0770MediumArrayTreeDepth-First Search

Relay Pairs Through Each Hub

Tracked in this browser only

Trains the technique from

LeetCode 3067Count Pairs of Connectable Servers in a Weighted Tree Network

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 campus network wires n relay nodes, numbered 0 to n - 1, into a tree. The wiring arrives as a weighted edge list links, where links[i] = [a, b, w] means a cable of length w runs between node a and node b. Since the network is a tree, links holds n - 1 cables and exactly one route runs between any two nodes. The distance between two nodes is the total cable length along that route.

The gear runs at a fixed cadence hop. Fix a node c and call it the hub. An unordered pair of nodes {a, b} relays through c when all of the following hold:

a, b and c are three different nodes;
the route from a to b passes through c;
the distance from a to c is a multiple of hop;
the distance from c to b is a multiple of hop.

Produce a list of length n whose entry at position c is how many unordered pairs relay through node c.

Examples

Example 1

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

From node 0 the branch through node 1 holds nodes 1 and 3 at distances 2 and 4, and the branch through node 2 holds nodes 2 and 4 at distances 2 and 6; all four distances are multiples of 2, so the four pairs taking one node from each branch all relay through node 0. Node 1 has nodes 0, 2 and 4 at distances 2, 4 and 8 on one side and node 3 at distance 2 on the other, giving three pairs, and node 2 works out the same way. Nodes 3 and 4 each have a single cable, so no route can enter and leave them.

Example 2

Input: links = [[0, 1, 3], [1, 2, 3], [0, 3, 1]], hop = 3
Output: [0, 1, 0, 0]

At node 1 the pair {0, 2} qualifies: node 0 is 3 away, node 2 is 3 away and the route between them runs through node 1. At node 0 the only other branch holds node 3 at distance 1, which is not a multiple of 3, so nothing pairs there. Nodes 2 and 3 are ends of the network.

Example 3

Input: links = [[0, 1, 5]], hop = 1
Output: [0, 0]

The network has two nodes, so no pair can supply two endpoints distinct from the hub and both counts are zero.

Constraints

1 <= links.length <= 999
links[i].length == 3
0 <= links[i][j] <= 10^6
1 <= hop <= 10^6
Writing n for links.length + 1, we have 2 <= n <= 1000 and every cable is [a_i, b_i, w_i] with 0 <= a_i, b_i < n and 1 <= w_i <= 10^6.
links is guaranteed to describe a tree.

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 relay_pairs(links: list[list[int]], hop: int) -> list[int]:

Java

public int[] relayPairs(int[][] links, int hop)

The intended approach

Time O(n^2) for n nodes: each hub walks the whole tree once.Space O(n) for the adjacency list and the walking stack, plus O(n) for the answer.

A route through the hub that avoids using the hub as an endpoint must leave the hub down one cable and arrive down a different one, so the two endpoints sit in different branches. Build an adjacency list, then for every hub walk each branch separately from an explicit stack, carrying the running distance from the hub, and count the branch's nodes whose distance is a multiple of the cadence. Multiplying each branch's count by the total of the branches already seen accumulates every cross-branch pair exactly once.

Pythonentry point: relay_pairs

def relay_pairs(links, hop):
    # For each hub, walk each of its branches with an explicit stack, so a chain of
    # a thousand nodes never touches the interpreter's call stack.
    # Time O(n^2), space O(n).
    n = len(links) + 1
    adj = [[] for _ in range(n)]
    for a, b, w in links:
        adj[a].append((b, w))
        adj[b].append((a, w))
    out = [0] * n
    for hub in range(n):
        seen_before = 0
        total = 0
        for start, first_w in adj[hub]:
            count = 0
            stack = [(start, first_w, hub)]
            while stack:
                node, dist, came_from = stack.pop()
                if dist % hop == 0:
                    count += 1
                for nxt, w in adj[node]:
                    if nxt != came_from:
                        stack.append((nxt, dist + w, node))
            total += seen_before * count
            seen_before += count
        out[hub] = total
    return out

The obvious version

Compute the distance between every pair of nodes by walking out from each node in turn, then for each hub try every unordered pair of other nodes and keep it when both distances to the hub are multiples of the cadence and the two distances add up to the pair's own distance, which is what says the route runs through the hub. Cubic in the node 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_relay_pairs(links, hop):
    n = len(links) + 1
    adj = [[] for _ in range(n)]
    for a, b, w in links:
        adj[a].append((b, w))
        adj[b].append((a, w))
    far = []
    for src in range(n):
        d = [-1] * n
        d[src] = 0
        stack = [src]
        while stack:
            u = stack.pop()
            for v, w in adj[u]:
                if d[v] == -1:
                    d[v] = d[u] + w
                    stack.append(v)
        far.append(d)
    out = [0] * n
    for hub in range(n):
        count = 0
        for a in range(n):
            if a == hub or far[a][hub] % hop != 0:
                continue
            for b in range(a + 1, n):
                if b == hub or far[hub][b] % hop != 0:
                    continue
                if far[a][hub] + far[hub][b] == far[a][b]:
                    count += 1
        out[hub] = count
    return out

How this problem was checked

Reference solution executed against 13 test cases — 13 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