LeetCode-in-Java

1921. Eliminate Maximum Number of Monsters

Medium

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:

Solution

import java.util.Arrays;

public class Solution {
    public int eliminateMaximum(int[] dist, int[] speed) {
        for (int i = 0; i < dist.length; i++) {
            dist[i] = (dist[i] - 1) / speed[i] + 1;
        }
        Arrays.sort(dist);
        int ans = 1;
        int time = 1;
        for (int i = 1; i < dist.length; i++) {
            if (dist[i] > time) {
                ans++;
                time++;
            } else {
                return ans;
            }
        }
        return ans;
    }
}