#2039

Medium

auto_awesomeGraph traversal with breadth-first search

LeetCode Problem Workspace

The Time When the Network Becomes Idle

Calculate the time when the network becomes idle, factoring in message resending and the BFS traversal method.

Problem Statement

You are given a network of servers labeled 0 to n - 1. There is a 2D array edges, where each edge represents a message channel between two servers. The servers must send messages to the master server (labeled 0) and wait for a reply. Each server can resend messages until the server receives a response or the message has been resent a certain number of times based on patience values. The goal is to determine the moment when no more messages are exchanged.

You must simulate the process where data servers send messages to the master server and wait for replies. Once the message is processed and the response is sent back, a server may resend the message depending on the patience value for that server. If the server does not receive the reply in time, it will not resend. The task is to calculate when the network becomes idle, i.e., when no more messages are exchanged between servers.

Examples

Example 1

Input: edges = [[0,1],[1,2]], patience = [0,2,1]

Output: 8

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

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.length
2 <= n <= 105
patience[0] == 0
1 <= patience[i] <= 105 for 1 <= i < n
1 <= edges.length <= min(105, n * (n - 1) / 2)
edges[i].length == 2
0 <= ui, vi < n
ui != vi
There are no duplicate edges.
Each server can directly or indirectly reach another server.

Solution Approach

Breadth-First Search (BFS)

To solve this problem, we use BFS to find the shortest time for messages to reach the master server. Each server is treated as a node, and the edges represent direct communication channels. The BFS traversal will help determine the minimum time for messages to travel between servers.

Simulate Message Resending

After determining the shortest time for messages to reach the master server, we simulate the message resending process based on the patience array. For each server, we check if the message has been resent based on the elapsed time and patience threshold, allowing us to model the network's message traffic.

Determine Idle Time

Once the message passing and resending simulations are complete, we calculate when the network becomes idle. This is the moment when no server is sending or receiving messages anymore, which occurs when all servers have received replies and no further messages are being sent.

Complexity Analysis

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

The time complexity depends on the BFS traversal, which is O(n + m), where n is the number of servers and m is the number of edges. The space complexity depends on the storage needed for the graph and the message states, which is also O(n + m). Therefore, both time and space complexities are linear in the size of the graph.

What Interviewers Usually Probe

Understanding BFS traversal and its application to graph problems is key to solving this problem.
The ability to simulate resending based on a time threshold shows the candidate's understanding of dynamic processes.
Being able to optimize for large input sizes, ensuring that the solution can handle up to 10^5 servers, demonstrates scalability.

Common Pitfalls or Variants

Common pitfalls

Forgetting to account for the patience array when determining whether to resend a message.
Not simulating the message traffic accurately, which could lead to incorrect idle time calculations.
Overlooking edge cases where multiple servers resend messages at different times due to varying patience values.

Follow-up variants

What if the patience value for all servers was the same?
How would the solution change if some servers had infinite patience and would resend indefinitely?
What if we used a different graph traversal method other than BFS, like DFS, to solve the problem?

FAQ

What is the BFS approach used for in this problem?

BFS is used to find the shortest time for messages to travel from data servers to the master server in the network.

How does the patience array affect the solution?

The patience array determines whether a server resends its message based on how much time has passed since it sent the initial message.

What happens when all servers receive their replies?

Once all servers have received their replies, the network becomes idle as no more messages are being passed or received.

What would happen if some servers never resend messages?

If servers never resend messages, the idle time would be determined by when the first set of messages completes the round-trip to the master server.

How can I optimize my solution for larger inputs?

To optimize for larger inputs, focus on efficient BFS traversal and avoid unnecessary recalculations of message paths or resend logic.

terminal

Solution

Solution 1: BFS

First, we construct an undirected graph $g$ based on the 2D array $edges$, where $g[u]$ represents all neighboring nodes of node $u$.

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class Solution:
    def networkBecomesIdle(self, edges: List[List[int]], patience: List[int]) -> int:
        g = defaultdict(list)
        for u, v in edges:
            g[u].append(v)
            g[v].append(u)
        q = deque([0])
        vis = {0}
        ans = d = 0
        while q:
            d += 1
            t = d * 2
            for _ in range(len(q)):
                u = q.popleft()
                for v in g[u]:
                    if v not in vis:
                        vis.add(v)
                        q.append(v)
                        ans = max(ans, (t - 1) // patience[v] * patience[v] + t + 1)
        return ans

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