What is the time complexity of Reorder Routes to Make All Paths Lead to the City Zero?

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

What data structures are used to solve Reorder Routes to Make All Paths Lead to the City Zero?

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

What are the key insights for solving Reorder Routes to Make All Paths Lead to the City Zero?

Insight 1: Treat the problem as a graph traversal problem. The cities are nodes, and the roads are edges. Insight 2: Representing edges with a 'cost' (0 or 1) is crucial. A cost of 1 indicates an original direction that goes away from city 0 (requires reorientation), and a cost of 0 indicates a direction towards a visited node (no reorientation needed).

Which companies ask LeetCode 1466. Reorder Routes to Make All Paths Lead to the City Zero in interviews?

This problem is frequently asked at DRW, Docusign.

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Reorder Routes to Make All Paths Lead to the City Zero - LeetCode 1466 Solution

Reorder Routes to Make All Paths Lead to the City Zero - Complete Solution Guide

Reorder Routes to Make All Paths Lead to the City Zero is LeetCode problem 1466, 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 are n cities numbered from 0 to n - 1 and n - 1 roads such that there is only one way to travel between two different cities (this network form a tree). Last year, The ministry of transport decided to orient the roads in one direction because they are too narrow. Roads are represented by connections where connections[i] = [a i , b i ] represents a road from city a i to city b i . This year, there will be a big event in the capital (city 0 ), and many people want to travel to this city. You

Detailed Explanation

The problem asks us to find the minimum number of roads we need to reorient in a given network of cities so that every city can reach city 0. We're given 'n' cities numbered from 0 to n-1, and 'connections' which is a list of roads. Each road is represented by a pair [ai, bi] meaning there's a directed road from city ai to city bi. The goal is to reorient some of these roads so that every city can reach city 0. The network forms a tree structure, guaranteeing that a path always exists between any two cities after reorienting. We need to return the smallest number of reorientations required.

Solution Approach

The solution uses Breadth-First Search (BFS) to traverse the graph starting from city 0. The graph is represented using an adjacency list, where each city has a list of its neighbors. Importantly, each edge in the adjacency list is stored with a cost of either 0 or 1. A cost of 1 means the edge initially points *away* from city 0, meaning it needs to be reoriented to allow travel *towards* city 0. A cost of 0 means the edge can be traversed in the opposite direction without any changes. As BFS progresses, the algorithm keeps track of the cumulative reorientation cost, which is the total number of reorientations needed.

Step-by-Step Algorithm

Step 1: Build an adjacency list representation of the graph. For each connection [u, v], add an edge from u to v with cost 1 (representing an original edge direction away from city 0) and an edge from v to u with cost 0 (representing a potential back-edge towards city 0).
Step 2: Initialize a queue with city 0 (the starting point of the BFS), and a 'visited' set to keep track of visited cities.
Step 3: Initialize a counter 'reorders' to 0 to store the total reorientation cost.
Step 4: While the queue is not empty:
Step 5: Dequeue a city 'u' from the queue.
Step 6: Iterate through all the neighbors 'v' of city 'u' in the adjacency list:
Step 7: If 'v' has not been visited:
Step 8: Add the 'cost' of the edge from 'u' to 'v' to the 'reorders' counter. This cost indicates whether or not the edge u->v needs to be reoriented (1 if it does, 0 if it doesn't).
Step 9: Mark 'v' as visited and enqueue it to the queue.
Step 10: After the BFS is complete, return the 'reorders' counter, which is the minimum number of roads that need to be reoriented.

Key Insights

Insight 1: Treat the problem as a graph traversal problem. The cities are nodes, and the roads are edges.
Insight 2: Representing edges with a 'cost' (0 or 1) is crucial. A cost of 1 indicates an original direction that goes away from city 0 (requires reorientation), and a cost of 0 indicates a direction towards a visited node (no reorientation needed).
Insight 3: Using BFS (Breadth-First Search) or DFS (Depth-First Search) can efficiently traverse the graph and calculate the minimum reorders.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: DRW, Docusign.

Reorder Routes to Make All Paths Lead to the City Zero | LeetCode 1466

Medium

Depth-First Search Breadth-First Search Graph

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

Solution Code

Available in:

import java.util.*;

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

        for (int[] connection : connections) {
            int u = connection[0];
            int v = connection[1];
            adj.get(u).add(new Pair(v, 1));
            adj.get(v).add(new Pair(u, 0));
        }

        Queue<Integer> q = new LinkedList<>();
        q.offer(0);
        Set<Integer> visited = new HashSet<>();
        visited.add(0);
        int reorders = 0;

        while (!q.isEmpty()) {
            int city = q.poll();
            for (Pair neighborPair : adj.get(city)) {
                int neighbor = neighborPair.node;
                int cost = neighborPair.cost;
                if (!visited.contains(neighbor)) {
                    visited.add(neighbor);
                    reorders += cost;
                    q.offer(neighbor);
                }
            }
        }

        return reorders;
    }

    static class Pair {
        int node;
        int cost;

        public Pair(int node, int cost) {
            this.node = node;
            this.cost = cost;
        }
    }
}

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Problem Info

Description

Problem 1466: Reorder Routes to Make All Paths Lead to the City Zero - Detailed explanation and solution approaches.

Topics

Depth-First Search Breadth-First Search Graph

Companies

DRWDocusign

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1466

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

10 steps

1
Step 1: Build an adjacency list representation of the graph. For each connection [u, v], add an edge from u to v with cost 1 (representing an original edge direction away from city 0) and an edge from v to u with cost 0 (representing a potential back-edge towards city 0).
2
Step 2: Initialize a queue with city 0 (the starting point of the BFS), and a 'visited' set to keep track of visited cities.
3
Step 3: Initialize a counter 'reorders' to 0 to store the total reorientation cost.
4
Step 4: While the queue is not empty:
5
Step 5: Dequeue a city 'u' from the queue.
6
Step 6: Iterate through all the neighbors 'v' of city 'u' in the adjacency list:
7
Step 7: If 'v' has not been visited:
8
Step 8: Add the 'cost' of the edge from 'u' to 'v' to the 'reorders' counter. This cost indicates whether or not the edge u->v needs to be reoriented (1 if it does, 0 if it doesn't).
9
Step 9: Mark 'v' as visited and enqueue it to the queue.
10
Step 10: After the BFS is complete, return the 'reorders' counter, which is the minimum number of roads that need to be reoriented.

💡

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 visits each node (city) and each edge (connection) at most once. The adjacency list stores the graph information, which takes O(n) space. The queue and visited set, used for BFS, can potentially store all nodes in the worst case, leading to O(n) space complexity.

📖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)

Instead of BFS, we can use DFS to traverse the graph. The logic for marking visited nodes and calculating the reorder cost remains the same.

Time: O(n)Space: O(n) in the worst case (due to the call stack during recursion), where n is the number of cities.

Trade-offs: DFS may use slightly more space than BFS in the worst case due to recursion depth, but both have the same asymptotic time complexity. BFS might be preferred if the graph is very wide, as it tends to use less memory in such cases.

✨

Key Insights

1Insight 1: Treat the problem as a graph traversal problem. The cities are nodes, and the roads are edges.
2Insight 2: Representing edges with a 'cost' (0 or 1) is crucial. A cost of 1 indicates an original direction that goes away from city 0 (requires reorientation), and a cost of 0 indicates a direction towards a visited node (no reorientation needed).
3Insight 3: Using BFS (Breadth-First Search) or DFS (Depth-First Search) can efficiently traverse the graph and calculate the minimum reorders.

📚

Key Terms Explained

For Students

Graph

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

Edge Cases

Edge case 1: n = 2, connections = [[0, 1]]. This is a simple case where only one edge needs checking. The code correctly handles it by adding edges in both directions to the adjacency list and incrementing reorders by 1 when necessary.
Edge case 2: n = 3, connections = [[1, 0], [2, 0]]. In this case, no roads need to be reoriented because both edges already point towards city 0. The code correctly assigns a cost of 0 when moving from 1 to 0 or 2 to 0, so reorders remains 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the reverse edges. It's crucial to add edges in both directions in the adjacency list, as the problem statement says the original configuration forms a tree, meaning there must be a path between every pair of nodes, but the given edges are directional. Failing to add reverse edges will result in not being able to traverse the entire graph.

✗Common mistake 2: Incorrectly assigning the reorder cost. The reorder cost should only be incremented when traversing the original direction of a connection (u -> v). Traversal in the opposite direction (v -> u) does not require reorientation and should have a cost of 0.

✅

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: Optimizing traffic flow in a city. Imagine a city where roads have preferred directions. You could use this algorithm to determine the minimum number of roads that need to have their direction changed to ensure efficient access to a central hub (e.g., the city center).

🏦Real-world scenario 2: Network routing. In a network, devices can be considered as cities, and connections between devices can be considered as roads. The algorithm can be used to minimize the number of connections that need to be reconfigured so that all devices can reach a central server.

💡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 27, 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

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

@AlgoMasterJul 14

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

@InterviewProJul 7

❓

Frequently Asked Questions

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Reorder Routes to Make All Paths Lead to the City Zero"?

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.

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