Find the Most Frequently Used Meeting Room

Problem Statement

You are given an integer n representing the number of meeting rooms numbered from 0 to n - 1, and a list of intervals meetings where meetings[i] = [start_i, end_i] represents a meeting scheduled to run from time start_i to end_i (exclusive).

Meetings are scheduled according to the following rules:

1. Meetings are processed in increasing order of their original start times. All original start times are unique.
2. Each meeting is assigned to the unused room with the lowest room number that is free at or before the meeting's scheduled start time.
3. If no room is free when a meeting is scheduled to start, the meeting is delayed until at least one room becomes free. The meeting retains its original duration (end_i - start_i).
4. When multiple rooms become free at the same time, the meeting is assigned to the room with the lowest room number.
5. If multiple meetings are delayed, they are scheduled in the order of their original start times.

Return the room number of the room that hosted the most meetings. If there is a tie, return the smallest room number.

Function Signature

def most_frequently_used_room(n: int, meetings: list[list[int]]) -> int:
    pass

Input and Output Format

• Input:
  • n: An integer representing the number of available rooms.
  • meetings: A list of lists of two integers [start, end] representing the original intervals.
• Output:
  • An integer representing the index of the most used room.

Examples

Example 1

Input: n = 2, meetings = [[0, 10], [1, 5], [2, 7], [3, 4]]
Output: 0
Explanation:
- At time 0, meeting [0, 10] starts in room 0.
- At time 1, meeting [1, 5] starts in room 1.
- At time 2, meeting [2, 7] is delayed since both rooms are occupied.
- At time 3, meeting [3, 4] is delayed.
- At time 5, room 1 becomes free. Meeting [2, 7] starts in room 1 for duration 5 (ends at time 10).
- At time 10, both room 0 and room 1 become free. Meeting [3, 4] takes room 0 (lowest index) for duration 1 (ends at time 11).
Room 0 held 2 meetings; room 1 held 2 meetings. Tie goes to room 0.

Example 2

Input: n = 3, meetings = [[1, 20], [2, 10], [3, 5], [4, 9], [6, 8]]
Output: 1
Explanation:
- At time 1, meeting [1, 20] starts in room 0.
- At time 2, meeting [2, 10] starts in room 1.
- At time 3, meeting [3, 5] starts in room 2.
- At time 4, meeting [4, 9] is delayed.
- At time 5, room 2 becomes free. Meeting [4, 9] starts in room 2 (duration 5, ends at 10).
- At time 6, meeting [6, 8] is delayed.
- At time 10, rooms 1 and 2 become free. Meeting [6, 8] starts in room 1 (lowest index, ends at 12).
Room 0 held 1 meeting, room 1 held 2 meetings, room 2 held 2 meetings. Tie between room 1 and 2 goes to room 1.

Constraints

• 1 <= n <= 100
• 1 <= len(meetings) <= 10^5
• meetings[i].length == 2
• 0 <= start_i < end_i <= 10^9
• All start_i are distinct.