0284MediumArraySortingHeap (Priority Queue)SimulationPrefix Sum

One-Way Tram Capacity Check

Tracked in this browser only

Trains the technique from

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 heritage tram makes one run along a line of numbered stops, starting at stop 0 and always moving to higher-numbered stops. It never doubles back.

bookings[i] = [party, board, leave] is a reserved party of party people who get on at stop board and get off at stop leave, with leave always after board. The tram is built to hold seats people at any one moment.

At a stop, everyone getting off steps down before anyone gets on, so a party that leaves at a stop hands its space to a party boarding at that same stop.

Return true when the single run can carry every reserved party, and false when some stretch of the line would put more than seats people on board.

Examples

Example 1

Input: bookings = [[2, 1, 5], [3, 5, 7]], seats = 3
Output: true

The party of 2 rides from stop 1 and steps down at stop 5, where the party of 3 gets on, so the tram carries 2 people and then 3, never above 3.

Example 2

Input: bookings = [[2, 1, 5], [3, 3, 7]], seats = 4
Output: false

From stop 3 to stop 5 both parties are on board at once, which is 5 people in a tram built for 4.

Example 3

Input: bookings = [[4, 0, 2], [4, 2, 4], [4, 4, 6]], seats = 4
Output: true

Each party steps down at the very stop where the next one gets on, so the tram carries 4 people on each leg of the run.

Constraints

1 <= bookings.length <= 1000
bookings[i].length == 3
1 <= party_i <= 100
0 <= board_i < leave_i <= 1000
1 <= seats <= 10^5

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 tram_run_fits(bookings: list[list[int]], seats: int) -> bool:

Java

public boolean tramRunFits(int[][] bookings, int seats)

The intended approach

Time O(m + S), S the 1001 possible stopsSpace O(S)

Keep one array indexed by stop and, for each booking, add the party size at its boarding stop and subtract the same size at its leaving stop. Running a total along the stops in order then gives the number of people on board along each leg, because the subtraction lands on the leaving stop itself and so is applied before that stop's boardings are counted. If any running total exceeds the seat count the run does not fit.

Pythonentry point: tram_run_fits

def tram_run_fits(bookings, seats):
    # Record each booking as a rise at the stop where the party boards and a matching
    # fall at the stop where it leaves.  Running those changes along the stops gives
    # the number of people aboard on each leg, and the fall being recorded at the
    # leaving stop itself is what lets a party board there in the same breath.
    # Time: O(m + S), S being the 1001 possible stops.  Space: O(S).
    stops = 1002
    delta = [0] * stops
    for party, board, leave in bookings:
        delta[board] += party
        delta[leave] -= party
    aboard = 0
    for s in range(stops):
        aboard += delta[s]
        if aboard > seats:
            return False
    return True

The obvious version

For every stop on the line, walk the whole booking list and add up the parties whose boarding stop is at or before it and whose leaving stop is after it. Report failure the first time such a total passes the seat count.

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_tram_run_fits(bookings, seats):
    for s in range(1001):
        load = 0
        for party, board, leave in bookings:
            if board <= s < leave:
                load += party
        if load > seats:
            return False
    return True

How this problem was checked

Reference solution executed against 9 test cases — 9 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