Eliminate Maximum Number of Monsters

lcpy gen -n 1921   # by problem number
lcpy gen -s eliminate_maximum_number_of_monsters   # by problem name

Input: dist = [1,3,4], speed = [1,1,1]
Output: 3
Explanation:
In the beginning, the distances of the monsters are [1,3,4]. You eliminate the first monster.
After a minute, the distances of the monsters are [X,2,3]. You eliminate the second monster.
After a minute, the distances of the monsters are [X,X,2]. You eliminate the third monster.
All 3 monsters can be eliminated.

Input: dist = [1,1,2,3], speed = [1,1,1,1]
Output: 1
Explanation:
In the beginning, the distances of the monsters are [1,1,2,3]. You eliminate the first monster.
After a minute, the distances of the monsters are [X,0,1,2], so you lose.
You can only eliminate 1 monster.

Input: dist = [3,2,4], speed = [5,3,2]
Output: 1
Explanation:
In the beginning, the distances of the monsters are [3,2,4]. You eliminate the first monster.
After a minute, the distances of the monsters are [X,0,2], so you lose.
You can only eliminate 1 monster.

class Solution:
    # Time: O(n log n)
    # Space: O(n)
    def eliminate_maximum(self, dist: list[int], speed: list[int]) -> int:
        arrival = sorted(-(-d // s) for d, s in zip(dist, speed, strict=True))
        for minute, time in enumerate(arrival):
            if time <= minute:
                return minute
        return len(arrival)