Simulate meeting-room bookings and return busiest room

Problem Statement

You are tasked with managing a set of meeting rooms in an office building. You have n rooms, indexed from 0 to n-1. All rooms start empty. Given a list of meetings where each meeting is defined by a [start, end] interval, assign each meeting to an available room according to these rules:

1. A meeting will take place if there is an available room. If multiple rooms are free, assign the meeting to the room with the lowest index.
2. If no room is free, the meeting that starts earliest must wait until a room becomes free.
3. When a room becomes free, it can be used for the next scheduled meeting in the queue.
4. If multiple meetings are scheduled to start at the same time, process them in the order they appear in the input list.

Your goal is to return the index of the room that held the most meetings. If there is a tie, return the room with the smallest index.

Function Signature

def busiest_room(n: int, meetings: List[List[int]]) -> int:

Examples

Example 1: Input: n = 2, meetings = [[0, 10], [1, 5], [2, 7], [3, 4]]

• Room 0 takes [0, 10]. Room 1 takes [1, 5].
• At time 5, Room 1 becomes free. Room 1 takes [2, 7].
• At time 7, Room 1 becomes free. Room 1 takes [3, 4].
• Output: 1 (Room 0 held 1, Room 1 held 3)

Constraints

• 1 <= n <= 50,000
• 1 <= meetings.length <= 100,000
• 0 <= start < end <= 10^9

Edge Cases

• All meetings fit in one room.
• More meetings than rooms (waitlist logic required).
• Meetings starting exactly when others end.