Problem: Count Number of Holes in Digits

You are given an integer N.
Each digit in the number contains a certain number of holes.

🔢 Hole Definition:

• Digits 0, 4, 6, 9 → have 1 hole
• Digit 8 → has 2 holes
• All other digits → have 0 holes

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🎯 Task

Calculate and print the total number of holes present in all digits of the given number.

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📥 Input Format

• A single integer N

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📤 Output Format

• Print a single integer representing the total number of holes

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📌 Constraints

• $1 \leq N \leq 10^9$1≤N≤109

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🧪 Example 1

Input:

649578

Output:

5

Explanation:

• 6 → 1 hole
• 4 → 1 hole
• 9 → 1 hole
• 5 → 0 holes
• 7 → 0 holes
• 8 → 2 holes

Total = 1 + 1 + 1 + 0 + 0 + 2 = 5

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🧪 Example 2

Input:

630

Output:

2

Explanation:

• 6 → 1 hole
• 3 → 0 holes
• 0 → 1 hole

Total = 2

C Program

#include <stdio.h>

int countHoles(int number) {
    int holes = 0;

    while (number > 0) {
        int digit = number % 10;

        if (digit == 0 || digit == 4 || digit == 6 || digit == 9) {
            holes += 1;
        } else if (digit == 8) {
            holes += 2;
        }

        number /= 10;
    }

    return holes;
}

int main() {
    int number;
    scanf("%d", &number);   // ❗ No print message

    printf("%d", countHoles(number));  // ❗ Only output

    return 0;
}