Seat Reservation Manager Leetcode Problem [Python Solution]

Welcome to another coding adventure! Today, we’re going to tackle a problem from LeetCode called the Seat Reservation Manager.

This is a great problem for beginners, as it helps you understand how to efficiently manage reservations in a system.

We’ll break down the problem, discuss the constraints, and then provide a Python solution.

Problem Overview

The problem statement asks us to design a system that manages the reservation state of n seats, which are numbered from 1 to n.

We need to implement the SeatManager class with the following methods:

1. SeatManager(int n): This method initializes a SeatManager object that will manage n seats, numbered from 1 to n.

All seats are initially available.

2. int reserve(): This method fetches the smallest-numbered unreserved seat, reserves it, and returns its number.
3. void unreserve(int seatNumber): This method unreserves the seat with the given seatNumber.

Example:

Let’s take an example to illustrate how this works:

Input
["SeatManager", "reserve", "reserve", "unreserve", "reserve", "reserve", "reserve", "reserve", "unreserve"]
[[5], [], [], [2], [], [], [], [], [5]]

Output
[null, 1, 2, null, 2, 3, 4, 5, null]

In this example, we initialize a SeatManager with 5 seats.

Then, we perform a series of reservations and unreservations.

The output shows the results of these operations.

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.

Now that we understand the problem, let’s dive into the solution.

Efficient Python Solution

We’ll solve this problem efficiently using a min-heap.

A min-heap is a great data structure for keeping track of the smallest available seat.

Here’s the Python code solution:

class SeatManager:
    def __init__(self, n: int):
        self.seats = [i for i in range(1, n + 1)]

    def reserve(self) -> int:
        return heapq.heappop(self.seats)

    def unreserve(self, seatNumber: int) -> None:
        heapq.heappush(self.seats, seatNumber)

The SeatManager class has three methods:

1. __init__(self, n): The constructor initializes the system with n seats.

We create a list self.seats containing seat numbers from 1 to n.

This list will serve as our min-heap.

2. reserve(self): This method reserves the smallest available seat by popping it from the min-heap using heapq.heappop.
3. unreserve(self, seatNumber): When a seat is unreserved, we push it back into the min-heap using heapq.heappush.

This maintains the min-heap property.

This solution is efficient and ensures that the smallest available seat is always reserved.

Let’s take a closer look at the reasoning behind this approach.

Reasoning Behind Our Efficient Approach

We use a min-heap to efficiently track the smallest available seat.

When a seat is reserved, we pop the smallest seat from the min-heap.

When a seat is unreserved, we push it back into the min-heap.

This way, the min-heap always contains the smallest available seat, making it an ideal data structure for this problem.

Time and Space Complexity

The time and space complexity of this solution is as follows:

• SeatManager.__init__: This method runs in O(n) time to initialize the min-heap, and it consumes O(n) space to store the list of seats.
• SeatManager.reserve: Reserving a seat with the min-heap operation takes O(log n) time, and it returns a seat number.

Thus, it has a time complexity of O(log n).

• SeatManager.unreserve: Unreserving a seat and pushing it back into the min-heap also takes O(log n) time.

Overall, our solution is efficient, with a space complexity of O(n) and a time complexity of O(log n) for both reservation and unreservation operations.

Constraints to Consider

We’ve designed this solution with the given constraints in mind:

• The number of seats (n) can be as large as 105, which is well within the capacity of our solution.
• The seat numbers range from 1 to n, making it a valid input range for our solution.
• The problem statement guarantees that there will always be at least one unreserved seat, so our code doesn’t need to handle edge cases where all seats are already reserved.

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Conclusion

In this blog post, we’ve tackled the Seat Reservation Manager problem on LeetCode.

We’ve provided an efficient Python solution that uses a min-heap to keep track of the smallest available seat.

This solution adheres to the given constraints and offers a clear explanation of the problem, the code, and the reasoning behind our approach.

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Coding is an exciting journey, and we’re here to help you along the way.

Question Link: Seat Reservation Manager LeetCode Problem

Thank you for joining us on this coding adventure, and we look forward to bringing you more solutions to interesting problems in the future.

Happy coding!