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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 $O(1)$.

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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

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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;
    }
}

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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;
    }
};

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

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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;
}

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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
    }
}

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int countDigits(int num) {
    int ans = 0;
    int cur = num;
    while (cur) {
        if (num % (cur % 10) == 0) {
            ans++;
        }
        cur /= 10;
    }
    return ans;
}

Solution 2

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function countDigits(num: number): number {
    let ans = 0;
    for (const s of num.toString()) {
        if (num % Number(s) === 0) {
            ans++;
        }
    }
    return ans;
}

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impl Solution {
    pub fn count_digits(num: i32) -> i32 {
        num.to_string()
            .chars()
            .filter(|&c| c != '0')
            .filter(|&c| num % (c.to_digit(10).unwrap() as i32) == 0)
            .count() as i32
    }
}

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