LeetCode 2269: Find the K-Beauty of a Number ·

Table of Contents

Meta Data
#

Difficulty: easy First Attempt: 2026-01-04

Submissions: Runtime Error → Accepted

Approach
#

Convert num to a string; each length-k substring is a fixed window. After converting back to an integer:

Skip it if it is 0 (0 divides nothing and cannot be used as a divisor).
Otherwise check num % sub == 0.

Leading zeros are allowed: int("04") is 4.

My Attempts
#

Attempt 1 ended with return len(res), but res was an integer counter, so len() raised TypeError. return res passed.

Solution
#

class Solution:
    def divisorSubstrings(self, num: int, k: int) -> int:
        s = str(num)
        count = 0
        for right in range(k - 1, len(s)):
            sub = int(s[right - k + 1:right + 1])
            if sub and num % sub == 0:             # 0 is never a divisor
                count += 1
        return count

num has at most 10 digits, so converting each slice is negligible; for strict O(n), roll the value with sub = sub * 10 + new_digit - old_digit * 10^k.

Complexity
#

Time O(n · k) with n digits, space O(n).

Mistakes to Avoid
#

Dividing by a substring equal to 0: ZeroDivisionError.
Calling len() on an integer.

Interview Explanation
#

I turn the number into a string, convert every length-k substring back to an integer, and count it if it is non-zero and divides num. With at most 10 digits slicing is fast enough; to be strict, the substring value can be rolled forward instead.