1845. Seat Reservation Manager
Description
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 aSeatManager
object that will managen
seats numbered from1
ton
. 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 givenseatNumber
.
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 thatseatNumber
will be reserved. - At most
105
calls in total will be made toreserve
andunreserve
.
Solutions
Solution 1: Priority Queue (Min-Heap)
We define a priority queue (min-heap) \(\textit{q}\) to store all the available seat numbers. Initially, we add all seat numbers from \(1\) to \(n\) into \(\textit{q}\).
When calling the reserve
method, we pop the top element from \(\textit{q}\), which is the smallest available seat number.
When calling the unreserve
method, we add the seat number back into \(\textit{q}\).
In terms of time complexity, the initialization time complexity is \(O(n)\) or \(O(n \times \log n)\), and the time complexity of the reserve
and unreserve
methods is both \(O(\log n)\). The space complexity is \(O(n)\).
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 |
|