Leetcode #728: Self Dividing Numbers
In this guide, we solve Leetcode #728 Self Dividing Numbers in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
A self-dividing number is a number that is divisible by every digit it contains. For example, 128 is a self-dividing number because 128 % 1 == 0, 128 % 2 == 0, and 128 % 8 == 0.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Math
Intuition
There is a mathematical invariant or formula that directly leads to the result.
Using math avoids unnecessary loops and reduces complexity.
Approach
Derive the formula or update rule, then compute the answer directly.
Handle edge cases like overflow or zero carefully.
Steps:
- Identify the math relationship.
- Compute the result with a loop or formula.
- Handle edge cases.
Example
Input: left = 1, right = 22
Output: [1,2,3,4,5,6,7,8,9,11,12,15,22]
Python Solution
class Solution:
def selfDividingNumbers(self, left: int, right: int) -> List[int]:
def check(x: int) -> bool:
y = x
while y:
if y % 10 == 0 or x % (y % 10):
return False
y //= 10
return True
return [x for x in range(left, right + 1) if check(x)]
Complexity
The time complexity is , where is the number of elements in the interval , and , which is the maximum value in the interval. The space complexity is O(1).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.