Leetcode #625: Minimum Factorization
In this guide, we solve Leetcode #625 Minimum Factorization 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
Given a positive integer num, return the smallest positive integer x whose multiplication of each digit equals num. If there is no answer or the answer is not fit in 32-bit signed integer, return 0.
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Greedy, Math
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
Input: num = 48
Output: 68
Python Solution
class Solution:
def smallestFactorization(self, num: int) -> int:
if num < 2:
return num
ans, mul = 0, 1
for i in range(9, 1, -1):
while num % i == 0:
num //= i
ans = mul * i + ans
mul *= 10
return ans if num < 2 and ans <= 2**31 - 1 else 0
Complexity
The time complexity is O(n log n). The space complexity is O(1) to O(n).
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.