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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 using namespace std;    // Function to store the primes void sieve(int maxEle, int prime[]) {     prime = prime = 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);        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 = prime = 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, prime = 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 =  * (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 = prime = 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



Output:

2 11 7

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