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. 

Algorithm:

Step 1: Start
Step 2: Create the function sieve(), which accepts a prime[] array and maxEle (the array’s maximum element) as inputs.
a. As they are not primes, set prime[0] and prime[1] to 1.
b. Iterate from 2 to the square root of maxEle using a for a loop. If the current index I is not a prime number, use another                   for loop to set prime[j] to 1 for each j that is a multiple of I marking all of its multiples as not primes.
Step 3: Establish the digitSum() function, which receives an integer n as input and returns the sum of its digits.
Step 4: Create the function printAdditivePrime(), which accepts as input an integer array arr[] of size n.
a. Use the max element() function from the algorithm library to determine the maximum element maxEle in arr[].
b. Use memset to declare an integer array prime[] with a size of maxEle + 1 and initialize it to all 0s ().
c. To add 0s and 1s to the prime[] array, use the sieve() function.
d. Iterate through each element of arr[ using a for a loop.
e. If the current element is a prime (prime[arr[i]] == 0), use the digitSum() function to get its digit sum.
f. Check if the sum of the digits is a prime number (prime[sum] == 0). Print the current element if it is.
Step 5: End

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.