# 1921. Eliminate Maximum Number of Monsters

## 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 third 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

## Solutions

• 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;
}
}

• 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;
}
};

• class Solution:
def eliminateMaximum(self, dist: List[int], speed: List[int]) -> int:
times = sorted((d - 1) // s for d, s in zip(dist, speed))
for i, t in enumerate(times):
if t < i:
return i
return len(times)


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

• function eliminateMaximum(dist: number[], speed: number[]): number {
const n = dist.length;
const times = new Array(n).fill(0);
for (let i = 0; i < n; ++i) {
times[i] = Math.floor((dist[i] - 1) / speed[i]);
}
times.sort((a, b) => a - b);
for (let i = 0; i < n; ++i) {
if (times[i] < i) {
return i;
}
}
return n;
}


• /**
* @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;
};


• public 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];
}
Array.Sort(times);
for (int i = 0; i < n; ++i) {
if (times[i] < i) {
return i;
}
}
return n;
}
}