1174MediumArrayHash TableHeap (Priority Queue)

The Seat the Watched Guest Takes

Tracked in this browser only

Trains the technique from

LeetCode 1942The Number of the Smallest Unoccupied Chair

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A waiting room has an endless run of seats numbered 0, 1, 2 and onwards. visits[i] = [arrives, leaves] says when guest i comes in and when they go.

A guest coming in takes the free seat with the smallest number. A seat falls free the moment its guest goes, so a guest arriving exactly then may take it.

No two guests arrive at the same moment. Return the number of the seat taken by guest watched.

Examples

Example 1

Input: visits = [[1, 2], [2, 3], [3, 4]], watched = 2
Output: 0

Each guest goes exactly as the next comes in, so seat 0 is free every time and everyone sits there.

Example 2

Input: visits = [[1, 10], [2, 10]], watched = 1
Output: 1

Both guests stay until the same moment, so the second has to take the next seat up.

Example 3

Input: visits = [[1, 100], [2, 5], [3, 4], [6, 50]], watched = 3
Output: 1

By the time the last guest arrives, seats 1 and 2 have both fallen free, and the smaller number wins even though seat 2 fell free first.

Constraints

2 <= visits.length <= 10^4
visits[i].length == 2
1 <= visits[i][0] <= visits[i][1] <= 10^5
visits[i][0] < visits[i][1]
0 <= watched <= visits.length - 1
no two guests arrive at the same moment

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def smallest_chair(visits: list[list[int]], watched: int) -> int:

Java

public int smallestChair(int[][] visits, int watched)

The intended approach

Time O(n log n) over the guests.Space O(n) for the two collections.

Guests have to be handled in the order they arrive, not the order they are listed, so sort the visits by arrival first while keeping track of which guest is which. Then two collections carry the room. One holds the seats standing empty, always giving up the smallest number first, which is exactly what a guest takes. The other holds the seats in use, always giving up the one whose guest goes soonest. On each arrival, move every seat whose guest has already gone across to the empty collection, then hand the arriving guest the smallest empty seat, or a brand new seat when none is empty. A seat falls free at the very moment its guest goes, so the move across uses at or before the arrival rather than strictly before. Once the watched guest is seated the answer is known and nothing further need be done.

Pythonentry point: smallest_chair

def smallest_chair(visits, watched):
    import heapq
    order = sorted(range(len(visits)), key=lambda i: visits[i][0])
    empty = []
    busy = []
    fresh = 0
    for i in order:
        arrives = visits[i][0]
        leaves = visits[i][1]
        while busy and busy[0][0] <= arrives:
            when, seat = heapq.heappop(busy)
            heapq.heappush(empty, seat)
        if empty:
            seat = heapq.heappop(empty)
        else:
            seat = fresh
            fresh += 1
        if i == watched:
            return seat
        heapq.heappush(busy, (leaves, seat))
    return -1

The obvious version

Keep a plain list saying when each seat next falls free, and for every arriving guest scan it from seat zero upwards for the first seat already free, adding a new seat at the end when none is. Costs the guest count squared.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_smallest_chair(visits, watched):
    order = sorted(range(len(visits)), key=lambda i: visits[i][0])
    until = []
    for i in order:
        arrives = visits[i][0]
        leaves = visits[i][1]
        seat = -1
        for s in range(len(until)):
            if until[s] <= arrives:
                seat = s
                break
        if seat < 0:
            until.append(leaves)
            seat = len(until) - 1
        else:
            until[seat] = leaves
        if i == watched:
            return seat
    return -1

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in