How do you solve Seat Reservation Manager on LeetCode?

Seat Reservation Manager (LeetCode #1845) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(n) space complexity. Key insight: Using a priority queue allows for efficient seat management.

What is the brute force solution for Seat Reservation Manager?

The brute force approach for Seat Reservation Manager is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we maintain a list of seats and check each time we reserve or unreserve a seat. This is simple but inefficient as it requires scanning through the list to find available seats. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Seat Reservation Manager?

Seat Reservation Manager uses the following concepts: Design, Heap (Priority Queue). The recommended patterns to study are: Heap, Priority Queue.

What are the interview tips for Seat Reservation Manager?

Explain your thought process clearly as you code. Consider edge cases, such as unreserving a seat that was never reserved. Discuss the trade-offs of different data structures before jumping into code.

What are common mistakes when solving Seat Reservation Manager?

Not using a data structure that efficiently supports both operations. Overcomplicating the logic for reserving and unreserving seats.

#1845

Seat Reservation Manager

Medium

DesignHeap (Priority Queue)HeapPriority Queue

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log n)
Space	O(1)	O(n)

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Intuition

Time O(log n)Space O(n)

Using a min-heap (priority queue) allows us to efficiently manage the available seats. We can quickly fetch the smallest available seat and also efficiently unreserve seats.

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Algorithm

3 steps

1Step 1: Initialize a min-heap to keep track of available seats.
2Step 2: For the reserve() method, pop the smallest seat from the heap, mark it as reserved, and return its number.
3Step 3: For the unreserve(seatNumber) method, push the seat number back into the heap to mark it as available.

solution.py12 lines

1import heapq
2
3class SeatManager:
4    def __init__(self, n: int):
5        self.available_seats = list(range(1, n + 1))
6        heapq.heapify(self.available_seats)
7
8    def reserve(self) -> int:
9        return heapq.heappop(self.available_seats)
10
11    def unreserve(self, seatNumber: int) -> None:
12        heapq.heappush(self.available_seats, seatNumber)

ℹ

Complexity note: The time complexity is O(log n) for both reserve and unreserve operations due to the heap operations. The space complexity is O(n) for storing the seats in the heap.

1Using a priority queue allows for efficient seat management.
2Min-heaps are ideal for problems requiring frequent access to the smallest element.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.