LeetCode #3067 — MEDIUM

Count Pairs of Connectable Servers in a Weighted Tree Network

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an unrooted weighted tree with n vertices representing servers numbered from 0 to n - 1, an array edges where edges[i] = [a_i, b_i, weight_i] represents a bidirectional edge between vertices a_i and b_i of weight weight_i. You are also given an integer signalSpeed.

Two servers a and b are connectable through a server c if:

a < b, a != c and b != c.
The distance from c to a is divisible by signalSpeed.
The distance from c to b is divisible by signalSpeed.
The path from c to b and the path from c to a do not share any edges.

Return an integer array count of length n where count[i] is the number of server pairs that are connectable through the server i.

Example 1:

Input: edges = [[0,1,1],[1,2,5],[2,3,13],[3,4,9],[4,5,2]], signalSpeed = 1
Output: [0,4,6,6,4,0]
Explanation: Since signalSpeed is 1, count[c] is equal to the number of pairs of paths that start at c and do not share any edges.
In the case of the given path graph, count[c] is equal to the number of servers to the left of c multiplied by the servers to the right of c.

Example 2:

Input: edges = [[0,6,3],[6,5,3],[0,3,1],[3,2,7],[3,1,6],[3,4,2]], signalSpeed = 3
Output: [2,0,0,0,0,0,2]
Explanation: Through server 0, there are 2 pairs of connectable servers: (4, 5) and (4, 6).
Through server 6, there are 2 pairs of connectable servers: (4, 5) and (0, 5).
It can be shown that no two servers are connectable through servers other than 0 and 6.

Constraints:

2 <= n <= 1000
edges.length == n - 1
edges[i].length == 3
0 <= a_i, b_i < n
edges[i] = [a_i, b_i, weight_i]
1 <= weight_i <= 10⁶
1 <= signalSpeed <= 10⁶
The input is generated such that edges represents a valid tree.

Step 01

Brute Force Baseline

Problem summary: You are given an unrooted weighted tree with n vertices representing servers numbered from 0 to n - 1, an array edges where edges[i] = [ai, bi, weighti] represents a bidirectional edge between vertices ai and bi of weight weighti. You are also given an integer signalSpeed. Two servers a and b are connectable through a server c if: a < b, a != c and b != c. The distance from c to a is divisible by signalSpeed. The distance from c to b is divisible by signalSpeed. The path from c to b and the path from c to a do not share any edges. Return an integer array count of length n where count[i] is the number of server pairs that are connectable through the server i.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Tree

Example 1

[[0,1,1],[1,2,5],[2,3,13],[3,4,9],[4,5,2]]
1

Example 2

[[0,6,3],[6,5,3],[0,3,1],[3,2,7],[3,1,6],[3,4,2]]
3

Core Insight

What unlocks the optimal approach

Take each node as the root of the tree, run DFS, and save for each node <code>i</code>, the number of nodes in the subtree rooted at <code>i</code> whose distance to the root is divisible by <code>signalSpeed</code>.
If the root has <code>m</code> children named <code>c1, c2, …, cm</code> that respectively have <code>num[c1], num[c2], …, num[cm]</code> nodes in their subtrees whose distance is divisible by signalSpeed. Then, there are <code>((S - num[ci]) * num[ci]) / 2</code>that are connectable through the root that we have fixed, where <code>S</code> is the sum of <code>num[ci]</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3067: Count Pairs of Connectable Servers in a Weighted Tree Network
class Solution {
    private int signalSpeed;
    private List<int[]>[] g;

    public int[] countPairsOfConnectableServers(int[][] edges, int signalSpeed) {
        int n = edges.length + 1;
        g = new List[n];
        this.signalSpeed = signalSpeed;
        Arrays.setAll(g, k -> new ArrayList<>());
        for (var e : edges) {
            int a = e[0], b = e[1], w = e[2];
            g[a].add(new int[] {b, w});
            g[b].add(new int[] {a, w});
        }
        int[] ans = new int[n];
        for (int a = 0; a < n; ++a) {
            int s = 0;
            for (var e : g[a]) {
                int b = e[0], w = e[1];
                int t = dfs(b, a, w);
                ans[a] += s * t;
                s += t;
            }
        }
        return ans;
    }

    private int dfs(int a, int fa, int ws) {
        int cnt = ws % signalSpeed == 0 ? 1 : 0;
        for (var e : g[a]) {
            int b = e[0], w = e[1];
            if (b != fa) {
                cnt += dfs(b, a, ws + w);
            }
        }
        return cnt;
    }
}

// Accepted solution for LeetCode #3067: Count Pairs of Connectable Servers in a Weighted Tree Network
func countPairsOfConnectableServers(edges [][]int, signalSpeed int) []int {
	n := len(edges) + 1
	type pair struct{ x, w int }
	g := make([][]pair, n)
	for _, e := range edges {
		a, b, w := e[0], e[1], e[2]
		g[a] = append(g[a], pair{b, w})
		g[b] = append(g[b], pair{a, w})
	}
	var dfs func(a, fa, ws int) int
	dfs = func(a, fa, ws int) int {
		cnt := 0
		if ws%signalSpeed == 0 {
			cnt++
		}
		for _, e := range g[a] {
			b, w := e.x, e.w
			if b != fa {
				cnt += dfs(b, a, ws+w)
			}
		}
		return cnt
	}
	ans := make([]int, n)
	for a := 0; a < n; a++ {
		s := 0
		for _, e := range g[a] {
			b, w := e.x, e.w
			t := dfs(b, a, w)
			ans[a] += s * t
			s += t
		}
	}
	return ans
}

# Accepted solution for LeetCode #3067: Count Pairs of Connectable Servers in a Weighted Tree Network
class Solution:
    def countPairsOfConnectableServers(
        self, edges: List[List[int]], signalSpeed: int
    ) -> List[int]:
        def dfs(a: int, fa: int, ws: int) -> int:
            cnt = 0 if ws % signalSpeed else 1
            for b, w in g[a]:
                if b != fa:
                    cnt += dfs(b, a, ws + w)
            return cnt

        n = len(edges) + 1
        g = [[] for _ in range(n)]
        for a, b, w in edges:
            g[a].append((b, w))
            g[b].append((a, w))
        ans = [0] * n
        for a in range(n):
            s = 0
            for b, w in g[a]:
                t = dfs(b, a, w)
                ans[a] += s * t
                s += t
        return ans

// Accepted solution for LeetCode #3067: Count Pairs of Connectable Servers in a Weighted Tree Network
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3067: Count Pairs of Connectable Servers in a Weighted Tree Network
// class Solution {
//     private int signalSpeed;
//     private List<int[]>[] g;
// 
//     public int[] countPairsOfConnectableServers(int[][] edges, int signalSpeed) {
//         int n = edges.length + 1;
//         g = new List[n];
//         this.signalSpeed = signalSpeed;
//         Arrays.setAll(g, k -> new ArrayList<>());
//         for (var e : edges) {
//             int a = e[0], b = e[1], w = e[2];
//             g[a].add(new int[] {b, w});
//             g[b].add(new int[] {a, w});
//         }
//         int[] ans = new int[n];
//         for (int a = 0; a < n; ++a) {
//             int s = 0;
//             for (var e : g[a]) {
//                 int b = e[0], w = e[1];
//                 int t = dfs(b, a, w);
//                 ans[a] += s * t;
//                 s += t;
//             }
//         }
//         return ans;
//     }
// 
//     private int dfs(int a, int fa, int ws) {
//         int cnt = ws % signalSpeed == 0 ? 1 : 0;
//         for (var e : g[a]) {
//             int b = e[0], w = e[1];
//             if (b != fa) {
//                 cnt += dfs(b, a, ws + w);
//             }
//         }
//         return cnt;
//     }
// }

// Accepted solution for LeetCode #3067: Count Pairs of Connectable Servers in a Weighted Tree Network
function countPairsOfConnectableServers(edges: number[][], signalSpeed: number): number[] {
    const n = edges.length + 1;
    const g: [number, number][][] = Array.from({ length: n }, () => []);
    for (const [a, b, w] of edges) {
        g[a].push([b, w]);
        g[b].push([a, w]);
    }
    const dfs = (a: number, fa: number, ws: number): number => {
        let cnt = ws % signalSpeed === 0 ? 1 : 0;
        for (const [b, w] of g[a]) {
            if (b != fa) {
                cnt += dfs(b, a, ws + w);
            }
        }
        return cnt;
    };
    const ans: number[] = Array(n).fill(0);
    for (let a = 0; a < n; ++a) {
        let s = 0;
        for (const [b, w] of g[a]) {
            const t = dfs(b, a, w);
            ans[a] += s * t;
            s += t;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2)

Space

O(n)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.