// Function to return the sum of digits
function digitSum($n)
{
    $sum = 0;
    while ($n)
    {
        $sum += $n % 10;
        $n = $n / 10;
    }
    return $sum;
}

// Function to print additive primes
function printAdditivePrime($arr, $n)
{

    $maxEle = max($arr);

    $prime = array_fill(0, $maxEle + 1, 0);
    sieve($maxEle, $prime);

    for ($i = 0; $i < $n; $i++)
    {

        // If the number is prime
        if ($prime[$arr[$i]] == 0)
        {
            $sum = digitSum($arr[$i]);

            // Check if it's digit sum is prime
            if ($prime[$sum] == 0)
                print($arr[$i] . " ");
        }
    }
}

// Driver code
$a = array(2, 4, 6, 11, 12, 18, 7);
$n = count($a);

printAdditivePrime($a, $n);

// This code is contributed by chandan_jnu
?>

Javascript

<script>

// Javascript implementation of the above approach

// Function to store the primes
function sieve(maxEle, prime)
{
    prime[0] = prime[1] = 1;

    for (var i = 2; i * i <= maxEle; i++) {
        if (!prime[i]) {
            for (var j = 2 * i; j <= maxEle; j += i)
                prime[j] = 1;
        }
    }
}

// Function to return the sum of digits
function digitSum(n)
{
    var sum = 0;
    while (n) {
        sum += n % 10;
        n = parseInt(n / 10);
    }
    return sum;
}

// Function to print additive primes
function printAdditivePrime(arr, n)
{

    var maxEle = arr.reduce((a,b)=> Math.max(a,b));

    var prime = Array(maxEle + 1).fill(0);
    sieve(maxEle, prime);

    for (var i = 0; i < n; i++) {

        // If the number is prime
        if (prime[arr[i]] == 0) {
            var sum = digitSum(arr[i]);

            // Check if it's digit sum is prime
            if (prime[sum] == 0)
                document.write( arr[i] + " ");
        }
    }
}

// Driver code
var a = [ 2, 4, 6, 11, 12, 18, 7 ];
var n = a.length;
printAdditivePrime(a, n);


</script>

Output:

2 11 7

Time Complexity: O(max*log(log(max))) where max is the maximum element in the array.

Auxiliary Space: O(maxEle), where maxEle is the largest element of the array a.