Introduction

In this blog post, we will explore how to solve the LeetCode problem "1845" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's a breakdown of the approach, complexity, and the Python code:

• Key Approach: Use a min-heap (priority queue) to store the available seat numbers. The reserve() operation retrieves the smallest available seat from the heap. The unreserve() operation adds the seat back to the heap.
• Efficiency: Using a min-heap ensures that the smallest available seat can be retrieved efficiently.
• Runtime & Storage Complexity:
  • Runtime: reserve() and unreserve() both take O(log n) time, where n is the number of seats. Initialization takes O(1) time.
  • Storage: O(n) to store all the seats in the heap in the worst-case scenario (when all seats are unreserved).

Code

    import heapq

class SeatManager:
    def __init__(self, n: int):
        self.available_seats = list(range(1, n + 1))
        heapq.heapify(self.available_seats)

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

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