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2520 - Count the Digits That Divide a Number (Easy)

https://leetcode.com/problems/count-the-digits-that-divide-a-number/

Problem Statement​

Given an integer num, return the number of digits in num that dividenum.

An integer val divides nums if nums % val == 0.

Example 1:

Input: num = 7
Output: 1
Explanation: 7 divides itself, hence the answer is 1.

Example 2:

Input: num = 121
Output: 2
Explanation: 121 is divisible by 1, but not 2. Since 1 occurs twice as a digit, we return 2.

Example 3:

Input: num = 1248
Output: 4
Explanation: 1248 is divisible by all of its digits, hence the answer is 4.

Constraints:

  • 1 <= num <= 10^9
  • num does not contain 0 as one of its digits.

Approach 1: Convert to String​

Written by @wingkwong
class Solution {
public:
    int countDigits(int num) {
        int ans = 0;
        // convert num to string
        string s = to_string(num);
        // for each character
        for (auto c : s) {
            // convert it to number
            int d = c - '0';
            // check if it divisible
            ans += num % d == 0;
        }
        return ans;
    }
};

Approach 2: Check each digit​

Written by @wingkwong
class Solution {
public:
    int countDigits(int num) {
        int ans = 0, n = num;
        while (n > 0) {
            // check each digit using remainder
            ans += num % (n % 10) == 0;
            // move to the next digit
            n /= 10;
        }
        return ans;
    }
};