README_EN.md

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 meters 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 meters per minute.

The monsters start moving at minute 0. You have a weapon that you can choose to use at the start of every minute, including minute 0. You cannot use the weapon in the middle of a minute. The weapon can eliminate any monster that is still alive. You lose when any monster reaches your city. If a monster reaches the city exactly at the start of a minute, it counts as a loss, and the game ends before you can use your weapon in that minute.

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:
At the start of minute 0, the distances of the monsters are [1,3,4], you eliminate the first monster.
At the start of minute 1, the distances of the monsters are [X,2,3], you don't do anything.
At the start of minute 2, the distances of the monsters are [X,1,2], you eliminate the second monster.
At the start of minute 3, the distances of the monsters are [X,X,1], you eliminate the third monster.
All 3 monsters can be eliminated.

Example 2:

Input: dist = [1,1,2,3], speed = [1,1,1,1]
Output: 1
Explanation:
At the start of minute 0, the distances of the monsters are [1,1,2,3], you eliminate the first monster.
At the start of minute 1, 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:
At the start of minute 0, the distances of the monsters are [3,2,4], you eliminate the first monster.
At the start of minute 1, 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

Solutions

Python3

class Solution:
    def eliminateMaximum(self, dist: List[int], speed: List[int]) -> int:
        n = len(dist)
        times = [(dist[i] - 1) // speed[i] for i in range(n)]
        times.sort()
        for i in range(n):
            if times[i] < i:
                return i
        return n

Java

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

JavaScript

/**
 * @param {number[]} dist
 * @param {number[]} speed
 * @return {number}
 */
var eliminateMaximum = function (dist, speed) {
    let arr = [];
    for (let i = 0; i < dist.length; i++) {
        arr[i] = dist[i] / speed[i];
    }
    arr.sort((a, b) => a - b);
    let ans = 0;
    while (arr[0] > ans) {
        arr.shift();
        ++ans;
    }
    return ans;
};

C++

class Solution {
public:
    int eliminateMaximum(vector<int>& dist, vector<int>& speed) {
        int n = dist.size();
        vector<int> times;
        for (int i = 0; i < n; ++i) {
            times.push_back((dist[i] - 1) / speed[i]);
        }
        sort(times.begin(), times.end());
        for (int i = 0; i < n; ++i) {
            if (times[i] < i) {
                return i;
            }
        }
        return n;
    }
};

Go

func eliminateMaximum(dist []int, speed []int) int {
	n := len(dist)
	times := make([]int, n)
	for i := 0; i < n; i++ {
		times[i] = (dist[i] - 1) / speed[i]
	}
	sort.Ints(times)
	for i := 0; i < n; i++ {
		if times[i] < i {
			return i
		}
	}
	return n
}

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