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.

## Solution: TreeSet

Time complexity: O(nlogn)
Space complexity: O(n)

## C++

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