Count of Prime digits in a Number

Given an integer N, the task is to count the number of prime digits in N.

Examples:

Input: N = 12
Output: 1
Explanation:
Digits of the number – {1, 2}
But, only 2 is prime number.

Input: N = 1032
Output: 2
Explanation:
Digits of the number – {1, 0, 3, 2}
3 and 2 are prime number

Approach: The idea is to iterate through all the digits of the number and check whether the digit is a prime or not. Since there are only four possible prime numbers in the range [0, 9] and every digit for sure lies in this range, we only need to check the number of digits equal to either of the elements in the set {2, 3, 5, 7}.



Below is the implementation of this approach:

CPP

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// C++ program to count the number of
// prime digits in a number
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the count of
// prime digits in a number
int countDigit(int n)
{
    int temp = n, count = 0;
  
    // Loop to compute all the digits
    // of the number
    while (temp != 0) {
  
        // Finding every digit of the
        // given number
        int d = temp % 10;
  
        temp /= 10;
  
        // Checking if digit is prime or not
        // Only 2, 3, 5 and 7 are prime
        // one-digit number
        if (d == 2 || d == 3
            || d == 5 || d == 7)
            count++;
    }
  
    return count;
}
  
// Driver code
int main()
{
    int n = 1234567890;
  
    cout << countDigit(n) << endl;
    return 0;
}

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Java

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// Java program to count the number of
// prime digits in a number
class GFG {
      
    // Function to find the count of
    // prime digits in a number
    static int countDigit(int n)
    {
        int temp = n, count = 0;
      
        // Loop to compute all the digits
        // of the number
        while (temp != 0) {
      
            // Finding every digit of the
            // given number
            int d = temp % 10;
      
            temp /= 10;
      
            // Checking if digit is prime or not
            // Only 2, 3, 5 and 7 are prime
            // one-digit number
            if (d == 2 || d == 3
                || d == 5 || d == 7)
                count++;
        }
      
        return count;
    }
      
    // Driver code
    public static void main (String[] args)
    {
        int n = 1234567890;
      
        System.out.println(countDigit(n)) ;
  
    }
}
  
// This code is contributed by AnkitRai01

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Python3

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# Python3 program to count the number of
# prime digits in a number
  
# Function to find the count of
# prime digits in a number
def countDigit(n):
    temp = n
    count = 0
  
    # Loop to compute all the digits
    # of the number
    while (temp != 0):
  
        # Finding every digit of the
        # given number
        d = temp % 10
  
        temp //= 10
  
        # Checking if digit is prime or not
        # Only 2, 3, 5 and 7 are prime
        # one-digit number
        if (d == 2 or d == 3 or d == 5 or d == 7):
            count += 1
  
    return count
  
# Driver code
if __name__ == '__main__':
    n = 1234567890
  
    print(countDigit(n))
  
# This code is contributed by mohit kumar 29

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C#

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// C# program to count the number of
// prime digits in a number
using System;
  
class GFG {
      
    // Function to find the count of
    // prime digits in a number
    static int countDigit(int n)
    {
        int temp = n, count = 0;
      
        // Loop to compute all the digits
        // of the number
        while (temp != 0) {
      
            // Finding every digit of the
            // given number
            int d = temp % 10;
      
            temp /= 10;
      
            // Checking if digit is prime or not
            // Only 2, 3, 5 and 7 are prime
            // one-digit number
            if (d == 2 || d == 3
                || d == 5 || d == 7)
                count++;
        }
      
        return count;
    }
      
    // Driver code
    public static void Main (string[] args)
    {
        int n = 1234567890;
      
        Console.WriteLine(countDigit(n)) ;
    }
}
  
// This code is contributed by AnkitRai01

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Output:

4

Time Complexity: O(N), where N is the length of the number.

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Improved By : mohit kumar 29, AnkitRai01