leetcode - 1921. Eliminate Maximum Number of Monsters

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Description

You are playing a video game where you are defending your city from a group of n monsters. You are given a 0-indexed integer array dist of size n, where dist[i] is the initial distance in kilometers of the ith monster from the city.

The monsters walk toward the city at a constant speed. The speed of each monster is given to you in an integer array speed of size n, where speed[i] is the speed of the ith monster in kilometers per minute.

You have a weapon that, once fully charged, can eliminate a single monster. However, the weapon takes one minute to charge. The weapon is fully charged at the very start.

You lose when any monster reaches your city. If a monster reaches the city at the exact moment the weapon is fully charged, it counts as a loss, and the game ends before you can use your weapon.

Return the maximum number of monsters that you can eliminate before you lose, or n if you can eliminate all the monsters before they reach the city.

Example 1:

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 thrid monster.
All 3 monsters can be eliminated.

Example 2:

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.

Example 3:

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.

Constraints:

n == dist.length == speed.length
1 <= n <= 105
1 <= dist[i], speed[i] <= 10^5

Solution

Solved after hints.

Sort the monsters by arrival time, start from the earliest monster.

Time complexity: o ( n log ⁡ n ) o(n\log n) o(nlogn)
Space complexity: o ( n ) o(n) o(n)

Code

class Solution:
    def eliminateMaximum(self, dist: List[int], speed: List[int]) -> int:
        monster_time = [dist[i] / speed[i] for i in range(len(dist))]
        monster_time.sort()
        cur_time = 0
        cnt = 0
        for each_monster_time in monster_time:
            if cur_time >= each_monster_time:
                return cnt
            cnt += 1
            cur_time += 1
        return cnt

本文标签： EliminateLeetCodeMonstersNumberMaximum