What is the time complexity of Reachable Nodes In Subdivided Graph?

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

What is the space complexity of Reachable Nodes In Subdivided Graph?

The space complexity is O(N+E).

What data structures are used to solve Reachable Nodes In Subdivided Graph?

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

What are the key insights for solving Reachable Nodes In Subdivided Graph?

Insight 1: The problem is essentially a shortest path problem on a modified graph. We need to find the shortest distance from node 0 to every other node. Insight 2: Dijkstra's algorithm is well-suited for finding the shortest paths from a single source node to all other nodes in a graph with non-negative edge weights.

Which companies ask LeetCode 882. Reachable Nodes In Subdivided Graph in interviews?

This problem is frequently asked at PhonePe.

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Reachable Nodes In Subdivided Graph - LeetCode 882 Solution

Reachable Nodes In Subdivided Graph - Complete Solution Guide

Reachable Nodes In Subdivided Graph is LeetCode problem 882, 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 graph (the "original graph" ) with n nodes labeled from 0 to n - 1 . You decide to subdivide each edge in the graph into a chain of nodes, with the number of new nodes varying between each edge. The graph is given as a 2D array of edges where edges[i] = [u i , v i , cnt i ] indicates that there is an edge between nodes u i and v i in the original graph, and cnt i is the total number of new nodes that you will subdivide the edge into. Note that cnt i == 0 means you wil

Detailed Explanation

The problem asks us to find the number of reachable nodes from node 0 in a subdivided graph. We are given the original graph as a list of edges, where each edge is represented by [u, v, cnt]. This means there's an edge between nodes 'u' and 'v' that will be subdivided into 'cnt' new nodes and 'cnt+1' new edges. After subdividing the graph, we need to determine how many nodes (both original and new) are reachable from node 0 within a maximum number of moves, 'maxMoves'. The output is the total count of reachable nodes.

Solution Approach

The solution uses Dijkstra's algorithm to find the shortest distances from node 0 to all other original nodes in the graph. After computing these shortest distances, it iterates through the original edges and calculates the number of subdivision nodes reachable from each endpoint of the edge within the 'maxMoves' limit. The final count is the sum of reachable original nodes and reachable subdivision nodes.

Step-by-Step Algorithm

Step 1: Build an adjacency list representation of the graph. The weight of an edge (u, v) is cnt + 1, where cnt is the number of subdivision nodes between u and v.
Step 2: Initialize a distance map/array 'dist' to store the shortest distances from node 0 to all other nodes. Set all distances to infinity initially, except for dist[0], which is set to 0.
Step 3: Use a priority queue (min-heap) to implement Dijkstra's algorithm. The priority queue stores nodes ordered by their distances from the source node.
Step 4: While the priority queue is not empty, extract the node 'u' with the smallest distance from the queue.
Step 5: For each neighbor 'v' of node 'u', calculate the distance from node 0 to 'v' through 'u'. If this distance is shorter than the current distance to 'v', update dist[v] and add 'v' to the priority queue.
Step 6: After Dijkstra's algorithm completes, count the number of original nodes 'i' for which dist[i] <= maxMoves.
Step 7: Iterate through the original edges and for each edge (u, v, cnt), calculate the number of subdivision nodes reachable from 'u' and 'v'. The number of reachable nodes is min(cnt, reach_from_u + reach_from_v), where reach_from_u = max(0, maxMoves - dist[u]) and reach_from_v = max(0, maxMoves - dist[v]).
Step 8: Add the number of reachable subdivision nodes to the result and return the final count.

Key Insights

Insight 1: The problem is essentially a shortest path problem on a modified graph. We need to find the shortest distance from node 0 to every other node.
Insight 2: Dijkstra's algorithm is well-suited for finding the shortest paths from a single source node to all other nodes in a graph with non-negative edge weights.
Insight 3: After finding the shortest distances to the original nodes, we need to carefully count the reachable subdivision nodes by considering the remaining moves from the two endpoint nodes of each edge.

Complexity Analysis

Time Complexity: O(ElogN)

Space Complexity: O(N+E)

Topics

This problem involves: Graph, Heap (Priority Queue), Shortest Path.

Companies

Asked at: PhonePe.

Reachable Nodes In Subdivided Graph | LeetCode 882

Hard

Graph Heap (Priority Queue)Shortest Path

Time: O(ElogN)Space: O(N+E)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int reachableNodes(int[][] edges, int maxMoves, int n) {
        Map<Integer, List<int[]>> 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];
            int cnt = edge[2];
            int weight = cnt + 1;
            adj.get(u).add(new int[]{v, weight});
            adj.get(v).add(new int[]{u, weight});
        }

        Map<Integer, Integer> dist = new HashMap<>();
        for (int i = 0; i < n; i++) {
            dist.put(i, Integer.MAX_VALUE);
        }
        dist.put(0, 0);
        PriorityQueue<int[]> pq = new PriorityQueue<>(Comparator.comparingInt(a -> a[0])); // (distance, node)
        pq.offer(new int[]{0, 0});

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

            if (d > dist.get(u)) {
                continue;
            }

            for (int[] neighbor : adj.get(u)) {
                int v = neighbor[0];
                int weight = neighbor[1];
                if (dist.get(u) + weight < dist.get(v)) {
                    dist.put(v, dist.get(u) + weight);
                    pq.offer(new int[]{dist.get(v), v});
                }
            }
        }

        // Count reachable original nodes
        int result = 0;
        for (int i = 0; i < n; i++) {
            if (dist.get(i) <= maxMoves) {
                result++;
            }
        }

        // Count reachable subdivision nodes
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            int cnt = edge[2];
            int reach_from_u = Math.max(0, maxMoves - dist.get(u));
            int reach_from_v = Math.max(0, maxMoves - dist.get(v));

            result += Math.min(cnt, reach_from_u + reach_from_v);
        }

        return result;
    }
}

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Problem Info

Description

Problem 882: Reachable Nodes In Subdivided Graph - Detailed explanation and solution approaches.

Topics

Graph Heap (Priority Queue)Shortest Path

Companies

PhonePe

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:882

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

8 steps

1
Step 1: Build an adjacency list representation of the graph. The weight of an edge (u, v) is cnt + 1, where cnt is the number of subdivision nodes between u and v.
2
Step 2: Initialize a distance map/array 'dist' to store the shortest distances from node 0 to all other nodes. Set all distances to infinity initially, except for dist[0], which is set to 0.
3
Step 3: Use a priority queue (min-heap) to implement Dijkstra's algorithm. The priority queue stores nodes ordered by their distances from the source node.
4
Step 4: While the priority queue is not empty, extract the node 'u' with the smallest distance from the queue.
5
Step 5: For each neighbor 'v' of node 'u', calculate the distance from node 0 to 'v' through 'u'. If this distance is shorter than the current distance to 'v', update dist[v] and add 'v' to the priority queue.
6
Step 6: After Dijkstra's algorithm completes, count the number of original nodes 'i' for which dist[i] <= maxMoves.
7
Step 7: Iterate through the original edges and for each edge (u, v, cnt), calculate the number of subdivision nodes reachable from 'u' and 'v'. The number of reachable nodes is min(cnt, reach_from_u + reach_from_v), where reach_from_u = max(0, maxMoves - dist[u]) and reach_from_v = max(0, maxMoves - dist[v]).
8
Step 8: Add the number of reachable subdivision nodes to the result and return the final count.

💡

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(ElogN)

💾

Space Complexity

O(N+E)

📝

Why?

The time complexity is dominated by Dijkstra's algorithm's priority queue operations. The space complexity accounts for the adjacency list representation of the graph and the distance map.

📖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 the shortest path problem. However, it's generally less efficient than Dijkstra's algorithm for graphs with non-negative edge weights.

Time: O(V*E), where V is the number of vertices and E is the number of edges.Space: O(V)

Trade-offs: Bellman-Ford is more suitable for graphs with negative edge weights, but it's less efficient than Dijkstra's algorithm for graphs with non-negative edge weights. Therefore, it is not an ideal choice for this problem.

✨

Key Insights

1Insight 1: The problem is essentially a shortest path problem on a modified graph. We need to find the shortest distance from node 0 to every other node.
2Insight 2: Dijkstra's algorithm is well-suited for finding the shortest paths from a single source node to all other nodes in a graph with non-negative edge weights.
3Insight 3: After finding the shortest distances to the original nodes, we need to carefully count the reachable subdivision nodes by considering the remaining moves from the two endpoint nodes of each edge.

📚

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. Dijkstra's algorithm correctly handles disconnected graphs. Nodes that are not reachable from node 0 will have a distance of infinity, and they will not be counted as reachable.
Edge case 2: maxMoves is smaller than any edge weight. This will lead to only the starting node 0 being reachable in some cases, which the solution correctly identifies. Edge case where n = 1, the code correctly returns 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the reachable subdivision nodes after calculating shortest distances to the original nodes. The problem requires counting both original and subdivided nodes.

✗Common mistake 2: Not correctly implementing Dijkstra's algorithm. It's crucial to use a priority queue to efficiently select the node with the smallest distance at each step.

✗Common mistake 3: Integer overflow when calculating distances. While not explicitly checked in these solutions, it's a potential pitfall if distances become very large. Using a long data type would be safer in such cases.

✗Common mistake 4: Not handling the maxMoves = 0 case properly. If maxMoves is 0, only the starting node 0 is reachable.

✅

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. Finding the shortest path between two nodes in a network, where the edges represent network links and the weights represent latency or cost.

🏦Real-world scenario 2: Transportation planning. Determining the number of reachable locations from a starting point within a certain travel time or distance, given a road network with varying road lengths.

💡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

Before coding, decide between BFS (shortest path, level-order) and DFS (exhaustive search, connected components). Draw the graph on paper first.

🎯 Interview Tip

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

⚠️ Watch Out

Forgetting to mark nodes as visited, causing infinite loops. Also watch for disconnected components.

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: November 10, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

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

@CodeNinjaJul 7

❓

Frequently Asked Questions

How do I represent a graph in code?

Use adjacency list (HashMap or array of lists) for sparse graphs - it's space-efficient O(V+E). Use adjacency matrix for dense graphs or when you need O(1) edge lookup. For weighted graphs, store (neighbor, weight) pairs in adjacency lists.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Reachable Nodes In Subdivided Graph"?

For this hard-level Graph problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether heap (priority queue) techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?