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Kotlin-based LeetCode algorithm problem solutions, regularly updated
1845\. Seat Reservation Manager
Medium
Design a system that manages the reservation state of `n` seats that are numbered from `1` to `n`.
Implement the `SeatManager` class:
* `SeatManager(int n)` Initializes a `SeatManager` object that will manage `n` seats numbered from `1` to `n`. All seats are initially available.
* `int reserve()` Fetches the **smallest-numbered** unreserved seat, reserves it, and returns its number.
* `void unreserve(int seatNumber)` Unreserves the seat with the given `seatNumber`.
**Example 1:**
**Input** ["SeatManager", "reserve", "reserve", "unreserve", "reserve", "reserve", "reserve", "reserve", "unreserve"] [[5], [], [], [2], [], [], [], [], [5]]
**Output:** [null, 1, 2, null, 2, 3, 4, 5, null]
**Explanation:**
SeatManager seatManager = new SeatManager(5); // Initializes a SeatManager with 5 seats.
seatManager.reserve(); // All seats are available, so return the lowest numbered seat, which is 1.
seatManager.reserve(); // The available seats are [2,3,4,5], so return the lowest of them, which is 2.
seatManager.unreserve(2); // Unreserve seat 2, so now the available seats are [2,3,4,5].
seatManager.reserve(); // The available seats are [2,3,4,5], so return the lowest of them, which is 2.
seatManager.reserve(); // The available seats are [3,4,5], so return the lowest of them, which is 3.
seatManager.reserve(); // The available seats are [4,5], so return the lowest of them, which is 4.
seatManager.reserve(); // The only available seat is seat 5, so return 5.
seatManager.unreserve(5); // Unreserve seat 5, so now the available seats are [5].
**Constraints:**
* 1 <= n <= 105
* `1 <= seatNumber <= n`
* For each call to `reserve`, it is guaranteed that there will be at least one unreserved seat.
* For each call to `unreserve`, it is guaranteed that `seatNumber` will be reserved.
* At most 105
calls **in total** will be made to `reserve` and `unreserve`.