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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 $O(1)$.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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
      }
  }
  
  

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