Question

2520. Count the Digits That Divide a Number

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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:
  def countDigits(self, num: int) -> int:
    ans, x = 0, num
    while x:
      x, val = divmod(x, 10)
      ans += num % val == 0
    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 {
public:
  int countDigits(int num) {
    int ans = 0;
    for (int x = num; x > 0; x /= 10) {
      if (num % (x % 10) == 0) {
        ++ans;
      }
    }
    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
  }
}

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

Solution 2

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

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
  }
}