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## [Solution] Meeting Rooms III solution leetcode

**Meeting Rooms III solution leetcode** – 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] = [start`

means that a meeting will be held during the _{i}, end_{i}]**half-closed** time interval `[start`

. All the values of _{i}, end_{i})`start`

are _{i}**unique**.

## [Solution] Meeting Rooms III solution leetcode

Meetings are allocated to rooms in the following manner:

- Each meeting will take place in the unused room with the
**lowest**number. - 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. - 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`

.

## [Solution] Meeting Rooms III solution leetcode

Input:n = 2, meetings = [[0,10],[1,5],[2,7],[3,4]]Output:0Explanation:- 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.

## [Solution] Meeting Rooms III solution leetcode

Input:n = 3, meetings = [[1,20],[2,10],[3,5],[4,9],[6,8]]Output:1Explanation:- 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.

## [Solution] Meeting Rooms III solution leetcode

`1 <= n <= 100`

`1 <= meetings.length <= 10`

^{5}`meetings[i].length == 2`

`0 <= start`

_{i}< end_{i}<= 5 * 10^{5}- All the values of
`start`

are_{i}**unique**.