Count prime numbers in range [L, R] whose single digit sum is also prime

Given two integers L and R. The task is to count the prime numbers in the range [L, R], whose single sum is also a prime number. 
A single sum is obtained by adding the digits of a number until a single digit is left.

Examples

Input: L = 5, R = 20 
Output: 3
Explanation: Prime numbers in the range L = 5 to R = 20 are {5, 7, 11, 13, 17, 19}
Their “single sum” of digits is {5, 7, 2, 4, 8, 1}.  
Only {5, 7, 2} are prime. Hence the answer is 3.

Input: L = 1, R = 10  
Output: 4
Explanation: Prime numbers in the range L = 1 to R = 10 are {2, 3, 5, 7}.  
Their “single sum” of digits is {2, 3, 5, 7}.  
Since all the numbers are prime, hence the answer is 4.

Approach:  The naive approach is to iterate for each number in the range [L, R] and check if the number is prime or not. If the number is prime, find the single sum of its digits and again check whether the single sum is prime or not. If the single sum is prime, then increment the counter and print the current element in the range [L, R].

Below is the implementation of the above approach.

C++14

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check whether number
// is prime or not
bool isPrime(int n)
{
    // Corner case
    if (n <= 1)
        return false;

    // Check from 2 to square root of n
    for (int i = 2; i <= sqrt(n); i++)
        if (n % i == 0)
            return false;

    return true;
}

// Function to find single digit sum
int SingleDigitSum(int& n)
{
    if (n <= 9)
        return n;
    return (n % 9 == 0) ? 9 : n % 9;
}

// Function to find single digit primes
int countSingleDigitPrimes(int l, int r)
{
    int count = 0, i;

    for (i = l; i <= r; i++) {
        if (isPrime(i)
            && isPrime(SingleDigitSum(i))) {
            count++;
        }
    }
    return count;
}

// Driver Code
int main()
{
    // Input range
    int L = 1, R = 10;

    // Function Call
    cout << countSingleDigitPrimes(L, R);

    return 0;
}

Java

// Java program to implement
// the above approach
class GFG
{

  // Function to check whether number
  // is prime or not
  static boolean isPrime(int n)
  {

    // Corner case
    if (n <= 1)
      return false;

    // Check from 2 to square root of n
    for (int i = 2; i <= Math.sqrt(n); i++)
      if (n % i == 0)
        return false;

    return true;
  }

  // Function to find single digit sum
  static int SingleDigitSum(int n)
  {
    if (n <= 9)
      return n;
    return (n % 9 == 0) ? 9 : n % 9;
  }

  // Function to find single digit primes
  static int countSingleDigitPrimes(int l, int r)
  {
    int count = 0, i;

    for (i = l; i <= r; i++) {
      if (isPrime(i)
          && isPrime(SingleDigitSum(i))) {
        count++;
      }
    }
    return count;
  }

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

    // Input range
    int L = 1, R = 10;

    // Function Call
    System.out.println(countSingleDigitPrimes(L, R));

  }
}

// This code is contributed by gfgking

Python3

# Python program for above approach
import math

# Function to check whether number
# is prime or not
def isPrime(n):

    # Corner case
    if n <= 1:
        return False

    # Check from 2 to square root of n
    for i in range(2, math.floor(math.sqrt(n)) + 1):
        if n % i == 0:
            return False

    return True

# Function to find single digit sum
def SingleDigitSum(n):
    if n <= 9:
        return n
    return 9 if (n % 9 == 0) else n % 9

# Function to find single digit primes
def countSingleDigitPrimes(l, r):
    count = 0
    i = None

    for i in range(l, r + 1):
        if isPrime(i) and isPrime(SingleDigitSum(i)):
            count += 1
    return count

# Driver Code

# Input range
L = 1
R = 10

# Function Call
print(countSingleDigitPrimes(L, R))

# This code is contributed by gfgking

C#

// C# program to implement
// the above approach
using System;
class GFG
{

  // Function to check whether number
  // is prime or not
  static bool isPrime(int n)
  {

    // Corner case
    if (n <= 1)
      return false;

    // Check from 2 to square root of n
    for (int i = 2; i <= Math.Sqrt(n); i++)
      if (n % i == 0)
        return false;

    return true;
  }

  // Function to find single digit sum
  static int SingleDigitSum(int n)
  {
    if (n <= 9)
      return n;
    return (n % 9 == 0) ? 9 : n % 9;
  }

  // Function to find single digit primes
  static int countSingleDigitPrimes(int l, int r)
  {
    int count = 0, i;

    for (i = l; i <= r; i++) {
      if (isPrime(i)
          && isPrime(SingleDigitSum(i))) {
        count++;
      }
    }
    return count;
  }

  // Driver Code
  public static void Main()
  {

    // Input range
    int L = 1, R = 10;

    // Function Call
    Console.Write(countSingleDigitPrimes(L, R));

  }
}

// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
    // JavaScript program for above approach

    // Function to check whether number
    // is prime or not
    const isPrime = (n) => {
        // Corner case
        if (n <= 1)
            return false;

        // Check from 2 to square root of n
        for (let i = 2; i <= Math.sqrt(n); i++)
            if (n % i == 0)
                return false;

        return true;
    }

    // Function to find single digit sum
    const SingleDigitSum = (n) => {
        if (n <= 9)
            return n;
        return (n % 9 == 0) ? 9 : n % 9;
    }

    // Function to find single digit primes
    const countSingleDigitPrimes = (l, r) => {
        let count = 0, i;

        for (i = l; i <= r; i++) {
            if (isPrime(i)
                && isPrime(SingleDigitSum(i))) {
                count++;
            }
        }
        return count;
    }

    // Driver Code

    // Input range
    let L = 1, R = 10;

    // Function Call
    document.write(countSingleDigitPrimes(L, R));

// This code is contributed by rakeshsahni

</script>

Output

4

Time Complexity: O((R – L)*N^(1/2)) where N is the prime number in the range [L, R]. 
Auxiliary Space: O(1)