Count Digits That Are Both Prime and Even

The program must accept an integer N as the input. The task is to count and print the number of digits that are both prime and even in the given number. If no such digits are found, print -1.

Note: Only 2 is both prime and even.

Boundary Condition(s):
1 <= N <= 10^9

Input Format:
The first line contains the integer N.

Output Format:
The first line contains either the count of the digits that are both prime and even or -1.

Example Input/Output 1:
Input:
12345

Output:
1

Explanation:
The only digit that is both prime and even in 12345 is 2. Hence the count is 1.

Example Input/Output 2:
Input:
34567

Output:
-1

Explanation:
There are no digits that are both prime and even in 34567.

n = input().strip()
count = n.count('2')
print(count if count > 0 else -1)

#include<stdio.h>

int main() {
    char n[12];
    scanf("%s", n);
    int count = 0;
    for(int i = 0; n[i]; i++) {
        if(n[i] == '2') count++;
    }
    printf("%d", (count > 0 ? count : -1));
    return 0;
}

#include <iostream>
#include <string>
using namespace std;

int main() {
    string n;
    cin >> n;
    int count = 0;
    for(char c : n) {
        if(c == '2') count++;
    }
    cout << (count > 0 ? count : -1);
    return 0;
}

import java.util.Scanner;

public class CountPrimeEvenDigits {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String n = sc.next();
        int count = 0;
        for(char c : n.toCharArray()) {
            if(c == '2') count++;
        }
        System.out.println(count > 0 ? count : -1);
    }
}