# 2520. Count the Digits That Divide a Number

## Description

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

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 <= 109
• num does not contain 0 as one of its digits.

## Solutions

Solution 1: Enumeration

We directly enumerate each digit $val$ of the integer $num$, and if $val$ can divide $num$, we add one to the answer.

After the enumeration, we return the answer.

The time complexity is $O(\log num)$, and the space complexity is $O(1)$.

• class Solution {
public int countDigits(int num) {
int ans = 0;
for (int x = num; x > 0; x /= 10) {
if (num % (x % 10) == 0) {
++ans;
}
}
return ans;
}
}

• class Solution {
public:
int countDigits(int num) {
int ans = 0;
for (int x = num; x > 0; x /= 10) {
if (num % (x % 10) == 0) {
++ans;
}
}
return ans;
}
};

• class Solution:
def countDigits(self, num: int) -> int:
ans, x = 0, num
while x:
x, val = divmod(x, 10)
ans += num % val == 0
return ans

• func countDigits(num int) (ans int) {
for x := num; x > 0; x /= 10 {
if num%(x%10) == 0 {
ans++
}
}
return
}

• function countDigits(num: number): number {
let ans = 0;
for (let x = num; x; x = (x / 10) | 0) {
if (num % (x % 10) === 0) {
++ans;
}
}
return ans;
}

• impl Solution {
pub fn count_digits(num: i32) -> i32 {
let mut ans = 0;
let mut cur = num;
while cur != 0 {
if num % (cur % 10) == 0 {
ans += 1;
}
cur /= 10;
}
ans
}
}