We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsDisclaimersPrivacyTermsContactAbout Us

The Time When the Network Becomes Idle

Difficulty: Medium

Problem Description

There is a network of n servers, labeled from 0 to n - 1. You are given a 2D integer array edges, where edges[i] = [u_i, v_i] indicates there is a message channel between servers u_i and v_i, 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:

  • If it has not, it will resend the message periodically. The data server i will resend the message every patience[i] second(s).
  • Otherwise, no more resending will occur 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.

Key Insights

  • Each data server sends a message to the master server at the start.
  • Messages travel optimally, and replies are sent back through the same path.
  • Servers will resend messages based on their patience level if they don't receive a reply.
  • The network becomes idle when all servers have received replies and are no longer resending messages.

Space and Time Complexity

Time Complexity: O(n + e), where n is the number of servers and e is the number of edges in the graph. Space Complexity: O(n + e), for storing the graph and the distance from each server to the master.

Solution

To solve this problem, we can:

  1. Represent the network as a graph using an adjacency list.
  2. Use a Breadth-First Search (BFS) to calculate the shortest distance from each server to the master server (server 0).
  3. For each data server, calculate the time it will take for the first reply to arrive and the time when they will resend messages based on their patience level.
  4. Determine the maximum time at which the last server will receive its final reply to find when the network becomes idle.

We will use a queue for BFS and a list to store the distances.

Code Solutions

from collections import deque

def networkIdleTime(edges, patience):
    # Create adjacency list for the graph
    graph = [[] for _ in range(len(patience))]
    for u, v in edges:
        graph[u].append(v)
        graph[v].append(u)

    # BFS to find the shortest distance from the master server (0)
    dist = [-1] * len(patience)
    dist[0] = 0
    queue = deque([0])

    while queue:
        node = queue.popleft()
        for neighbor in graph[node]:
            if dist[neighbor] == -1:  # Not visited
                dist[neighbor] = dist[node] + 1
                queue.append(neighbor)

    # Calculate the time until the network becomes idle
    max_time = 0
    for i in range(1, len(patience)):
        round_trip_time = dist[i] * 2  # Time to send and receive the message
        # Calculate the last time this server resends the message
        last_send_time = (round_trip_time - 1) // patience[i] * patience[i]
        max_time = max(max_time, last_send_time + round_trip_time)

    return max_time + 1  # Add 1 for the idle time
← Back to All Questions