3270. Find the Key of the Numbers

#Problem

You are given three positive integers num1, num2, and num3.

The key of num1, num2, and num3 is defined as a four-digit number such that:

• Initially, if any number has less than four digits, it is padded with leading zeros.
• The iᵗʰ digit (1 <= i <= 4) of the key is generated by taking the smallest digit among the iᵗʰ digits of num1, num2, and num3.

Return the key of the three numbers without leading zeros (if any).

https://leetcode.cn/problems/find-the-key-of-the-numbers/

Example 1:

Input: num1 = 1, num2 = 10, num3 = 1000
Output: 0
Explanation:
On padding, num1 becomes "0001", num2 becomes "0010", and num3 remains "1000".

• The 1ˢᵗ digit of the key is min(0, 0, 1).
• The 2ⁿᵈ digit of the key is min(0, 0, 0).
• The 3ʳᵈ digit of the key is min(0, 1, 0).
• The 4ᵗʰ digit of the key is min(1, 0, 0).

Hence, the key is "0000", i.e. 0.

Example 2:

Input: num1 = 987, num2 = 879, num3 = 798
Output: 777

Example 3:

Input: num1 = 1, num2 = 2, num3 = 3
Output: 1

Constraints:

• 1 <= num1, num2, num3 <= 9999

#Test Cases

class Solution:
    def generateKey(self, num1: int, num2: int, num3: int) -> int:

import pytest

from solution import Solution


@pytest.mark.parametrize('num1, num2, num3, expected', [
    (1, 10, 1000, 0),
    (987, 879, 798, 777),
    (1, 2, 3, 1),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, num1, num2, num3, expected):
    assert sol.generateKey(num1, num2, num3) == expected

#Thoughts

#Code

class Solution:
    def generateKey(self, num1: int, num2: int, num3: int) -> int:
        nums = [num1, num2, num3]
        key = 0
        base = 1
        for _ in range(4):
            d = 9
            for j in range(3):
                nums[j], r = divmod(nums[j], 10)
                if r < d: d = r
            key += d * base
            base *= 10

        return key