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1845. Seat Reservation Manager

Medium67.1% acceptance4 companies

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.

Asked at 4 companies

Hints open one at a time

Hint 1
You need a data structure that maintains the states of the seats. This data structure should also allow you to get the first available seat and flip the state of a seat in a reasonable time.
Hint 2
You can let the data structure contain the available seats. Then you want to be able to get the lowest element and erase an element, in a reasonable time.
Hint 3
Ordered sets support these operations.
7:42
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1class SeatManager:
2
3 def __init__(self, n: int):
4
5
6 def reserve(self) -> int:
7
8
9 def unreserve(self, seatNumber: int) -> None:
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11
12
13# Your SeatManager object will be instantiated and called as such:
14# obj = SeatManager(n)
15# param_1 = obj.reserve()
16# obj.unreserve(seatNumber)
Starter codePython 3Spaces: 4
Examples
Input
["SeatManager","reserve","reserve","unreserve","reserve","reserve","reserve","reserve","unreserve"]
[[5],[],[],[2],[],[],[],[],[5]]
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