What is the time complexity of Find Edges in Shortest Paths?

The optimal solution has a time complexity of O(m + n + m*log(n)).

What is the space complexity of Find Edges in Shortest Paths?

The space complexity is O(n + m).

What data structures are used to solve Find Edges in Shortest Paths?

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

What are the key insights for solving Find Edges in Shortest Paths?

Insight 1: We need to find the shortest distance between node 0 and node n-1. Dijkstra's algorithm is the standard approach for finding shortest paths in a weighted graph. Insight 2: Once the shortest distance between node 0 and node n-1 is known, we can determine if an edge (u, v, w) lies on any shortest path by checking if dist(0, u) + w + dist(v, n-1) == shortest_distance or dist(0, v) + w + dist(u, n-1) == shortest_distance. This leverages the optimal substructure property of shortest paths.

Which companies ask LeetCode 3123. Find Edges in Shortest Paths in interviews?

This problem is frequently asked at WeRide.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Find Edges in Shortest Paths - LeetCode 3123 Solution

Find Edges in Shortest Paths - Complete Solution Guide

Find Edges in Shortest Paths is LeetCode problem 3123, a Hard 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 an undirected weighted graph of n nodes numbered from 0 to n - 1 . The graph consists of m edges represented by a 2D array edges , where edges[i] = [a i , b i , w i ] indicates that there is an edge between nodes a i and b i with weight w i . Consider all the shortest paths from node 0 to node n - 1 in the graph. You need to find a boolean array answer where answer[i] is true if the edge edges[i] is part of at least one shortest path. Otherwise, answer[i] is false . Return the arra

Detailed Explanation

The problem asks us to find all edges that lie on at least one shortest path between node 0 and node n-1 in a given undirected weighted graph. The input is the number of nodes 'n' and a list of edges, where each edge is represented by a tuple (u, v, w), denoting an edge between nodes u and v with weight w. The output is a boolean array 'answer' of the same length as the 'edges' array. answer[i] is true if edge edges[i] belongs to at least one shortest path from node 0 to n-1, and false otherwise. The graph may not be connected, and there are no repeated edges.

Solution Approach

The solution uses Dijkstra's algorithm to find the shortest distances from node 0 to all other nodes (dist_from_0) and from node n-1 to all other nodes (dist_from_n_1). Then, it iterates through each edge and checks if that edge can be part of a shortest path between node 0 and node n-1. The core idea is that an edge (u, v, w) is on a shortest path from 0 to n-1 if and only if the shortest distance from 0 to u, plus the weight of the edge (u, v), plus the shortest distance from v to n-1 equals the shortest distance from 0 to n-1 (and similarly for the path from 0 to v, through edge (v, u) to n-1).

Step-by-Step Algorithm

Step 1: Build an adjacency list representation of the graph from the given edge list.
Step 2: Use Dijkstra's algorithm to compute the shortest distances from node 0 to all other nodes. Store the result in dist_from_0.
Step 3: Use Dijkstra's algorithm to compute the shortest distances from node n-1 to all other nodes. Store the result in dist_from_n_1.
Step 4: Calculate the shortest distance between node 0 and node n-1, using the result from dist_from_0.
Step 5: Create a boolean array 'answer' to store the results. Initialize all elements to false.
Step 6: Iterate through each edge (u, v, w) in the 'edges' list. For each edge, check if dist_from_0[u] + w + dist_from_n_1[v] == shortest_path_len OR dist_from_0[v] + w + dist_from_n_1[u] == shortest_path_len. If either condition is true, then the edge is part of at least one shortest path, so set answer[i] to true.
Step 7: Return the 'answer' array.

Key Insights

Insight 1: We need to find the shortest distance between node 0 and node n-1. Dijkstra's algorithm is the standard approach for finding shortest paths in a weighted graph.
Insight 2: Once the shortest distance between node 0 and node n-1 is known, we can determine if an edge (u, v, w) lies on any shortest path by checking if dist(0, u) + w + dist(v, n-1) == shortest_distance or dist(0, v) + w + dist(u, n-1) == shortest_distance. This leverages the optimal substructure property of shortest paths.
Insight 3: The graph may be disconnected. Thus the shortest distance between 0 and n-1 might be infinite. This must be checked.

Complexity Analysis

Time Complexity: O(m + n + m*log(n))

Space Complexity: O(n + m)

Topics

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

Companies

Asked at: WeRide.

Find Edges in Shortest Paths | LeetCode 3123

Hard

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

Time: O(m + n + m*log(n))Space: O(n + m)

Solution Code

Available in:

import java.util.*;

class Solution {
    public boolean[] findAnswer(int n, int[][] edges) {
        Map<Integer, List<int[]>> graph = new HashMap<>();
        for (int i = 0; i < n; i++) {
            graph.put(i, new ArrayList<>());
        }
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            int w = edge[2];
            graph.get(u).add(new int[]{v, w});
            graph.get(v).add(new int[]{u, w});
        }

        class DijkstraResult {
            public int[] distances;

            public DijkstraResult(int[] distances) {
                this.distances = distances;
            }
        }
        
        class Dijkstra {
            public DijkstraResult dijkstra(int start_node, int n, Map<Integer, List<int[]>> graph) {
                int[] distances = new int[n];
                Arrays.fill(distances, Integer.MAX_VALUE);
                distances[start_node] = 0;
                PriorityQueue<int[]> pq = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
                pq.offer(new int[]{0, start_node});

                while (!pq.isEmpty()) {
                    int[] curr = pq.poll();
                    int dist = curr[0];
                    int u = curr[1];
                    if (dist > distances[u]) {
                        continue;
                    }

                    for (int[] neighbor : graph.get(u)) {
                        int v = neighbor[0];
                        int weight = neighbor[1];
                        if (distances[u] != Integer.MAX_VALUE && distances[u] + weight < distances[v]) {
                            distances[v] = distances[u] + weight;
                            pq.offer(new int[]{distances[v], v});
                        }
                    }
                }
                return new DijkstraResult(distances);
            }
        }
        Dijkstra dijkstraInstance = new Dijkstra();
        DijkstraResult dist_from_0_result = dijkstraInstance.dijkstra(0, n, graph);
        int[] dist_from_0 = dist_from_0_result.distances;
        DijkstraResult dist_from_n_1_result = dijkstraInstance.dijkstra(n - 1, n, graph);
        int[] dist_from_n_1 = dist_from_n_1_result.distances;


        int shortest_path_len = dist_from_0[n - 1];

        int m = edges.length;
        boolean[] answer = new boolean[m];

        if (shortest_path_len == Integer.MAX_VALUE) {
            return answer;
        }

        for (int i = 0; i < m; i++) {
            int u = edges[i][0];
            int v = edges[i][1];
            int w = edges[i][2];
            if (dist_from_0[u] != Integer.MAX_VALUE && dist_from_n_1[v] != Integer.MAX_VALUE && dist_from_0[u] + w + dist_from_n_1[v] == shortest_path_len) {
                answer[i] = true;
            } else if (dist_from_0[v] != Integer.MAX_VALUE && dist_from_n_1[u] != Integer.MAX_VALUE && dist_from_0[v] + w + dist_from_n_1[u] == shortest_path_len) {
                answer[i] = true;
            }
        }

        return answer;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 3123: Find Edges in Shortest Paths - Detailed explanation and solution approaches.

Topics

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

Companies

WeRide

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3123

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 an adjacency list representation of the graph from the given edge list.
2
Step 2: Use Dijkstra's algorithm to compute the shortest distances from node 0 to all other nodes. Store the result in dist_from_0.
3
Step 3: Use Dijkstra's algorithm to compute the shortest distances from node n-1 to all other nodes. Store the result in dist_from_n_1.
4
Step 4: Calculate the shortest distance between node 0 and node n-1, using the result from dist_from_0.
5
Step 5: Create a boolean array 'answer' to store the results. Initialize all elements to false.
6
Step 6: Iterate through each edge (u, v, w) in the 'edges' list. For each edge, check if dist_from_0[u] + w + dist_from_n_1[v] == shortest_path_len OR dist_from_0[v] + w + dist_from_n_1[u] == shortest_path_len. If either condition is true, then the edge is part of at least one shortest path, so set answer[i] to true.
7
Step 7: Return the 'answer' array.

💡

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(m + n + m*log(n))

💾

Space Complexity

O(n + m)

📝

Why?

The time complexity is dominated by the two calls to Dijkstra's algorithm. The space complexity comes from storing the graph and the distance arrays used within Dijkstra. In the Java implementation, since `Integer.MAX_VALUE` is used, there is a potential risk of integer overflow. In the C++ implementation, `long long` is used to address this overflow.

📖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 can also be used to find the shortest paths. However, it has a time complexity of O(n*m), which is generally slower than Dijkstra's algorithm for sparse graphs.

Time: O(n*m)Space: O(n)

Trade-offs: Bellman-Ford can handle negative edge weights, whereas Dijkstra's algorithm cannot. However, since the problem statement does not mention negative edge weights, Dijkstra's algorithm is more efficient.

Floyd-Warshall Algorithm

Floyd-Warshall computes shortest paths between all pairs of nodes in the graph. Can be used here. However, its complexity is higher.

Time: O(n^3)Space: O(n^2)

Trade-offs: Suitable when shortest paths between all pairs of nodes are needed. For this specific problem, it is less efficient since we are only interested in paths originating from node 0 and node n-1.

✨

Key Insights

1Insight 1: We need to find the shortest distance between node 0 and node n-1. Dijkstra's algorithm is the standard approach for finding shortest paths in a weighted graph.
2Insight 2: Once the shortest distance between node 0 and node n-1 is known, we can determine if an edge (u, v, w) lies on any shortest path by checking if dist(0, u) + w + dist(v, n-1) == shortest_distance or dist(0, v) + w + dist(u, n-1) == shortest_distance. This leverages the optimal substructure property of shortest paths.
3Insight 3: The graph may be disconnected. Thus the shortest distance between 0 and n-1 might be infinite. This must be checked.

📚

Key Terms Explained

For Students

Graph

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

Edge Cases

Edge case 1: Disconnected graph. If there is no path from node 0 to node n-1, the shortest distance will be infinity. The solution handles this by checking if shortest_path_len == infinity. If so, it returns an answer array filled with false values.
Edge case 2: Single node graph. The provided constraints state that n >= 2 so this is impossible. If n=1, the algorithm should still work, but will need extra handling.
Edge case 3: No edges. The algorithm handles this as a special case of a disconnected graph.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer Overflow. In languages like Java and C++, using `Integer.MAX_VALUE` for infinity can lead to integer overflow when adding edge weights. Using `long` type or `Long.MAX_VALUE` resolves this. C++'s implementation uses `long long`, but careful usage is needed in Java.

✗Common mistake 2: Not considering both directions of the edge. Since the graph is undirected, it's important to consider both dist_from_0[u] + w + dist_from_n_1[v] and dist_from_0[v] + w + dist_from_n_1[u] when determining if an edge lies on a shortest path.

✗Common mistake 3: Incorrectly implementing Dijkstra's algorithm. A proper implementation of Dijkstra's algorithm, using a priority queue to select the next node to visit, is crucial for the correct solution and the desired time complexity.

✅

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 routing. Determining which links are part of the shortest paths between two nodes in a communication network can help optimize network performance and ensure reliable data transmission.

🏦Real-world scenario 2: Transportation planning. Finding the most critical roads (edges) in a city's road network for commuting between two locations can help in infrastructure planning and traffic 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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: August 3, 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.

@CodeNinjaJul 13

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

@TechLeadJul 6

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

@DevExpert92Jun 29

❓

Frequently Asked Questions

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Find Edges in Shortest Paths"?

For this hard-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?