There is a network of n servers, labeled from 0 to n - 1. You are given a 2D integer array edges, where edges[i] = [ui, vi] indicates there is a message channel between servers ui and vi, and they can pass any number of messages to each other directly in one second. You are also given a 0-indexed integer array patience of length n.
All servers are connected, i.e., a message can be passed from one server to any other server(s) directly or indirectly through the message channels.
The server labeled 0 is the master server. The rest are data servers. Each data server needs to send its message to the master server for processing and wait for a reply. Messages move between servers optimally, so every message takes the least amount of time to arrive at the master server. The master server will process all newly arrived messages instantly and send a reply to the originating server via the reversed path the message had gone through.
At the beginning of second 0, each data server sends its message to be processed. Starting from second 1, at the beginning of every second, each data server will check if it has received a reply to the message it sent (including any newly arrived replies) from the master server:
i will resend the message every patience[i] second(s), i.e., the data server i will resend the message if patience[i] second(s) have elapsed since the last time the message was sent from this server.The network becomes idle when there are no messages passing between servers or arriving at servers.
Return the earliest second starting from which the network becomes idle.
Example 1:
Input: edges = [[0,1],[1,2]], patience = [0,2,1] Output: 8 Explanation: At (the beginning of) second 0, - Data server 1 sends its message (denoted 1A) to the master server. - Data server 2 sends its message (denoted 2A) to the master server. At second 1, - Message 1A arrives at the master server. Master server processes message 1A instantly and sends a reply 1A back. - Server 1 has not received any reply. 1 second (1 < patience[1] = 2) elapsed since this server has sent the message, therefore it does not resend the message. - Server 2 has not received any reply. 1 second (1 == patience[2] = 1) elapsed since this server has sent the message, therefore it resends the message (denoted 2B). At second 2, - The reply 1A arrives at server 1. No more resending will occur from server 1. - Message 2A arrives at the master server. Master server processes message 2A instantly and sends a reply 2A back. - Server 2 resends the message (denoted 2C). ... At second 4, - The reply 2A arrives at server 2. No more resending will occur from server 2. ... At second 7, reply 2D arrives at server 2. Starting from the beginning of the second 8, there are no messages passing between servers or arriving at servers. This is the time when the network becomes idle.
Example 2:
Input: edges = [[0,1],[0,2],[1,2]], patience = [0,10,10] Output: 3 Explanation: Data servers 1 and 2 receive a reply back at the beginning of second 2. From the beginning of the second 3, the network becomes idle.
Constraints:
n == patience.length2 <= n <= 105patience[0] == 01 <= patience[i] <= 105 for 1 <= i < n1 <= edges.length <= min(105, n * (n - 1) / 2)edges[i].length == 20 <= ui, vi < nui != viApproach: Use Breadth-First Search (BFS) to find the shortest path from each server to the master server (server 0). Each server needs to send a message to and receive a reply back from the master server. For this, calculate the round-trip time for each server. If the patience time of a server is less than twice its one-way distance, the server will resend messages until it receives a reply. Determine the time when the network becomes idle by considering all these factors.
This Python code utilizes a queue to perform BFS starting from the master server. The level of each node in the BFS corresponds to the shortest distance. For each server, calculate the time taken for a message roundtrip and determine if additional messages are sent based on the patience time of the server. Finally, determine the maximum time required for the network to become idle and add 1 to account for zero-based indexing.
Java
Time Complexity: O(E + V), where V is the number of servers and E is the number of edges. BFS traverses each node and edge once.
Space Complexity: O(V + E) to store the graph and additional arrays for distance and queue operations.
Approach: Use Depth-First Search (DFS) to simulate a BFS-like level exploration without maintaining an explicit queue. By recursively visiting each node and its neighbors, similar to BFS, compute the shortest paths. Despite DFS being inherently stack-based, we can emulate BFS by careful control of recursion depth. This alternative can offer a recursive way of computation, matching typical BFS outcomes on tree-like graphs.
In C++, we take an alternative route with DFS to mimic the shortest path evaluation usually done by BFS. By tracking current depth in recursion and updating path lengths accordingly, this solution efficiently determines the idle time, just as the BFS would.
Time Complexity: O(V + E) as we visit each edge and node similarly to BFS, ensuring linear traversal in interconnected graphs.
Space Complexity: O(V + E), including the internal stack usage depth required for recursion.
| Approach | Complexity |
|---|---|
| BFS to Find Shortest Path and Calculate Idle Time | Time Complexity: O(E + V), where V is the number of servers and E is the number of edges. BFS traverses each node and edge once. Space Complexity: O(V + E) to store the graph and additional arrays for distance and queue operations. |
| DFS Mimicking BFS for Recursive Traversal | Time Complexity: O(V + E) as we visit each edge and node similarly to BFS, ensuring linear traversal in interconnected graphs. Space Complexity: O(V + E), including the internal stack usage depth required for recursion. |
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