LeetCode 1921. Eliminate Maximum Number of Monsters Solution in Java, C++, Python & More | Explanation + Code

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

Solution 1: Sorting + Greedy

We use the 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:

times[i] = \left\lfloor ^dist[i]-1⁄_speed[i] \right\rfloor

Next, we sort the times array in ascending order.

Then, we traverse the times array. For the i-th monster, if times[i] ≥ 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 × log n), and the space complexity is O(n). Here, n is the length of the array.

PythonJavaC++GoTypeScriptJavaScriptC#

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)(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)

/** * @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; };(code-box)

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; } }(code-box)