What is the time complexity of The Time When the Network Becomes Idle?

The optimal solution has a time complexity of O(n + E).

What is the space complexity of The Time When the Network Becomes Idle?

The space complexity is O(n + E).

What data structures are used to solve The Time When the Network Becomes Idle?

This problem uses Array, Breadth-First Search, Graph. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving The Time When the Network Becomes Idle?

Insight 1: The shortest distance from each data server to the master server is crucial to calculate the round trip time (RTT). Insight 2: The 'patience' value determines how often a server resends its message. Calculating the last time a server sends its message based on its patience and RTT is essential.

Which companies ask LeetCode 2039. The Time When the Network Becomes Idle in interviews?

This problem is frequently asked at Atlassian, Deutsche Bank.

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The Time When the Network Becomes Idle - LeetCode 2039 Solution

The Time When the Network Becomes Idle - Complete Solution Guide

The Time When the Network Becomes Idle is LeetCode problem 2039, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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] = [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 se

Detailed Explanation

The problem describes a network of 'n' servers where server 0 is the master server and the rest are data servers. The data servers need to send messages to the master server and receive replies. The goal is to find the earliest time when the network becomes idle, meaning no messages are being transmitted or received. Data servers resend their messages every 'patience[i]' seconds until they receive a reply from the master server. The network's topology is defined by the 'edges' array, which specifies the connections between the servers. Messages travel along the shortest paths.

Solution Approach

The solution uses Breadth-First Search (BFS) to calculate the shortest distance (number of hops) from each data server to the master server (server 0). Then, for each data server, it computes the round trip time (RTT) which is twice the shortest distance to the master server. Using RTT and patience level, it determines the last arrival time of the reply to that server. Finally, it finds the maximum of all the last arrival times and adds 1 to it. This gives the time when the entire network becomes idle.

Step-by-Step Algorithm

Step 1: Build the adjacency list representation of the network graph from the given 'edges'.
Step 2: Use BFS, starting from the master server (server 0), to compute the shortest distance from the master server to each of the data servers. Store the distances in an array.
Step 3: Iterate through the data servers (servers 1 to n-1). For each server, calculate the Round Trip Time (RTT) as twice the distance from the server to the master server.
Step 4: Determine the last time the server sent a message. This depends on whether the RTT is less than or equal to the patience value. If RTT <= patience, the message is only sent once initially. Otherwise, determine the last send time using formula ((rtt-1) / patience[i]) * patience[i]
Step 5: Compute the last arrival time of the reply for each server. This is the last send time + RTT.
Step 6: Find the maximum of the last arrival times across all data servers.
Step 7: Return the maximum last arrival time + 1, which is the time when the network becomes idle.

Key Insights

Insight 1: The shortest distance from each data server to the master server is crucial to calculate the round trip time (RTT).
Insight 2: The 'patience' value determines how often a server resends its message. Calculating the last time a server sends its message based on its patience and RTT is essential.
Insight 3: The network becomes idle only after all servers have received replies and no messages are in transit. Therefore, the maximum last arrival time among all the servers determines the idle time.

Complexity Analysis

Time Complexity: O(n + E)

Space Complexity: O(n + E)

Topics

This problem involves: Array, Breadth-First Search, Graph.

Companies

Asked at: Atlassian, Deutsche Bank.

The Time When the Network Becomes Idle | LeetCode 2039

Medium

Array Breadth-First Search Graph

Time: O(n + E)Space: O(n + E)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int networkBecomesIdle(int[][] edges, int[] patience) {
        int n = patience.length;
        Map<Integer, List<Integer>> adj = new HashMap<>();
        for (int i = 0; i < n; i++) {
            adj.put(i, new ArrayList<>());
        }
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            adj.get(u).add(v);
            adj.get(v).add(u);
        }

        int[] dist = new int[n];
        Arrays.fill(dist, -1);
        Queue<Pair<Integer, Integer>> queue = new LinkedList<>();
        queue.offer(new Pair<>(0, 0));
        dist[0] = 0;

        while (!queue.isEmpty()) {
            Pair<Integer, Integer> p = queue.poll();
            int u = p.getKey();
            int d = p.getValue();
            for (int v : adj.get(u)) {
                if (dist[v] == -1) {
                    dist[v] = d + 1;
                    queue.offer(new Pair<>(v, d + 1));
                }
            }
        }

        int max_last_arrival = 0;
        for (int i = 1; i < n; i++) {
            int rtt = 2 * dist[i];
            int p = patience[i];

            int last_arrival_time;
            if (rtt <= p) {
                last_arrival_time = rtt;
            } else {
                int last_send_time = ((rtt - 1) / p) * p;
                last_arrival_time = last_send_time + rtt;
            }

            max_last_arrival = Math.max(max_last_arrival, last_arrival_time);
        }

        return max_last_arrival + 1;
    }

    static class Pair<K, V> {
        private final K key;
        private final V value;

        public Pair(K key, V value) {
            this.key = key;
            this.value = value;
        }

        public K getKey() {
            return key;
        }

        public V getValue() {
            return value;
        }
    }
}

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Problem Info

Description

Problem 2039: The Time When the Network Becomes Idle - Detailed explanation and solution approaches.

Topics

Array Breadth-First Search Graph

Companies

AtlassianDeutsche Bank

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2039

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Build the adjacency list representation of the network graph from the given 'edges'.
2
Step 2: Use BFS, starting from the master server (server 0), to compute the shortest distance from the master server to each of the data servers. Store the distances in an array.
3
Step 3: Iterate through the data servers (servers 1 to n-1). For each server, calculate the Round Trip Time (RTT) as twice the distance from the server to the master server.
4
Step 4: Determine the last time the server sent a message. This depends on whether the RTT is less than or equal to the patience value. If RTT <= patience, the message is only sent once initially. Otherwise, determine the last send time using formula ((rtt-1) / patience[i]) * patience[i]
5
Step 5: Compute the last arrival time of the reply for each server. This is the last send time + RTT.
6
Step 6: Find the maximum of the last arrival times across all data servers.
7
Step 7: Return the maximum last arrival time + 1, which is the time when the network becomes idle.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n + E)

💾

Space Complexity

O(n + E)

📝

Why?

The time complexity is determined by the BFS, which visits each node and edge once. The space complexity is determined by the adjacency list, distance array, and the BFS queue. The adjacency list stores the graph structure, the distance array keeps track of the shortest path from the root, and queue is needed for BFS.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dijkstra's Algorithm

Dijkstra's Algorithm could be used to find the shortest paths from the master server to all other servers. While BFS is sufficient for unweighted graphs, Dijkstra's is suitable for weighted graphs.

Time: O(E + n log n) with a priority queue.Space: O(n + E)

Trade-offs: Dijkstra's is more general but adds complexity to the code, and isn't needed for the given problem since the edges are unweighted. BFS is simpler and more efficient in this scenario.

✨

Key Insights

1Insight 1: The shortest distance from each data server to the master server is crucial to calculate the round trip time (RTT).
2Insight 2: The 'patience' value determines how often a server resends its message. Calculating the last time a server sends its message based on its patience and RTT is essential.
3Insight 3: The network becomes idle only after all servers have received replies and no messages are in transit. Therefore, the maximum last arrival time among all the servers determines the idle time.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: The graph is disconnected. The problem statement guarantees that the graph is connected, so this case is handled by the problem constraints.
Edge case 2: Patience value is larger than RTT. The calculation for last_send_time needs to account for this case where the server doesn't resend the message. The code correctly handles it.
Edge case 3: patience[i] is equal to 1. The last send time formula ((rtt - 1) / patience[i]) * patience[i] should work correctly in this situation. The provided solution does handle this case correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating RTT (forgetting to multiply the distance by 2).

✗Common mistake 2: Not accounting for the patience value when calculating the last message send time, leading to an incorrect last arrival time.

✗Common mistake 3: Not adding 1 to the maximum last arrival time to determine the idle time.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Network monitoring. Determining when a network becomes idle can help optimize resource allocation and identify potential bottlenecks.

🏦Real-world scenario 2: Distributed systems. Coordinating tasks in a distributed system requires understanding message propagation delays and resending mechanisms, similar to this problem. Understanding when all tasks have completed and the system is idle is crucial for system management.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 30, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "The Time When the Network Becomes Idle"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether breadth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?