What is the time complexity of Network Delay Time?

The optimal solution has a time complexity of O(ElogV).

What is the space complexity of Network Delay Time?

The space complexity is O(V+E).

What data structures are used to solve Network Delay Time?

This problem uses Depth-First Search, Breadth-First Search, Graph, Heap (Priority Queue), Shortest Path. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Network Delay Time?

Insight 1: The problem can be modeled as a shortest path problem on a directed graph. Insight 2: Dijkstra's algorithm is well-suited for finding the shortest paths from a single source node to all other nodes in a weighted graph.

Which companies ask LeetCode 743. Network Delay Time in interviews?

This problem is frequently asked at Akuna Capital, Netflix.

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Network Delay Time - LeetCode 743 Solution

Network Delay Time - Complete Solution Guide

Network Delay Time is LeetCode problem 743, 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

You are given a network of n nodes, labeled from 1 to n . You are also given times , a list of travel times as directed edges times[i] = (u i , v i , w i ) , where u i is the source node, v i is the target node, and w i is the time it takes for a signal to travel from source to target. We will send a signal from a given node k . Return the minimum time it takes for all the n nodes to receive the signal . If it is impossible for all the n nodes to receive the signal, return -1 . Example 1: Input:

Detailed Explanation

The problem asks us to find the minimum time required for a signal to reach all nodes in a network. The network is represented as a directed graph where nodes are numbered from 1 to n, and edges represent travel times between nodes. We are given a starting node 'k' from which the signal originates. If it's impossible for the signal to reach all nodes, we return -1.

Solution Approach

The provided solutions use Dijkstra's algorithm to find the shortest time it takes for a signal to reach each node from the starting node 'k'. The algorithm maintains a priority queue to explore nodes in order of their shortest known time from the starting node. It also keeps track of the shortest time it takes to reach each node. After running Dijkstra's, the algorithm checks if all nodes have been reached (i.e., their shortest time is known). If so, it returns the maximum of these shortest times. Otherwise, it returns -1.

Step-by-Step Algorithm

Step 1: Build the graph represented as an adjacency list where the key is the source node and the value is a list of its neighbors along with the travel time to each neighbor.
Step 2: Initialize a priority queue (min-heap) to store nodes to visit, prioritized by the time it takes to reach them from the source node 'k'. Initially, the priority queue contains only the starting node 'k' with a time of 0.
Step 3: Initialize a dictionary/hash map 'dist' to store the shortest time it takes to reach each node from the source node. This can be used as a visited set as well.
Step 4: While the priority queue is not empty, extract the node with the smallest time from the priority queue.
Step 5: If the extracted node has already been visited (i.e., its shortest time is already known and stored in 'dist'), skip it.
Step 6: Mark the extracted node as visited and store its shortest time in 'dist'.
Step 7: Iterate through all the neighbors of the extracted node. For each neighbor, calculate the time it takes to reach the neighbor from the source node (which is the time to reach the current node plus the travel time from the current node to the neighbor).
Step 8: If the neighbor has not been visited, add it to the priority queue with its calculated time.
Step 9: After the priority queue is empty, check if all nodes have been visited (i.e., if the size of 'dist' is equal to 'n').
Step 10: If all nodes have been visited, find the maximum time in 'dist'. This is the minimum time it takes for all nodes to receive the signal. Return this maximum time.
Step 11: If not all nodes have been visited, it means that some nodes are not reachable from the starting node. Return -1.

Key Insights

Insight 1: The problem can be modeled as a shortest path problem on a directed graph.
Insight 2: Dijkstra's algorithm is well-suited for finding the shortest paths from a single source node to all other nodes in a weighted graph.
Insight 3: The problem requires us to check if all nodes are reachable from the starting node 'k' and return the maximum of the shortest path times to all reachable nodes.

Complexity Analysis

Time Complexity: O(ElogV)

Space Complexity: O(V+E)

Topics

This problem involves: Depth-First Search, Breadth-First Search, Graph, Heap (Priority Queue), Shortest Path.

Companies

Asked at: Akuna Capital, Netflix.

Network Delay Time | LeetCode 743

Medium

Depth-First Search Breadth-First Search Graph Heap (Priority Queue)Shortest Path

Time: O(ElogV)Space: O(V+E)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int networkDelayTime(int[][] times, int n, int k) {
        Map<Integer, List<int[]>> graph = new HashMap<>();
        for (int[] time : times) {
            int u = time[0];
            int v = time[1];
            int w = time[2];
            graph.computeIfAbsent(u, x -> new ArrayList<>()).add(new int[]{v, w});
        }

        PriorityQueue<int[]> pq = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
        pq.offer(new int[]{0, k});
        Map<Integer, Integer> dist = new HashMap<>();

        while (!pq.isEmpty()) {
            int[] curr = pq.poll();
            int time = curr[0];
            int node = curr[1];

            if (dist.containsKey(node)) {
                continue;
            }

            dist.put(node, time);

            if (graph.containsKey(node)) {
                for (int[] neighborInfo : graph.get(node)) {
                    int neighbor = neighborInfo[0];
                    int travel_time = neighborInfo[1];
                    if (!dist.containsKey(neighbor)) {
                        pq.offer(new int[]{time + travel_time, neighbor});
                    }
                }
            }
        }

        if (dist.size() == n) {
            int maxTime = 0;
            for (int time : dist.values()) {
                maxTime = Math.max(maxTime, time);
            }
            return maxTime;
        } else {
            return -1;
        }
    }
}

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Problem Info

Description

Problem 743: Network Delay Time - Detailed explanation and solution approaches.

Topics

Depth-First Search Breadth-First Search Graph Heap (Priority Queue)Shortest Path

Companies

Akuna CapitalNetflix

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:743

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

11 steps

1
Step 1: Build the graph represented as an adjacency list where the key is the source node and the value is a list of its neighbors along with the travel time to each neighbor.
2
Step 2: Initialize a priority queue (min-heap) to store nodes to visit, prioritized by the time it takes to reach them from the source node 'k'. Initially, the priority queue contains only the starting node 'k' with a time of 0.
3
Step 3: Initialize a dictionary/hash map 'dist' to store the shortest time it takes to reach each node from the source node. This can be used as a visited set as well.
4
Step 4: While the priority queue is not empty, extract the node with the smallest time from the priority queue.
5
Step 5: If the extracted node has already been visited (i.e., its shortest time is already known and stored in 'dist'), skip it.
6
Step 6: Mark the extracted node as visited and store its shortest time in 'dist'.
7
Step 7: Iterate through all the neighbors of the extracted node. For each neighbor, calculate the time it takes to reach the neighbor from the source node (which is the time to reach the current node plus the travel time from the current node to the neighbor).
8
Step 8: If the neighbor has not been visited, add it to the priority queue with its calculated time.
9
Step 9: After the priority queue is empty, check if all nodes have been visited (i.e., if the size of 'dist' is equal to 'n').
10
Step 10: If all nodes have been visited, find the maximum time in 'dist'. This is the minimum time it takes for all nodes to receive the signal. Return this maximum time.
11
Step 11: If not all nodes have been visited, it means that some nodes are not reachable from the starting node. Return -1.

💡

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(ElogV)

💾

Space Complexity

O(V+E)

📝

Why?

The time complexity is determined by the number of edges and vertices because we iterate over the edges in the graph to find the shortest path. The priority queue ensures we are always visiting the next closest unvisited node. The space complexity comes from storing the graph's adjacency list (O(E)) and the distances to each node from the source (O(V)).

📖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

Bellman-Ford Algorithm

Bellman-Ford algorithm can also be used to solve this problem. It can handle negative edge weights (which are not present in this problem, but it's a more general algorithm).

Time: O(VE), where V is the number of vertices and E is the number of edges.Space: O(V), for storing distances to each node.

Trade-offs: Bellman-Ford is generally slower than Dijkstra's algorithm when there are no negative edge weights. However, it can handle negative edge weights, which Dijkstra's cannot. If the graph is dense (E is close to V^2), Bellman-Ford might be comparable to Dijkstra's with a poorly implemented priority queue. The provided problem states that edge weights are non-negative, so Dijkstra's is a better choice here.

Floyd-Warshall Algorithm

Floyd-Warshall algorithm finds the shortest paths between all pairs of nodes in a graph.

Time: O(V^3), where V is the number of vertices.Space: O(V^2), to store the distance matrix.

Trade-offs: Floyd-Warshall is not suitable for this problem because it finds all-pairs shortest paths, while we only need the shortest paths from a single source node 'k'. Its time and space complexities are higher than Dijkstra's for this specific problem.

✨

Key Insights

1Insight 1: The problem can be modeled as a shortest path problem on a directed graph.
2Insight 2: Dijkstra's algorithm is well-suited for finding the shortest paths from a single source node to all other nodes in a weighted graph.
3Insight 3: The problem requires us to check if all nodes are reachable from the starting node 'k' and return the maximum of the shortest path times to all reachable nodes.

📚

Key Terms Explained

For Students

Graph

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

Edge Cases

Edge case 1: The graph is disconnected, meaning some nodes are unreachable from the starting node 'k'. The solution correctly handles this by checking if the number of visited nodes ('dist' size) is equal to the total number of nodes 'n'. If not, it returns -1.
Edge case 2: The starting node 'k' cannot reach any other node. In this case, 'dist' will only contain 'k', and the solution will correctly return -1 because the number of nodes in 'dist' will be less than 'n'.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where some nodes are unreachable from the starting node. This results in an incorrect maximum time being returned, or potentially an infinite loop.

✗Common mistake 2: Using a naive implementation of Dijkstra's algorithm without a priority queue, leading to a time complexity of O(V^2) or O(E*V) which may be too slow for larger graphs.

✗Common mistake 3: Incorrectly building the graph. For example, confusing the source and destination nodes, or not properly storing the travel times. This leads to wrong shortest paths being calculated.

✅

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: Routing network packets. The network delay time problem is directly applicable to finding the minimum time it takes for a packet to reach all nodes in a network, which is crucial for efficient network routing.

🏦Real-world scenario 2: Delivery logistics. Determining the shortest time to deliver goods to all locations from a central depot can be modeled as a network delay time problem. The nodes represent locations, and the edges represent travel times between locations.

💡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

Always test your solution with edge cases: empty input, single element, and boundary values.

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: June 24, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaMay 29

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

@TechLeadMay 22

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

@DevExpert92May 15

❓

Frequently Asked Questions

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 "Network Delay Time"?

For this medium-level Depth-First Search 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?