What is the brute force approach and complexity for Seat Reservation Manager?

Boolean array of taken seats; reserve scans from seat 1 for the first free one with time complexity O(n) and space complexity O(n).

What is the optimal approach and complexity for Seat Reservation Manager?

A counter covers the never-used seats; a min-heap holds only the seats that were given back with optimal time complexity O(log m) and space complexity O(m).

How do you identify the pattern for Seat Reservation Manager?

Look for key signals in the prompt: "smallest-numbered free seat" indicates We need the minimum of a set that keeps changing → a min-heap of free seats, not a scan.; "unreserve can give back any taken seat" indicates A returned seat can be smaller than seats still in use, so it has to be stored for reuse.; "n up to 10^5" indicates Don't build the whole structure up front. A counter covers the never-used block for free..

Seat Reservation Manager

Medium·LeetCode #1845 Pro OnlySlight Variation

LeetCode

Problem Statement

A hall has n seats numbered from 1 to n. Build a class SeatManager that hands out seats.

SeatManager(int n) starts with all n seats free.
int reserve() takes the smallest-numbered free seat, marks it taken, and returns its number.
void unreserve(int seatNumber) makes a taken seat free again.

Input

The number n, then a series of reserve() and unreserve(seatNumber) calls.

Output

For each reserve() call, the seat number it gave out. unreserve returns nothing.

Example 1

Input: n = 5; reserve(), reserve(), unreserve(2), reserve(), reserve(), reserve(), reserve(), unreserve(5)

Output: 1, 2, 2, 3, 4, 5

Seats 1 and 2 go out first. Seat 2 is given back, so the next reserve takes 2 again, because it is the smallest free seat. Then 3, 4 and 5 go out in order. Giving back 5 leaves it free.

Example 2

Input: n = 3; reserve(), reserve(), reserve(), unreserve(3), unreserve(1), reserve()

Output: 1, 2, 3, 1

After unreserve(3) and unreserve(1), seats 1 and 3 are free. The smallest one, 1, comes out first, even though seat 3 was given back earlier.

Constraints

• 1 ≤ n ≤ 10^5
• 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 10^5 calls in total will be made to reserve and unreserve

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“smallest-numbered free seat”

→ We need the minimum of a set that keeps changing → a min-heap of free seats, not a scan.

“unreserve can give back any taken seat”

→ A returned seat can be smaller than seats still in use, so it has to be stored for reuse.

“n up to 10^5”

→ Don't build the whole structure up front. A counter covers the never-used block for free.

Approach Complexity Ladder

Brute Force — one flag per seat, scan from 1O(n)O(n)

Boolean array of taken seats; reserve scans from seat 1 for the first free one

Better — min-heap of every free seatO(log n)O(n)

Put all n seats in a min-heap at the start; reserve pops the top, unreserve pushes back

Optimal — counter + min-heap of given-back seatsO(log m)O(m)

A counter covers the never-used seats; a min-heap holds only the seats that were given back

Brute Force Intuition

Keep taken[s] for every seat. reserve scans from seat 1 and takes the first seat whose flag is false. unreserve clears that seat's flag.

Recognition rule for this sub-pattern:

Items keep arriving; maintain the heap so each query is O(log K) or O(1).

Remember this · Heap — Top K

Trigger

"K largest / smallest / most frequent / closest", kth element, streaming top-K.

Think

Do I really need everything sorted, or only the best K?

Key question

Min-heap or max-heap of size K — which one lets me kick out the worst candidate?

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