Leetcode #2039: The Time When the Network Becomes Idle
In this guide, we solve Leetcode #2039 The Time When the Network Becomes Idle in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Breadth-First Search, Graph, Array
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
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.
Python Solution
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
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.