# 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

 // C++ program to count the number of // prime digits in a number    #include 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; }

## Java

 // 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

## Python3

 # 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

## C#

 // 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

Output:

4

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

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Improved By : mohit kumar 29, AnkitRai01

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