3622 - Check Divisibility by Digit Sum and Product (Easy)

Problem Link

https://leetcode.com/problems/check-divisibility-by-digit-sum-and-product

Problem Statement

You are given a positive integer n. Determine whether n is divisible by the sumof the following two values:

• The digit sum of n (the sum of its digits).
• The digit product of n (the product of its digits).

Return true if n is divisible by this sum; otherwise, return false.

Example 1:

Input: n = 99

Output: true

Explanation:

Since 99 is divisible by the sum (9 + 9 = 18) plus product (9 * 9 = 81) of its digits (total 99), the output is true.

Example 2:

Input: n = 23

Output: false

Explanation:

Since 23 is not divisible by the sum (2 + 3 = 5) plus product (2 * 3 = 6) of its digits (total 11), the output is false.

Constraints:

• 1 <= n <= 1e6

Approach 1: Simulation

We can simply simulate digit extraction using modulo and integer division. The helper function ds calculates the digit sum by repeatedly adding n % 10, while dp calculates the digit product by multiplying each extracted digit. After processing a digit, n //= 10 removes it from the number. Once you get the idea, the code can be further simplified by combining them together.

Finally, the code adds the digit sum and digit product and checks whether the original $n$ is divisible by the result. It returns true when the remainder is zero and false otherwise. The time complexity is $O(log n)$, and the space complexity is $O(1)$.

class Solution:
    def checkDivisibility(self, n: int) -> bool:
        def ds(n):
            res = 0
            while n > 0:
                res += n % 10
                n //= 10
            return res
        def dp(n):
            res = 1
            while n > 0:
                res *= n % 10
                n //= 10
            return res
        return n % (dp(n) + ds(n)) == 0