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Kotlin-based LeetCode algorithm problem solutions, regularly updated
2402\. Meeting Rooms III
Hard
You are given an integer `n`. There are `n` rooms numbered from `0` to `n - 1`.
You are given a 2D integer array `meetings` where meetings[i] = [starti, endi]
means that a meeting will be held during the **half-closed** time interval [starti, endi)
. All the values of starti
are **unique**.
Meetings are allocated to rooms in the following manner:
1. Each meeting will take place in the unused room with the **lowest** number.
2. If there are no available rooms, the meeting will be delayed until a room becomes free. The delayed meeting should have the **same** duration as the original meeting.
3. When a room becomes unused, meetings that have an earlier original **start** time should be given the room.
Return _the **number** of the room that held the most meetings._ If there are multiple rooms, return _the room with the **lowest** number._
A **half-closed interval** `[a, b)` is the interval between `a` and `b` **including** `a` and **not including** `b`.
**Example 1:**
**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.
**Example 2:**
**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.
**Constraints:**
* `1 <= n <= 100`
* 1 <= meetings.length <= 105
* `meetings[i].length == 2`
* 0 <= starti < endi <= 5 * 105
* All the values of starti
are **unique**.