Class Solution
Medium
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 != candb != c.- The distance from
ctoais divisible bysignalSpeed. - The distance from
ctobis divisible bysignalSpeed. - The path from
ctoband the path fromctoado 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 <= 1000edges.length == n - 1edges[i].length == 30 <= ai, bi < nedges[i] = [ai, bi, weighti]1 <= weighti <= 1061 <= signalSpeed <= 106- The input is generated such that
edgesrepresents a valid tree.
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Constructor Summary
Constructors -
Method Summary
Modifier and TypeMethodDescriptionint[]countPairsOfConnectableServers(int[][] edges, int signalSpeed)
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Constructor Details
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Solution
public Solution()
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Method Details
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countPairsOfConnectableServers
public int[] countPairsOfConnectableServers(int[][] edges, int signalSpeed)
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