#3067

Medium

auto_awesomeBinary-tree traversal and state tracking

LeetCode Problem Workspace

Count Pairs of Connectable Servers in a Weighted Tree Network

This problem involves counting pairs of connectable servers in a weighted tree network using binary-tree traversal and DFS.

array tree depth first search

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] = [ai, bi, weighti] represents a bidirectional edge between vertices ai and bi with weight weighti. You are also given an integer signalSpeed.

Two servers a and b are connectable through a server c if the total distance from a to c and from b to c does not exceed the signalSpeed. Your task is to return an integer array count of length n, where count[i] represents the number of pairs of servers that are connectable through the server i.

Examples

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]

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]

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 <= ai, bi < n
edges[i] = [ai, bi, weighti]
1 <= weighti <= 106
1 <= signalSpeed <= 106
The input is generated such that edges represents a valid tree.

Solution Approach

Binary-Tree Traversal

The solution begins by treating each node as the root of the tree. Perform a DFS traversal for each node to calculate the number of servers in its subtree that are within the signal range. For each node, track the number of connectable server pairs through that server.

Subtree Size Calculation

During the DFS traversal, compute the size of the subtree rooted at each node and the distances of all other nodes in that subtree. For each node, use the subtree sizes to count the pairs of servers that can be connected by the current server.

Efficient Pair Counting

After gathering the subtree sizes and distances, calculate the number of connectable server pairs efficiently by iterating over all pairs in the subtree and checking if the sum of distances is within the signalSpeed. This approach ensures that you don't check all possible pairs, minimizing computational cost.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this solution depends on the approach taken, particularly the number of DFS traversals required. In the worst case, the solution will run in O(n^2) due to n DFS calls, each performing calculations for all n nodes. Space complexity is O(n) for storing the tree structure, distances, and subtree sizes during the DFS traversal.

What Interviewers Usually Probe

Look for a solution that efficiently calculates connectable pairs using DFS traversal and subtree size tracking.
Candidates should demonstrate how they optimize the pair-counting process to avoid checking all possible pairs.
Be aware of the need to consider the signalSpeed constraint when connecting servers through intermediate nodes.

Common Pitfalls or Variants

Common pitfalls

Forgetting to account for the signalSpeed when determining connectable pairs through each server.
Inefficient pair counting that leads to excessive computation, especially for larger inputs.
Not properly managing subtree size and distance calculations during the DFS, resulting in incorrect pair counts.

Follow-up variants

Handling cases where signalSpeed is very large or small compared to the tree's structure.
Exploring alternative methods for DFS traversal that avoid redundant calculations for each server.
Optimizing space complexity by reducing the need to store all distances or subtree sizes at once.

FAQ

What is the primary approach to solving the Count Pairs of Connectable Servers in a Weighted Tree Network problem?

The primary approach involves performing depth-first search (DFS) from each node in the tree, tracking subtree sizes, and calculating the number of connectable server pairs using the signalSpeed.

How do I efficiently count server pairs that are connectable through a given server?

Efficient counting is achieved by tracking distances within the subtree of each server and ensuring that pairs of servers do not exceed the signalSpeed constraint.

What is the time complexity of solving this problem?

The time complexity can be O(n^2) due to performing DFS for each node, where n is the number of servers in the tree.

What are the common pitfalls to avoid in this problem?

Common pitfalls include not properly accounting for the signalSpeed and inefficient pair counting that leads to unnecessary computations.

What variations of this problem might I encounter in interviews?

Variations may involve different signalSpeed values, optimized DFS traversal techniques, or minimizing space complexity by storing less information during traversal.

terminal

Solution

Solution 1: Enumeration + DFS

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.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

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

Continue Practicing

#2920 maximum points after collecting coins from all nodes #3327 check if dfs strings are palindromes #3331 find subtree sizes after changes