Problem Description

In this problem, you are given a set of servers connected by weighted bidirectional edges, which together form an undirected tree graph. Each server is labeled with a unique number from 0 to n-1, where n is the number of servers. The connections between them are defined by the edges, where each edge is represented by a triplet [ai, bi, weighti] indicating a connection between server ai and bi with the weight weighti. In addition, you have a value signalSpeed that is used as a divisor to determine connectivity.

The task is to determine for every server in the tree how many pairs of servers (a and b) can be connectable through it. Two servers are considered connectable through a third server c if all of the following conditions are met:

• a < b and both a and b are distinct from c.
• The distance from server c to a is exactly divisible by the signalSpeed.
• The distance from server c to b is also exactly divisible by the signalSpeed.
• The paths from c to a and c to b do not have common edges.

The result should be an array count, where count[i] represents the number of server pairs that are connectable through server i.

Flowchart Walkthrough

Let’s analyze LeetCode problem 3067 using the provided algorithm flowchart which can be explored further through the Flowchart. Here's a detailed breakdown using the specified flowchart:

1. Is it a graph?

   • Yes: The problem describes a network of servers, which can be represented as a graph.

2. Is it a tree?

   • Yes: The problem specifically mentions that the network forms a weighted tree.

3. DFS

   • Since we have a tree and depth-first search is an efficient way to explore trees, we proceed with DFS to explore the server network and utilize it to count the connectable pairs of servers.

By traversing the servers using the DFS algorithm, we can explore the tree structure while maintaining relevant computational states required to count pairs of connectable servers by comparing attributes like connections or weights, as the problem might suggest.

Conclusion: DFS is the appropriate algorithm to use for this problem, according to the flowchart, due to the structure of the data (a weighted tree) and the need to explore connections recursively between servers.

Intuition

To solve this problem, we devise an approach that leverages Depth-First Search (DFS). We will need to count the number of servers that are at a distance divisible by signalSpeed from each server, and then we will need to determine how many unique pairs of servers can be formed through that server.

The first step is to represent the network of servers using an adjacency list g, which will allow us to traverse the graph efficiently and find the neighbors of any given server.

Next, for each server a, we can consider it as a potential intermediate connecting server. Starting from a, we visit all directly connected neighbors b using DFS. While traversing the graph, we keep accumulating the count of nodes reachable from a via b at distances that are divisible by signalSpeed. We perform this count by invoking dfs(b, a, w) where b is the neighbor node, a is the current node, and w is the weight of the edge connecting them.

As we move through the tree, we keep a cumulative count s of nodes reachable from previous neighbors. When we calculate the count t for a new neighbor, we update the number of connectable server pairs for server a by adding s * t to it. This represents the number of new unique pairs that can be formed by combining servers reachable from previous neighbors with servers reachable from the current neighbor. After evaluating all neighbors of a, we will have the total number of connectable server pairs through a.

After iterating through all servers in this manner, we obtain the desired count of connectable server pairs for each server, which we return as the result.

Learn more about Tree and Depth-First Search patterns.

Solution Approach

The solution approach uses Depth-First Search (DFS) which is a common algorithm for traversing or searching through tree structures. DFS is a good fit for this problem because it allows us to explore all the paths from a given server to its descendent servers, thereby calculating distances and checking divisibility by signalSpeed.

Here's how the algorithm is implemented:

1. Construct an adjacency list g; for each edge [a, b, w] in the edges list, add an entry such that g[a] includes the tuple (b, w), representing a neighbor b and the weight of the edge w from a to b, and vice versa since it's an undirected graph.

2. Create an array ans initialized with zeros to store the resulting counts of connectable server pairs for each server.

3. For each server a, we want to find the number of connectable server pairs that include a as an intermediate server.

4. Initialize a count s to 0, representing the cumulative count of servers found through previous neighbors of server a.

5. For each neighbor b with edge weight w connected to a:

   • Perform a DFS to calculate the number t of servers connectable starting from b such that the distance to a (accumulated in a variable ws within the DFS) is divisible by signalSpeed.
   • The DFS function dfs(a, fa, ws) receives the current server a, its parent server fa, and the current weight sum ws. The function returns the number of servers reachable from a with the sum of edge weights divisible by signalSpeed.

6. Use the count t from the DFS to update the answer for the current server a. The number of new connectable pairs is s * t because each server counted by s can form a pair with each server counted by t. Add this number to ans[a].

7. Update the cumulative count s by adding t to it, for the next iterations. This step is crucial because it accumulates the total number of servers that can be connected through a from all its neighbors.

8. After computing this for all neighbors b of a server a, you end up with the total count of server pairs that can connect through a, stored in ans[a].

9. Repeat steps 4-8 for each server in the graph to fill the array ans with the total count of connectable server pairs for all servers.

10. Return the array ans which now contains the required result for each server in the tree.

This algorithm works efficiently due to the tree's properties, which ensure there are no cycles, and each distinct path from a node a to any other nodes in its subtree can be explored exactly once by DFS without revisiting nodes, thus avoiding redundant calculations, and enabling the solution approach to work in polynomial time complexity.

Example Walkthrough

Let's consider a small example with 4 servers and signalSpeed = 2 to illustrate the solution approach.

Servers (number of servers, n): 4
signalSpeed: 2
edges: [[0, 1, 2], [1, 2, 4], [1, 3, 2]]

Our tree graph for servers will look like this:

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

Step 1: Construct the adjacency list g from the edges:

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

Step 2: Initialize an array ans with zeros [0, 0, 0, 0] to hold the counts for each server.

Step 3: We'll iterate over each server to find the number of connectable server pairs.

Step 4-7: For server 0, the neighbor is server 1 with an edge weight of 2, we perform a DFS starting from server 1. Since we are only interested in servers that are connectable through server 0, we find servers 2 and 3 are both reachable from 1 with distances divisible by signalSpeed = 2. Since server 1 has two connectable servers, there is only one pair (2, 3) that can be routed through it. So we add 1 to ans[0].

Step 4-7: For server 1, we see three potential paths (one through 0, one through 2, and one through 3). Through DFS for each neighbor, we find the following:

• Server 0 contributes 0 connectable servers.
• Server 2 contributes 1 connectable server (itself).
• Server 3 contributes 1 connectable server (itself). Combining the paths from 2 and 3, as 2 < 3, gives us 1 pair. Hence, we add 1 to ans[1].

Servers 2 and 3 have only one neighbor they connect to without common edges, which is server 1, and no server pairs (a, b) such that a < b are connectable through them. Hence, ans[2] and ans[3] remain 0.

Step 9: After iterating over all servers, the ans array values are finalized as [1, 1, 0, 0].

Step 10: The final output [1, 1, 0, 0] is returned, indicating that through server 0, 1 pair of servers can be connected, the same for server 1, and no connectable pairs through servers 2 and 3.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(n^2). This is because for each of the n nodes, there is a Depth-First Search (DFS) that is initiated. DFS itself may visit each node one time in the worst case, and since the DFS is occurring in a loop of n nodes, this results in a potential of n * (n-1) operations, hence the O(n^2).

Furthermore, within the DFS function, there is a check for whether ws % signalSpeed is equal to zero that occurs with each recursive call, but this does not add to the complexity in terms of n, as it is a constant-time operation.

The space complexity of the code is O(n). The primary space usage comes from the adjacency list g, which contains a list for each of the n nodes. Each list hold pairs ((b, w)) representing edges and their weights. There is also the ans array of size n that is kept throughout the execution. Temporary variables such as cnt and the stack frames due to recursive calls do not increase the overall space complexity, as they will at most be proportional to the depth of the recursion which is at most n, in the case of a path graph.

Additionally, the recursive nature of the DFS function does entail a call stack, which could have a depth of up to n in a worst-case scenario (such as a linear tree), but since the adjacency list is the dominant space-consuming structure and they are both of O(n) space complexity, the overall space complexity remains O(n).

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