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 i^th 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 i^th 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

Solution 1: Sorting + Greedy

We use the $\textit{times}$ array to record the latest time each monster can be eliminated. For the $i$-th monster, the latest time it can be eliminated is:

$$\textit{times}[i] = \left\lfloor \frac{\textit{dist}[i]-1}{\textit{speed}[i]} \right\rfloor$$

Next, we sort the $\textit{times}$ array in ascending order.

Then, we traverse the $\textit{times}$ array. For the $i$-th monster, if $\textit{times}[i] \geq i$, it means the $i$-th monster can be eliminated. Otherwise, it means the $i$-th monster cannot be eliminated, and we return $i$ immediately.

If all monsters can be eliminated, we return $n$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array.

Python3

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)

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;
    }
}

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

TypeScript

function eliminateMaximum(dist: number[], speed: number[]): number {
    const n = dist.length;
    const times: number[] = 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;
}

JavaScript

/**
 * @param {number[]} dist
 * @param {number[]} speed
 * @return {number}
 */
var eliminateMaximum = function (dist, speed) {
    const n = dist.length;
    const times = 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;
};

C

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;
    }
}