Meeting Rooms III Python Solution with Tests

lcpy gen -n 2402   # by problem number
lcpy gen -s meeting_rooms_iii   # by problem name

Input: n = 2, meetings = [[0,10],[1,5],[2,7],[3,4]]
Output: 0
Explanation:
- At time 0, both rooms are not being used. The first meeting starts in room 0.
- At time 1, only room 1 is not being used. The second meeting starts in room 1.
- At time 2, both rooms are being used. The third meeting is delayed.
- At time 3, both rooms are being used. The fourth meeting is delayed.
- At time 5, the meeting in room 1 finishes. The third meeting starts in room 1 for the time period [5,10).
- At time 10, the meetings in both rooms finish. The fourth meeting starts in room 0 for the time period [10,11).
Both rooms 0 and 1 held 2 meetings, so we return 0.

Input: n = 3, meetings = [[1,20],[2,10],[3,5],[4,9],[6,8]]
Output: 1
Explanation:
- At time 1, all three rooms are not being used. The first meeting starts in room 0.
- At time 2, rooms 1 and 2 are not being used. The second meeting starts in room 1.
- At time 3, only room 2 is not being used. The third meeting starts in room 2.
- At time 4, all three rooms are being used. The fourth meeting is delayed.
- At time 5, the meeting in room 2 finishes. The fourth meeting starts in room 2 for the time period [5,10).
- At time 6, all three rooms are being used. The fifth meeting is delayed.
- At time 10, the meetings in rooms 1 and 2 finish. The fifth meeting starts in room 1 for the time period [10,12).
Room 0 held 1 meeting while rooms 1 and 2 each held 2 meetings, so we return 1.

import heapq


class Solution:
    # Time: O(m log m + m log n)
    # Space: O(n)
    def most_booked(self, n: int, meetings: list[list[int]]) -> int:
        meetings.sort(key=lambda meeting: meeting[0])

        available: list[int] = list(range(n))
        ongoing: list[tuple[int, int]] = []
        counts = [0] * n

        for start, end in meetings:
            while ongoing and ongoing[0][0] <= start:
                _, room = heapq.heappop(ongoing)
                heapq.heappush(available, room)

            if available:
                room = heapq.heappop(available)
                counts[room] += 1
                heapq.heappush(ongoing, (end, room))
            else:
                free_time, room = heapq.heappop(ongoing)
                counts[room] += 1
                duration = end - start
                heapq.heappush(ongoing, (free_time + duration, room))

        best_room = 0
        for room in range(1, n):
            if counts[room] > counts[best_room]:
                best_room = room

        return best_room