What is the time complexity of Minimum Edge Reversals So Every Node Is Reachable?

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

What data structures are used to solve Minimum Edge Reversals So Every Node Is Reachable?

This problem uses Dynamic Programming, Depth-First Search, Breadth-First Search, Graph. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Edge Reversals So Every Node Is Reachable?

Insight 1: The problem can be solved using a two-pass approach based on dynamic programming and graph traversal (BFS or DFS). Insight 2: The core idea is to first calculate the number of reversals needed for a specific root (node 0 in this case).

Which companies ask LeetCode 2858. Minimum Edge Reversals So Every Node Is Reachable in interviews?

This problem is frequently asked at MathWorks.

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Minimum Edge Reversals So Every Node Is Reachable - LeetCode 2858 Solution

Minimum Edge Reversals So Every Node Is Reachable - Complete Solution Guide

Minimum Edge Reversals So Every Node Is Reachable is LeetCode problem 2858, 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

There is a simple directed graph with n nodes labeled from 0 to n - 1 . The graph would form a tree if its edges were bi-directional. You are given an integer n and a 2D integer array edges , where edges[i] = [u i , v i ] represents a directed edge going from node u i to node v i . An edge reversal changes the direction of an edge, i.e., a directed edge going from node u i to node v i becomes a directed edge going from node v i to node u i . For every node i in the range [0, n - 1] , your task i

Detailed Explanation

The problem asks us to find the minimum number of edge reversals needed for each node in a directed graph (which would be a tree if the edges were undirected) such that it's possible to reach any other node starting from that node. For each node i, we need to find the minimum number of edge direction changes to make all other nodes reachable from i. The input is the number of nodes 'n' and a list of directed edges 'edges', where each edge is represented as [u, v] (from node u to node v). The output is an array 'answer' where answer[i] is the minimum number of reversals required for node i.

Solution Approach

The solution uses a two-pass approach. The first pass calculates the minimum edge reversals needed to reach all other nodes when node 0 is the root. This is done using a Breadth-First Search (BFS). During this BFS, we count how many edges point in the 'wrong' direction (v -> u instead of u -> v when exploring from node u). The second pass performs another BFS to 're-root' the tree. For each node, it utilizes the reversal count of its parent to determine its own reversal count. When moving from a parent node u to a child node v, we check the original edge direction. If the edge was u -> v, the reversal count for node v increases by 1 compared to node u because for node v, the edge direction is incorrect now. If the edge was v -> u, the reversal count for node v decreases by 1 compared to node u because the edge is now in the correct direction for node v.

Step-by-Step Algorithm

Step 1: Create an adjacency list representation of the graph from the given edges. Store the original edges in a set for quick lookup.
Step 2: Perform a BFS starting from node 0 to calculate the initial 'cost' (number of reversals) when node 0 is the root. Keep track of visited nodes to avoid cycles.
Step 3: During the BFS from node 0, for each edge (u, v), check if the original edge was (v, u). If it was, increment the 'cost'.
Step 4: Store the initial cost for node 0 in the 'answer' array: answer[0] = cost.
Step 5: Perform a second BFS starting from node 0 to calculate the costs for all other nodes.
Step 6: During the second BFS, for each edge (u, v), determine whether the edge was originally u -> v or v -> u.
Step 7: If the original edge was u -> v, then answer[v] = answer[u] + 1 (one more reversal is needed for node v to reach all nodes).
Step 8: If the original edge was v -> u, then answer[v] = answer[u] - 1 (one fewer reversal is needed for node v to reach all nodes).
Step 9: Return the 'answer' array.

Key Insights

Insight 1: The problem can be solved using a two-pass approach based on dynamic programming and graph traversal (BFS or DFS).
Insight 2: The core idea is to first calculate the number of reversals needed for a specific root (node 0 in this case).
Insight 3: Then, we can use this initial result to efficiently calculate the number of reversals needed for all other nodes by 're-rooting' the tree and updating the reversal count based on the direction of edges between parent and child nodes.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Dynamic Programming, Depth-First Search, Breadth-First Search, Graph.

Companies

Asked at: MathWorks.

Minimum Edge Reversals So Every Node Is Reachable | LeetCode 2858

Hard

Dynamic Programming Depth-First Search Breadth-First Search Graph

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int[] minEdgeReversals(int n, int[][] edges) {
        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj.add(new ArrayList<>());
        }

        Set<String> originalEdges = new HashSet<>(); // Use String representation "u,v"

        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            adj.get(u).add(v);
            adj.get(v).add(u);
            originalEdges.add(u + "," + v);
        }

        int[] answer = new int[n];

        // Pass 1: Calculate the answer for node 0 as the root.
        Queue<Integer> q1 = new LinkedList<>();
        Set<Integer> visited1 = new HashSet<>();
        q1.offer(0);
        visited1.add(0);
        int cost0 = 0;

        while (!q1.isEmpty()) {
            int u = q1.poll();
            for (int v : adj.get(u)) {
                if (!visited1.contains(v)) {
                    visited1.add(v);
                    if (originalEdges.contains(v + "," + u)) {
                        cost0++;
                    }
                    q1.offer(v);
                }
            }
        }

        answer[0] = cost0;

        // Pass 2: Rerooting. Calculate the answer for all other nodes based on their parent's answer.
        Queue<Integer> q2 = new LinkedList<>();
        Set<Integer> visited2 = new HashSet<>();
        q2.offer(0);
        visited2.add(0);

        while (!q2.isEmpty()) {
            int u = q2.poll();
            for (int v : adj.get(u)) {
                if (!visited2.contains(v)) {
                    visited2.add(v);
                    if (originalEdges.contains(u + "," + v)) {
                        answer[v] = answer[u] + 1;
                    } else {
                        answer[v] = answer[u] - 1;
                    }
                    q2.offer(v);
                }
            }
        }

        return answer;
    }
}

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Problem Info

Description

Problem 2858: Minimum Edge Reversals So Every Node Is Reachable - Detailed explanation and solution approaches.

Topics

Dynamic Programming Depth-First Search Breadth-First Search Graph

Companies

MathWorks

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2858

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

9 steps

1
Step 1: Create an adjacency list representation of the graph from the given edges. Store the original edges in a set for quick lookup.
2
Step 2: Perform a BFS starting from node 0 to calculate the initial 'cost' (number of reversals) when node 0 is the root. Keep track of visited nodes to avoid cycles.
3
Step 3: During the BFS from node 0, for each edge (u, v), check if the original edge was (v, u). If it was, increment the 'cost'.
4
Step 4: Store the initial cost for node 0 in the 'answer' array: answer[0] = cost.
5
Step 5: Perform a second BFS starting from node 0 to calculate the costs for all other nodes.
6
Step 6: During the second BFS, for each edge (u, v), determine whether the edge was originally u -> v or v -> u.
7
Step 7: If the original edge was u -> v, then answer[v] = answer[u] + 1 (one more reversal is needed for node v to reach all nodes).
8
Step 8: If the original edge was v -> u, then answer[v] = answer[u] - 1 (one fewer reversal is needed for node v to reach all nodes).
9
Step 9: 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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The BFS algorithm has a time complexity of O(V + E), where V is the number of vertices (nodes) and E is the number of edges. Since the graph is a tree, E = n - 1. Therefore, the time complexity of each BFS is O(n + n - 1) = O(n). The space complexity is dominated by the adjacency list which takes O(n) space, visited sets, answer array and the queue used in BFS, also taking O(n) space.

📖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

Depth-First Search (DFS)

The BFS can be replaced with a DFS for both the initial cost calculation and the re-rooting step. The logic remains the same, but instead of using a queue, recursion is used to traverse the graph.

Time: O(n)Space: O(n) in the worst case (skewed tree), due to the recursion stack. Can be O(log n) in the average case for balanced trees.

Trade-offs: DFS may have slightly higher space complexity in the worst case compared to BFS. Also DFS may cause stack overflow for huge graphs.

✨

Key Insights

1Insight 1: The problem can be solved using a two-pass approach based on dynamic programming and graph traversal (BFS or DFS).
2Insight 2: The core idea is to first calculate the number of reversals needed for a specific root (node 0 in this case).
3Insight 3: Then, we can use this initial result to efficiently calculate the number of reversals needed for all other nodes by 're-rooting' the tree and updating the reversal count based on the direction of edges between parent and child nodes.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Graph

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

Edge Cases

Edge case 1: n = 2, edges = [[0, 1]]. In this case, the solution will correctly compute answer[0] = 0 and answer[1] = 1.
Edge case 2: n = 2, edges = [[1, 0]]. In this case, the solution will correctly compute answer[0] = 1 and answer[1] = 0.
Edge case 3: A single node graph (though the constraint says n >= 2). The algorithm still should not crash, and the returned array will have correct value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the direction of the original edges when calculating the number of reversals. It's crucial to distinguish between u -> v and v -> u.

✗Common mistake 2: Incorrectly updating the reversal count during the re-rooting step. Forgetting to add 1 or subtract 1 based on the original edge direction will lead to incorrect results.

✗Common mistake 3: Failing to handle visited nodes correctly in BFS/DFS can lead to infinite loops.

✅

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 optimization. Minimizing the number of 'direction changes' needed to ensure efficient data flow within a network.

🏦Real-world scenario 2: Infrastructure planning. Determining the optimal direction of one-way streets or pipelines to minimize disruptions and ensure accessibility.

💡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

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

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 6, 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

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

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 "Minimum Edge Reversals So Every Node Is Reachable"?

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

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