LeetCode #1845 — MEDIUM

Seat Reservation Manager

Move from brute-force thinking to an efficient approach using design strategy.

Solve on LeetCode

The Problem

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Design

Example 1

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

Core Insight

What unlocks the optimal approach

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.
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.
Ordered sets support these operations.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1845: Seat Reservation Manager
class SeatManager {
    private PriorityQueue<Integer> q = new PriorityQueue<>();

    public SeatManager(int n) {
        for (int i = 1; i <= n; ++i) {
            q.offer(i);
        }
    }

    public int reserve() {
        return q.poll();
    }

    public void unreserve(int seatNumber) {
        q.offer(seatNumber);
    }
}

/**
 * Your SeatManager object will be instantiated and called as such:
 * SeatManager obj = new SeatManager(n);
 * int param_1 = obj.reserve();
 * obj.unreserve(seatNumber);
 */

// Accepted solution for LeetCode #1845: Seat Reservation Manager
type SeatManager struct {
	q hp
}

func Constructor(n int) SeatManager {
	q := hp{}
	for i := 1; i <= n; i++ {
		heap.Push(&q, i)
	}
	return SeatManager{q}
}

func (this *SeatManager) Reserve() int {
	return heap.Pop(&this.q).(int)
}

func (this *SeatManager) Unreserve(seatNumber int) {
	heap.Push(&this.q, seatNumber)
}

type hp struct{ sort.IntSlice }

func (h hp) Less(i, j int) bool { return h.IntSlice[i] < h.IntSlice[j] }
func (h *hp) Push(v any)        { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
	a := h.IntSlice
	v := a[len(a)-1]
	h.IntSlice = a[:len(a)-1]
	return v
}

/**
 * Your SeatManager object will be instantiated and called as such:
 * obj := Constructor(n);
 * param_1 := obj.Reserve();
 * obj.Unreserve(seatNumber);
 */

# Accepted solution for LeetCode #1845: Seat Reservation Manager
class SeatManager:
    def __init__(self, n: int):
        self.q = list(range(1, n + 1))

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

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


# Your SeatManager object will be instantiated and called as such:
# obj = SeatManager(n)
# param_1 = obj.reserve()
# obj.unreserve(seatNumber)

// Accepted solution for LeetCode #1845: Seat Reservation Manager
/**
 * [1845] Seat Reservation Manager
 *
 * 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 <= 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.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/seat-reservation-manager/
// discuss: https://leetcode.com/problems/seat-reservation-manager/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

struct SeatManager {
    hp: std::collections::BinaryHeap<i32>,
}

/**
 * `&self` means the method takes an immutable reference.
 * If you need a mutable reference, change it to `&mut self` instead.
 */
impl SeatManager {
    fn new(n: i32) -> Self {
        Self {
            hp: std::collections::BinaryHeap::from((-n..=-1).collect::<Vec<i32>>()),
        }
    }

    fn reserve(&mut self) -> i32 {
        let i = self.hp.pop().unwrap();

        -i
    }

    fn unreserve(&mut self, seat_number: i32) {
        self.hp.push(-seat_number);
    }
}

/**
 * Your SeatManager object will be instantiated and called as such:
 * let obj = SeatManager::new(n);
 * let ret_1: i32 = obj.reserve();
 * obj.unreserve(seatNumber);
 */

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_1845_example_1() {
        let mut seat_manager = SeatManager::new(5); // Initializes a SeatManager with 5 seats.

        assert_eq!(seat_manager.reserve(), 1); // All seats are available, so return the lowest numbered seat, which is 1.
        assert_eq!(seat_manager.reserve(), 2); // The available seats are [2,3,4,5], so return the lowest of them, which is 2.
        seat_manager.unreserve(2); // Unreserve seat 2, so now the available seats are [2,3,4,5].
        assert_eq!(seat_manager.reserve(), 2); // The available seats are [2,3,4,5], so return the lowest of them, which is 2.
        assert_eq!(seat_manager.reserve(), 3); // The available seats are [3,4,5], so return the lowest of them, which is 3.
        assert_eq!(seat_manager.reserve(), 4); // The available seats are [4,5], so return the lowest of them, which is 4.
        assert_eq!(seat_manager.reserve(), 5); // The only available seat is seat 5, so return 5.
        seat_manager.unreserve(5); // Unreserve seat 5, so now the available seats are [5].
    }
}

// Accepted solution for LeetCode #1845: Seat Reservation Manager
class SeatManager {
    private q = new MinPriorityQueue<number>();
    constructor(n: number) {
        for (let i = 1; i <= n; i++) {
            this.q.enqueue(i);
        }
    }

    reserve(): number {
        return this.q.dequeue();
    }

    unreserve(seatNumber: number): void {
        this.q.enqueue(seatNumber);
    }
}

/**
 * Your SeatManager object will be instantiated and called as such:
 * var obj = new SeatManager(n)
 * var param_1 = obj.reserve()
 * obj.unreserve(seatNumber)
 */

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(1) per op

Space

O(n)

Approach Breakdown

NAIVE

O(n) per op time

O(n) space

Use a simple list or array for storage. Each operation (get, put, remove) requires a linear scan to find the target element — O(n) per operation. Space is O(n) to store the data. The linear search makes this impractical for frequent operations.

OPTIMIZED DESIGN

O(1) per op time

O(n) space

Design problems target O(1) amortized per operation by combining data structures (hash map + doubly-linked list for LRU, stack + min-tracking for MinStack). Space is always at least O(n) to store the data. The challenge is achieving constant-time operations through clever structure composition.

Shortcut: Combine two data structures to get O(1) for each operation type. Space is always O(n).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.