Problem Description

You have an unrooted weighted tree with n servers (vertices) numbered from 0 to n - 1. The tree is defined by an array edges where each edges[i] = [ai, bi, weighti] represents a bidirectional edge between servers ai and bi with weight weighti. You're also given an integer signalSpeed.

Two servers a and b can be connectable through a server c if all of these conditions are met:

1. a < b (server a has a smaller number than server b)
2. a != c and b != c (the connecting server c must be different from both a and b)
3. The distance from server c to server a is divisible by signalSpeed
4. The distance from server c to server b is divisible by signalSpeed
5. The paths from c to a and from c to b don't share any edges (they branch out in different directions from c)

Your task is to return an array count of length n where count[i] represents the number of server pairs (a, b) that are connectable through server i.

In other words, for each server i, you need to count how many pairs of servers can communicate with each other using server i as an intermediate connection point, following the rules above. The distance between servers is the sum of edge weights along the path connecting them.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The problem explicitly involves an unrooted weighted tree with vertices (servers) and edges. A tree is a special type of graph.

Is it a tree?

• Yes: The problem statement clearly mentions we're working with an "unrooted weighted tree" with n vertices and n-1 edges (since edges array defines the tree structure).

DFS

• Yes: We arrive at DFS as the solution pattern.

Why DFS is appropriate for this problem:

1. Tree traversal: We need to explore paths from each potential intermediate server c to find all servers at specific distances (divisible by signalSpeed).

2. Path exploration: For each server i acting as the intermediate node, we need to:

   • Explore each branch (subtree) independently
   • Count servers in each branch whose distance from i is divisible by signalSpeed
   • Ensure paths don't share edges (naturally handled by tree structure where each branch is independent)

3. Distance calculation: DFS allows us to accumulate distances as we traverse, making it easy to check if the cumulative distance is divisible by signalSpeed.

4. Counting valid pairs: By using DFS from each neighbor of the intermediate server, we can count valid servers in each direction and multiply these counts to get the number of valid pairs.

Conclusion: The flowchart correctly leads us to use DFS for this tree-based problem where we need to explore paths, calculate distances, and count valid server pairs through each potential intermediate server.

Intuition

The key insight is to think about what makes two servers connectable through an intermediate server. For any intermediate server c, we need to find pairs of servers (a, b) where both are reachable from c at distances divisible by signalSpeed, and they lie in different branches of the tree rooted at c.

Since we're working with a tree, when we pick any server c as the intermediate point, the tree naturally splits into several subtrees (branches) connected to c. The crucial observation is that paths from c to servers in different branches will never share edges - this is guaranteed by the tree structure where there's only one path between any two nodes.

For each server i as the intermediate point, we can:

1. Explore each of its neighboring branches independently using DFS
2. Count how many servers in each branch are at a "valid" distance (divisible by signalSpeed) from i
3. Multiply the counts from different branches to get the number of valid pairs

For example, if server i has three branches with 2, 3, and 4 valid servers respectively, the number of connectable pairs through i would be 2×3 + 2×4 + 3×4 = 26 pairs.

The multiplication works because:

• We need to pick one server from one branch and another server from a different branch
• All such combinations form valid pairs (assuming a < b constraint is handled)

We accumulate the counts incrementally: as we process each branch, we multiply the current branch's count with the sum of all previous branches' counts, then add this branch's count to our running sum. This efficiently calculates all pair combinations without redundant work.

The DFS helper function recursively explores from a starting point, tracking the cumulative distance, and returns the count of servers at valid distances in that subtree.

Learn more about Tree and Depth-First Search patterns.

Solution Approach

First, we construct an adjacency list g based on the edges given in the problem, where g[a] represents all the neighbor nodes of node a and their corresponding edge weights. This is done by iterating through the edges array and adding bidirectional connections.

g = [[] for _ in range(n)]
for a, b, w in edges:
    g[a].append((b, w))
    g[b].append((a, w))

The core of the solution involves two main components:

1. DFS Helper Function

The dfs function explores from a given node and counts servers at valid distances:

• Parameters: a (current node), fa (parent node to avoid revisiting), ws (cumulative weight/distance)
• Returns: count of servers whose distance from the starting point is divisible by signalSpeed
• Base case: If ws % signalSpeed == 0, we found a valid server (count = 1), otherwise 0
• Recursively explore all neighbors except the parent, accumulating distances

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

2. Main Enumeration Logic

For each potential intermediate server a:

• Initialize a cumulative sum s = 0 to track servers found in previous branches
• For each neighbor b of a:
  • Call dfs(b, a, w) to count valid servers in that branch (starting with edge weight w)
  • The number of new pairs formed = s * t (current branch count × previous branches' total)
  • Add these pairs to ans[a]
  • Update s += t for the next iteration

for a in range(n):
    s = 0
    for b, w in g[a]:
        t = dfs(b, a, w)
        ans[a] += s * t
        s += t

The key insight in the accumulation pattern is that it efficiently counts all combinations of servers from different branches without double-counting. By multiplying the current branch's count with the accumulated sum of previous branches, we ensure each pair is counted exactly once.

Time Complexity: O(n²) - For each of the n nodes, we potentially visit all other nodes through DFS. Space Complexity: O(n) - For the adjacency list and recursion stack.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Example Input:

• n = 5 servers (numbered 0 to 4)
• edges = [[0,1,4], [1,2,2], [1,3,4], [3,4,2]]
• signalSpeed = 2

This creates a tree structure:

    0
    |
   (4)
    |
    1
   / \
 (2) (4)
 /     \
2       3
        |
       (2)
        |
        4

Step 1: Build Adjacency List

g[0] = [(1, 4)]
g[1] = [(0, 4), (2, 2), (3, 4)]
g[2] = [(1, 2)]
g[3] = [(1, 4), (4, 2)]
g[4] = [(3, 2)]

Step 2: Process Each Server as Intermediate Point

Let's focus on server 1 as the intermediate point:

Server 1 has 3 branches (neighbors): 0, 2, and 3.

Branch 1 (toward server 0):

• Call dfs(0, 1, 4) starting with edge weight 4
• At server 0: distance = 4, 4 % 2 = 0 ✓ (valid)
• No more neighbors to explore
• Returns count = 1

Branch 2 (toward server 2):

• Call dfs(2, 1, 2) starting with edge weight 2
• At server 2: distance = 2, 2 % 2 = 0 ✓ (valid)
• No more neighbors to explore
• Returns count = 1

Branch 3 (toward server 3):

• Call dfs(3, 1, 4) starting with edge weight 4
• At server 3: distance = 4, 4 % 2 = 0 ✓ (valid), count = 1
• Explore neighbor 4: dfs(4, 3, 6) with distance 4+2=6
• At server 4: distance = 6, 6 % 2 = 0 ✓ (valid), count = 1
• Returns total count = 2

Calculating Pairs for Server 1:

• Initially s = 0
• After branch 1: ans[1] += 0 * 1 = 0, update s = 1
• After branch 2: ans[1] += 1 * 1 = 1, update s = 2
• After branch 3: ans[1] += 2 * 2 = 4, update s = 4
• Total ans[1] = 5

The 5 connectable pairs through server 1 are:

• (0, 2): distances 4 and 2, both divisible by 2 ✓
• (0, 3): distances 4 and 4, both divisible by 2 ✓
• (0, 4): distances 4 and 6, both divisible by 2 ✓
• (2, 3): distances 2 and 4, both divisible by 2 ✓
• (2, 4): distances 2 and 6, both divisible by 2 ✓

Complete Solution: Processing all servers similarly:

• ans[0] = 0 (only one branch from server 0)
• ans[1] = 5 (as calculated above)
• ans[2] = 0 (only one branch from server 2)
• ans[3] = 2 (pairs: (1,4) and (2,4) through server 3)
• ans[4] = 0 (only one branch from server 4)

Final answer: [0, 5, 0, 2, 0]

The algorithm efficiently counts all valid server pairs by:

1. Using DFS to explore each branch and count servers at valid distances
2. Multiplying counts from different branches to get pair combinations
3. Accumulating results incrementally to avoid redundant calculations

Solution Implementation

Time and Space Complexity

Time Complexity: O(n²)

The algorithm iterates through each node as a potential intermediate server (outer loop runs n times). For each node a, it explores all its neighbors and performs a DFS from each neighbor. The DFS traverses the tree structure excluding the subtree containing node a, visiting at most n-1 nodes. Since we do this for each neighbor of every node, and in the worst case (like a star graph), we might traverse nearly all nodes multiple times. Across all iterations, each edge is traversed O(n) times, resulting in O(n²) total time complexity.

Space Complexity: O(n)

The space is used for:

• The adjacency list g which stores all edges, requiring O(n) space (since there are n-1 edges in a tree)
• The recursion stack for DFS, which in the worst case (a linear tree) can go up to depth n, requiring O(n) space
• The answer array ans of size n, requiring O(n) space

Therefore, the overall space complexity is O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrectly Counting the Intermediate Node Itself

A common mistake is including the intermediate node c when counting valid servers in each branch. According to the problem constraints, the intermediate node should NOT be counted as part of the connectable pairs since a != c and b != c.

Incorrect approach:

def dfs(a: int, fa: int, ws: int) -> int:
    # Wrong: Always counting current node when ws == 0
    cnt = 1 if ws == 0 else 0  
    for b, w in g[a]:
        if b != fa:
            cnt += dfs(b, a, ws + w)
    return cnt

Correct approach:

def dfs(a: int, fa: int, ws: int) -> int:
    # Correct: Only count if ws % signalSpeed == 0 AND ws != 0
    cnt = 1 if ws % signalSpeed == 0 and ws > 0 else 0
    for b, w in g[a]:
        if b != fa:
            cnt += dfs(b, a, ws + w)
    return cnt

2. Double-Counting Server Pairs

Another pitfall is counting the same pair (a, b) multiple times when iterating through branches. The solution uses a cumulative approach (s * t) to ensure each pair is counted exactly once, but developers might mistakenly try to count all combinations directly.

Incorrect approach:

for a in range(n):
    branches = []
    for b, w in g[a]:
        t = dfs(b, a, w)
        branches.append(t)
  
    # Wrong: This counts each pair twice (once as (i,j) and once as (j,i))
    for i in range(len(branches)):
        for j in range(len(branches)):
            if i != j:
                ans[a] += branches[i] * branches[j]

Correct approach:

for a in range(n):
    s = 0
    for b, w in g[a]:
        t = dfs(b, a, w)
        ans[a] += s * t  # Only pairs with previous branches
        s += t

3. Misunderstanding Distance Calculation

The distance should start from the edge weight when exploring branches, not from 0. Starting from 0 would incorrectly include the intermediate node in the distance calculation.

Incorrect approach:

for a in range(n):
    s = 0
    for b, w in g[a]:
        # Wrong: Starting distance from 0 ignores the first edge
        t = dfs(b, a, 0)  
        ans[a] += s * t
        s += t

Correct approach:

for a in range(n):
    s = 0
    for b, w in g[a]:
        # Correct: Start with the edge weight w
        t = dfs(b, a, w)
        ans[a] += s * t
        s += t