1921. Eliminate Maximum Number of Monsters - Explanation

1. Sorting

class Solution:
    def eliminateMaximum(self, dist: List[int], speed: List[int]) -> int:
        minReach = [math.ceil(d / s) for d, s in zip(dist, speed)]
        minReach.sort()

        res = 0
        for minute in range(len(minReach)):
            if minute >= minReach[minute]:
                return res
            res += 1

        return res

Time & Space Complexity

Time complexity: $O(n \log n)$
Space complexity: $O(n)$

2. Sorting (Overwrting Input Array)

class Solution:
    def eliminateMaximum(self, dist: List[int], speed: List[int]) -> int:
        for i in range(len(dist)):
            dist[i] = math.ceil(dist[i] / speed[i])

        dist.sort()
        for minute in range(len(dist)):
            if minute >= dist[minute]:
                return minute

        return len(dist)

Time & Space Complexity

Time complexity: $O(n \log n)$
Space complexity: $O(1)$ or $O(n)$ depending on the sorting algorithm.

3. Min-Heap

class Solution:
    def eliminateMaximum(self, dist: List[int], speed: List[int]) -> int:
        minHeap = []
        for i in range(len(dist)):
            heapq.heappush(minHeap, dist[i] / speed[i])

        res = 0
        while minHeap:
            if res >= heapq.heappop(minHeap):
                return res
            res += 1

        return res

Time & Space Complexity

Time complexity: $O(n \log n)$
Space complexity: $O(n)$