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LeetCode: 1921-Eliminate Maximum Number of Monsters 解題紀錄

Last Updated on 2023-11-07 by Clay

題目

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] <= 105

題目的情境為你正在玩一款遊戲,遊戲中你需要從怪物手中保衛城鎮,並且怪獸的初始位置都儲存在 dist 中,他們朝城鎮移動的速度為 speed

你的手上有武器可以擊殺怪物,但是一次你只能擊殺一支怪物(移動距離不限)—— 請返回你能擊殺的怪獸最大數量(INT)。


解題思路

最直覺的做法就是計算出『每個怪獸抵達城市所需要的時間』,排序後再每擊敗一隻怪獸增加一次時間 —— 直到我們還來不及擊敗怪獸、怪獸就已經抵達了城市。過程中自然是要統計好我們擊敗的怪獸數,畢竟這才是題目要的答案。

我最早開始是想要使用優先權佇列priority queue)去自動排序,但後來意識到每次加入元素就重新排序一次反而變成浪費時間;於是就很單純地使用陣列儲存怪獸抵達城市的時間,然後直接內建的 sort 排序了。


C++ 範例程式碼

class Solution {
public:
    int eliminateMaximum(vector<int>& dist, vector<int>& speed) {
        // Init
        int eliminate = 0;
        int turn = 1;
        std::vector<int> rounds;
        for (int i=0; i<dist.size(); ++i) {
            int round = (dist[i] + speed[i] - 1) / speed[i];
            rounds.emplace_back(round);
        }

        // Sort
        sort(rounds.begin(), rounds.end());

        // Eliminate monster
        for (int& round: rounds) {
            if (round < turn) {
                break;
            }

            ++eliminate;
            ++turn;
        }

        return eliminate;
    }
};



Python 範例程式碼

class Solution:
    def eliminateMaximum(self, dist: List[int], speed: List[int]) -> int:
        # Init
        eliminate = 0
        turns = 1
        rounds = [(dist[i] + speed[i] - 1) // speed[i] for i in range(len(dist))]

        # Sort
        rounds = sorted(rounds)

        # Eliminate monster
        for round in rounds:
            if round < turns:
                break

            eliminate += 1
            turns += 1

        return eliminate



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