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)

Another method:
 The approach for finding single-digit primes within a given range is to use the following observations:

• The only single-digit prime numbers are 2, 3, 5, and 7.
• The sum of the digits of a number is congruent to the number modulo 9.
• If a number is not divisible by 3, then its sum of digits is not divisible by 3.
• If a number is not divisible by 2 or 5, then its sum of digits is not divisible by 2 or 5.

Based on these observations, we can take the following approach:

• Check if the range includes any of the single-digit prime numbers (2, 3, 5, or 7) and count them.
• For each number in the range that is not a single-digit prime number, check if it is divisible by 2, 3, 5, or 7. If it is, skip it. Otherwise, calculate the sum of its digits and check if it is a single-digit prime number. If it is, increment the count.
• Return the count as the result.

Algorithm:

1. Declare the ‘countSingleDigitPrimes‘ function that will find the count of single-digit prime numbers in the given range.
2. Initialize a variable ‘count‘ to 0, which will keep track of the count of single-digit prime numbers in the range.
3. In the first for loop, iterate over numbers 2 to 7 (inclusive), and check if they are prime and if they lie in the given range. If they are both prime and in the range, increment the ‘count’ variable.
4. In the second for loop, iterate over numbers in the range that are greater than or equal to 8. For each number, check if it is divisible by 2, 3, 5, or 7, and continue to the next number if any of these conditions are true.
5. Calculate the sum of digits of the current number, and if it is 0, 1, 4, 7, or 9, continue to the next number.
6. Check if both the current number and its sum of digits are prime, and increment the ‘count’ variable if this condition is true.
7. Return the ‘count’ variable as the output of the ‘countSingleDigitPrimes’ function.

Below is the implementation of the above code:

C++

#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 primes
int countSingleDigitPrimes(int l, int r)
{
    int count = 0;

    // Check if range includes any single digit primes
    for (int i = 2; i <= 7 && i <= r; i++) {
        if (l <= i && isPrime(i)) {
            count++;
        }
    }

    // Check remaining numbers
    for (int i = max(l, 8); i <= r; i++) {
        if (i % 2 == 0 || i % 5 == 0) {
            continue;
        }
        if (i % 3 == 0 || i % 7 == 0) {
            continue;
        }
        int sum = i % 9;
        if (sum == 0 || sum == 1 || sum == 4 || sum == 7 || sum == 9) {
            continue;
        }
        if (isPrime(i) && isPrime(sum)) {
            count++;
        }
    }

    return count;
}

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

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

    return 0;
}

Java

import java.util.*;

public class Main {
  // Function to check whether number is prime or not
  public 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 primes
  public static int countSingleDigitPrimes(int l, int r)
  {
    int count = 0;

    // Check if range includes any single digit primes
    for (int i = 2; i <= 7 && i <= r; i++) {
      if (l <= i && isPrime(i)) {
        count++;
      }
    }

    // Check remaining numbers
    for (int i = Math.max(l, 8); i <= r; i++) {
      if (i % 2 == 0 || i % 5 == 0) {
        continue;
      }
      if (i % 3 == 0 || i % 7 == 0) {
        continue;
      }
      int sum = i % 9;
      if (sum == 0 || sum == 1 || sum == 4 || sum == 7
          || sum == 9) {
        continue;
      }
      if (isPrime(i) && isPrime(sum)) {
        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 Prajwal Kandekar

Python3

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, int(math.sqrt(n)) + 1):
        if n % i == 0:
            return False

    return True

# Function to find single digit primes


def countSingleDigitPrimes(l, r):
    count = 0

    # Check if range includes any single digit primes
    for i in range(2, 8):
        if l <= i <= r and isPrime(i):
            count += 1

    # Check remaining numbers
    for i in range(max(l, 8), r + 1):
        if i % 2 == 0 or i % 5 == 0:
            continue
        if i % 3 == 0 or i % 7 == 0:
            continue
        sum = i % 9
        if sum == 0 or sum == 1 or sum == 4 or sum == 7 or sum == 9:
            continue
        if isPrime(i) and isPrime(sum):
            count += 1

    return count


# Driver Code
L, R = 1, 10

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

C#

using System;

public class GFG {
  // Function to check whether number is prime or not
  public 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 primes
  public static int countSingleDigitPrimes(int l, int r)
  {
    int count = 0;

    // Check if range includes any single digit primes
    for (int i = 2; i <= 7 && i <= r; i++) {
      if (l <= i && isPrime(i)) {
        count++;
      }
    }

    // Check remaining numbers
    for (int i = Math.Max(l, 8); i <= r; i++) {
      if (i % 2 == 0 || i % 5 == 0) {
        continue;
      }
      if (i % 3 == 0 || i % 7 == 0) {
        continue;
      }
      int sum = i % 9;
      if (sum == 0 || sum == 1 || sum == 4 || sum == 7
          || sum == 9) {
        continue;
      }
      if (isPrime(i) && isPrime(sum)) {
        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 shivanisinghss2110

Javascript

// Function to check whether number
// is prime or not
function 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 primes
function countSingleDigitPrimes(l, r) {
let count = 0;

// Check if range includes any single digit primes
for (let i = 2; i <= 7 && i <= r; i++) {
if (l <= i && isPrime(i)) {
count++;
}
}

// Check remaining numbers
for (let i = Math.max(l, 8); i <= r; i++) {
if (i % 2 == 0 || i % 5 == 0) {
continue;
}
if (i % 3 == 0 || i % 7 == 0) {
continue;
}
let sum = i % 9;
if (sum == 0 || sum == 1 || sum == 4 || sum == 7 || sum == 9) {
continue;
}
if (isPrime(i) && isPrime(sum)) {
count++;
}
}

return count;
}

// Driver Code

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

// Function Call
console.log(countSingleDigitPrimes(L, R));

// This code is contributed by Vaibhav

Output:

4

Time Complexity: O((R – L + 1) * sqrt(R))

Space Complexity:  O(1)