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Fuel to Bring Every Warden In

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Trains the technique from

LeetCode 2477Minimum Fuel Cost to Report to the Capital

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 park has n depots numbered 0 through n - 1, joined by n - 1 two-way tracks. roads[i] = [a, b] is a track between depot a and depot b, and the layout is such that exactly one route of tracks joins any two depots.

One warden waits at every depot other than depot 0, and every warden must reach depot 0, where the ranger station stands. Each of those depots also holds one van, and a van carries at most seats wardens.

Wardens travel by van along the tracks. A van burns one litre of fuel for each track it drives along. Wardens may change vans freely whenever they meet at a depot, and a van may be abandoned at any depot.

Return the least fuel, in litres, that gets every warden to depot 0.

Examples

Example 1

Input: roads = [[0, 1], [0, 2], [0, 3]], seats = 5
Output: 3

Each of the three wardens drives their own van straight along one track to depot 0, burning one litre each.

Example 2

Input: roads = [[0, 1], [1, 2], [2, 3], [3, 4]], seats = 2
Output: 6

The depots form a line 0-1-2-3-4. The track between depots 3 and 4 is driven once, the one between 2 and 3 once, the one between 1 and 2 twice, and the one between 0 and 1 twice, for six litres in total.

Example 3

Input: roads = [[0, 1]], seats = 10
Output: 1

A single warden drives one track, burning one litre.

Constraints

roads.length == 0 or roads.length >= 1
1 <= roads.length <= 99999
roads[i].length == 2
0 <= roads[i][0] <= 99999
0 <= roads[i][1] <= 99999
The tracks join n = roads.length + 1 depots so that exactly one route links any two of them
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 total_fuel(roads: list[list[int]], seats: int) -> int:

Java

public long totalFuel(int[][] roads, int seats)

The intended approach

Time O(n) for n depots: one stack walk to order the depots, one sweep back over them.Space O(n) for the neighbour lists, the order and the branch sizes.

Because exactly one route joins any two depots, every warden beyond a given track must cross that track exactly once, so the number of crossings of a track is the number of wardens in the branch it cuts off. Carrying that many wardens over one track needs that count divided by the seat count, rounded up, however the vans are shuffled, and the choices on one track never constrain another. Settle an order in which each depot appears after everything beyond it using an explicit stack, then sweep that order backwards accumulating branch sizes and charging each track as it is left behind.

Pythonentry point: total_fuel

def total_fuel(roads, seats):
    # one van per depot, all converging on depot 0; a road is crossed by every
    # rider from the branch beyond it, so it costs ceil(riders / seats) litres
    n = len(roads) + 1
    if n == 1:
        return 0
    nbr = [[] for _ in range(n)]
    for a, b in roads:
        nbr[a].append(b)
        nbr[b].append(a)

    # settle an order in which every depot appears after its own branch
    order = []
    above = [-1] * n
    stack = [0]
    seen = [False] * n
    seen[0] = True
    while stack:
        here = stack.pop()
        order.append(here)
        for nxt in nbr[here]:
            if not seen[nxt]:
                seen[nxt] = True
                above[nxt] = here
                stack.append(nxt)

    riders = [1] * n
    litres = 0
    for here in reversed(order):
        if here == 0:
            continue
        litres += (riders[here] + seats - 1) // seats
        riders[above[here]] += riders[here]
    return litres

The obvious version

Work out the crossings directly: from each depot, walk the tracks to depot 0 and add one to a tally for every track used. Then charge each track its tally divided by the seat count, rounded up. Repeats the walk once per depot instead of accumulating branch sizes in a single sweep.

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_total_fuel(roads, seats):
    n = len(roads) + 1
    if n == 1:
        return 0
    nbr = [[] for _ in range(n)]
    for a, b in roads:
        nbr[a].append(b)
        nbr[b].append(a)
    # count crossings road by road: walk out from each depot to depot 0 and tally
    crossed = {}
    for start in range(1, n):
        seen = [False] * n
        seen[start] = True
        trail = {start: None}
        queue = [start]
        head = 0
        while head < len(queue):
            here = queue[head]
            head += 1
            if here == 0:
                break
            for nxt in nbr[here]:
                if not seen[nxt]:
                    seen[nxt] = True
                    trail[nxt] = here
                    queue.append(nxt)
        step = 0
        while trail[step] is not None or step != start:
            back = trail[step]
            if back is None:
                break
            key = (min(step, back), max(step, back))
            crossed[key] = crossed.get(key, 0) + 1
            step = back
    total = 0
    for count in crossed.values():
        total += -(-count // seats)
    return total

How this problem was checked

Reference solution executed against 11 test cases — 11 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

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