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The Longest Walk in a Pipe 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 network of junctions is joined by the pipes pipes, where pipes[i] = [a, b] is a two-way pipe between junction a and junction b. The network holds one more junction than it has pipes, and the junctions are numbered from 0 upwards. Every junction is reachable from every other, and there is no loop anywhere.

Return the greatest number of pipes on any walk between two junctions that never uses a pipe twice.

Examples

Example 1

Input: pipes = [[0, 1], [1, 2], [2, 3], [3, 4], [2, 5]]
Output: 4

Six junctions. The walk from 0 through 1, 2 and 3 to 4 uses four pipes, and so does the walk from 0 to 5 by way of 1 and 2 plus one more. Nothing longer exists, since junction 2 is where the two ends part company.

Example 2

Input: pipes = []
Output: 0

No pipes means a single junction, and a walk that goes nowhere uses none.

Example 3

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

Every junction hangs straight off junction 0, so the longest walk goes from one of them through 0 to another, using two pipes.

Constraints

0 <= pipes.length <= 9999
pipes[i].length == 2
0 <= pipes[i][j] <= pipes.length
The two ends of a pipe are different junctions.
The pipes reach every junction and hold no loop.

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 tree_diameter(pipes: list[list[int]]) -> int:

Java

public int treeDiameter(int[][] pipes)

The intended approach

Time O(n) over the junctions and pipes.Space O(n) for the neighbour lists and the distances.

Walk outwards from any junction, level by level, and note the junction reached last, which is one of the farthest away. That junction is always at one end of some longest walk. Walk outwards from it in turn, and how far the second walk reaches is the answer. Why the first walk lands on an end of a longest walk: take any longest walk and the junction the first walk found. The route from the starting junction to that junction has to meet the longest walk somewhere, because there is only one route between any two junctions when nothing loops. From that meeting point, the found junction is at least as far as either end of the longest walk, or the first walk would have stopped somewhere else, so swapping the found junction in for the nearer end leaves a walk just as long. Two outward walks, each touching every junction and pipe once.

Pythonentry point: tree_diameter

def tree_diameter(pipes):
    from collections import deque
    n = len(pipes) + 1
    near = [[] for _ in range(n)]
    for pair in pipes:
        near[pair[0]].append(pair[1])
        near[pair[1]].append(pair[0])
    def walk(start):
        far = [-1] * n
        far[start] = 0
        queue = deque([start])
        end = start
        while queue:
            here = queue.popleft()
            if far[here] > far[end]:
                end = here
            for there in near[here]:
                if far[there] < 0:
                    far[there] = far[here] + 1
                    queue.append(there)
        return end, far[end]
    first, _ = walk(0)
    _, span = walk(first)
    return span

The obvious version

Walk outwards from every junction in turn and keep the greatest distance any walk ever reaches. Costs one walk per junction rather than two walks in total.

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_tree_diameter(pipes):
    from collections import deque
    n = len(pipes) + 1
    near = [[] for _ in range(n)]
    for pair in pipes:
        near[pair[0]].append(pair[1])
        near[pair[1]].append(pair[0])
    best = 0
    for start in range(n):
        far = [-1] * n
        far[start] = 0
        queue = deque([start])
        while queue:
            here = queue.popleft()
            for there in near[here]:
                if far[there] < 0:
                    far[there] = far[here] + 1
                    queue.append(there)
        for value in far:
            if value > best:
                best = value
    return best

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

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