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 $O(n)$.
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