0664EasyGraph Theory

Hub Depot of the Courier Network

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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 courier company runs n depots labelled 1 through n. One of them is the hub: every other depot is joined to the hub by its own road, and there are no other roads, so the network holds exactly n - 1 roads and each road has the hub at one end.

roads lists those roads in no particular order. roads[i] = [a, b] records a road running between depot a and depot b, and the two ends of a road are always different depots.

Return the label of the hub depot.

Examples

Example 1

Input: roads = [[4, 2], [3, 4], [4, 5], [1, 4]]
Output: 4

There are four roads, so the network has five depots. Depot `4` is written on every road, and no other depot is.

Example 2

Input: roads = [[9, 7], [7, 2], [7, 10], [1, 7], [7, 5], [3, 7], [7, 8], [6, 7], [7, 4]]
Output: 7

Nine roads means ten depots. Depot `7` shares a road with each of the other nine depots, and every road listed has `7` on one side.

Example 3

Input: roads = [[1, 3], [2, 3]]
Output: 3

The smallest network of this shape has three depots and two roads. Depot `3` appears on both roads.

Constraints

3 <= n <= 10^5
roads.length == n - 1
roads[i].length == 2
1 <= a, b <= n
a != b
The roads describe a network of the shape above, so a hub always exists.

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 hub_depot(roads: list[list[int]]) -> int:

Java

public int hubDepot(int[][] roads)

The intended approach

Time O(1): two roads are read and each holds two labels.Space O(1).

Every road lists the hub as one of its two ends, so the hub is the only depot that can appear on both of the first two roads. Read the two ends of `roads[0]` and answer with whichever of them also appears in `roads[1]`. Nothing beyond the first two roads is examined.

Pythonentry point: hub_depot

def hub_depot(roads):
    # the hub is on every road, so it is the endpoint the first two roads share
    a, b = roads[0]
    return a if a in roads[1] else b

The obvious version

Tally how many roads touch each depot, then return the depot whose tally equals the number of roads, since only the hub sits on all of them.

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_hub_depot(roads):
    touches = {}
    for a, b in roads:
        touches[a] = touches.get(a, 0) + 1
        touches[b] = touches.get(b, 0) + 1
    wanted = len(roads)
    for depot, count in touches.items():
        if count == wanted:
            return depot
    return -1

How this problem was checked

Reference solution executed against 8 test cases — 8 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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