What is the time complexity of Bus Routes?

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

What is the space complexity of Bus Routes?

The space complexity is O(n*m).

What data structures are used to solve Bus Routes?

This problem uses Array, Hash Table, Breadth-First Search. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Bus Routes?

Insight 1: The problem can be modeled as a graph where bus stops are nodes, and an edge exists between two stops if they are on the same bus route. The number of buses taken corresponds to the number of edges traversed. Insight 2: Breadth-First Search (BFS) is suitable for finding the shortest path (in terms of bus rides) in an unweighted graph.

Which companies ask LeetCode 815. Bus Routes in interviews?

This problem is frequently asked at BitGo, Citadel, Coupang, Info Edge, PhonePe and 2 more companies.

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

Bus Routes - LeetCode 815 Solution

Bus Routes - Complete Solution Guide

Bus Routes is LeetCode problem 815, 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 array routes representing bus routes where routes[i] is a bus route that the i th bus repeats forever. For example, if routes[0] = [1, 5, 7] , this means that the 0 th bus travels in the sequence 1 -> 5 -> 7 -> 1 -> 5 -> 7 -> 1 -> ... forever. You will start at the bus stop source (You are not on any bus initially), and you want to go to the bus stop target . You can travel between bus stops by buses only. Return the least number of buses you must take to travel from source to t

Detailed Explanation

The problem asks us to find the minimum number of buses needed to travel from a source bus stop to a target bus stop, given a list of bus routes. Each route represents a bus that travels repeatedly through a list of stops. We start at the source, and we want to reach the target by taking buses. The goal is to find the fewest number of bus rides required. If it's impossible to reach the target, we return -1.

Solution Approach

The solution uses a Breadth-First Search (BFS) algorithm. We first build a mapping from each bus stop to the list of bus routes that pass through that stop. Then, we use a queue to perform BFS starting from the source stop. In each level of the BFS, we iterate through the stops and find all possible routes from that stop. We mark routes as visited so we don't reconsider the same routes. For each unvisited route, we add all the unvisited stops in that route to the queue for the next level of BFS. We continue until we find the target stop or the queue is empty (meaning the target is unreachable).

Step-by-Step Algorithm

Step 1: Create a dictionary `stop_to_routes` that maps each bus stop to a set of route indices that contain that stop. This allows us to quickly find which bus routes are available at a particular stop.
Step 2: Initialize a queue `q` with the source stop. Also initialize sets `visited_stops` and `visited_routes` to keep track of visited stops and routes, respectively.
Step 3: Perform a BFS. Initialize `buses_taken` to 0. Increment `buses_taken` for each level of BFS representing taking one more bus.
Step 4: In each BFS level, iterate through all the stops in the current level. For each `current_stop`, find all the routes it is part of.
Step 5: For each `route_idx` that passes through the `current_stop`, check if the route has been visited. If not, mark the route as visited.
Step 6: Iterate through all stops in the current `route_idx`. If a stop is the target, return `buses_taken`. If a stop has not been visited, add it to the queue and mark it as visited.
Step 7: If the queue becomes empty, it means there is no path from the source to the target. Return -1.

Key Insights

Insight 1: The problem can be modeled as a graph where bus stops are nodes, and an edge exists between two stops if they are on the same bus route. The number of buses taken corresponds to the number of edges traversed.
Insight 2: Breadth-First Search (BFS) is suitable for finding the shortest path (in terms of bus rides) in an unweighted graph.
Insight 3: To optimize the search, we should avoid revisiting the same bus route multiple times. We should also create a mapping from bus stop to the list of buses that stop there for efficient searching.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(n*m)

Topics

This problem involves: Array, Hash Table, Breadth-First Search.

Companies

Asked at: BitGo, Citadel, Coupang, Info Edge, PhonePe, Pinterest, Snap.

Bus Routes | LeetCode 815

Hard

Array Hash Table Breadth-First Search

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

Solution Code

Available in:

import java.util.*;

class Solution {
    public int numBusesToDestination(int[][] routes, int source, int target) {
        if (source == target) {
            return 0;
        }

        Map<Integer, Set<Integer>> stopToRoutes = new HashMap<>();
        for (int i = 0; i < routes.length; i++) {
            for (int stop : routes[i]) {
                if (!stopToRoutes.containsKey(stop)) {
                    stopToRoutes.put(stop, new HashSet<>());
                }
                stopToRoutes.get(stop).add(i);
            }
        }

        Queue<Integer> q = new LinkedList<>();
        q.offer(source);
        Set<Integer> visitedStops = new HashSet<>();
        visitedStops.add(source);
        Set<Integer> visitedRoutes = new HashSet<>();

        int busesTaken = 0;
        while (!q.isEmpty()) {
            busesTaken += 1;
            int stopsInLevel = q.size();
            for (int i = 0; i < stopsInLevel; i++) {
                int currentStop = q.poll();

                for (int routeIdx : stopToRoutes.get(currentStop)) {
                    if (visitedRoutes.contains(routeIdx)) {
                        continue;
                    }

                    visitedRoutes.add(routeIdx);

                    for (int nextStop : routes[routeIdx]) {
                        if (nextStop == target) {
                            return busesTaken;
                        }

                        if (!visitedStops.contains(nextStop)) {
                            visitedStops.add(nextStop);
                            q.offer(nextStop);
                        }
                    }
                }
            }
        }

        return -1;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 815: Bus Routes - Detailed explanation and solution approaches.

Topics

Array Hash Table Breadth-First Search

Companies

BitGoCitadelCoupangInfo EdgePhonePePinterestSnap

Stats

Difficulty:Hard

Languages:4

Companies:7

Problem #:815

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: Create a dictionary `stop_to_routes` that maps each bus stop to a set of route indices that contain that stop. This allows us to quickly find which bus routes are available at a particular stop.
2
Step 2: Initialize a queue `q` with the source stop. Also initialize sets `visited_stops` and `visited_routes` to keep track of visited stops and routes, respectively.
3
Step 3: Perform a BFS. Initialize `buses_taken` to 0. Increment `buses_taken` for each level of BFS representing taking one more bus.
4
Step 4: In each BFS level, iterate through all the stops in the current level. For each `current_stop`, find all the routes it is part of.
5
Step 5: For each `route_idx` that passes through the `current_stop`, check if the route has been visited. If not, mark the route as visited.
6
Step 6: Iterate through all stops in the current `route_idx`. If a stop is the target, return `buses_taken`. If a stop has not been visited, add it to the queue and mark it as visited.
7
Step 7: If the queue becomes empty, it means there is no path from the source to the target. 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(n*m)

💾

Space Complexity

O(n*m)

📝

Why?

The time complexity is governed by the BFS traversal. In the worst case, we explore all possible routes and stops. The space complexity is dominated by the `stop_to_routes` map and the data structures used for BFS (queue, visited sets).

📖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

Dijkstra's Algorithm

Dijkstra's algorithm could be used if we wanted to generalize the problem where different routes have different costs (e.g., some routes take longer than others). However, since all buses are treated equally in the original problem, BFS is sufficient and more efficient.

Time: O(E log V) where E is the number of edges and V is the number of vertices. In this case E and V depends on m*n, so it will be more costly.Space: O(V), where V is the number of vertices.

Trade-offs: Dijkstra's is more general but overkill for this problem. BFS is faster and simpler when all edges have equal weight.

✨

Key Insights

1Insight 1: The problem can be modeled as a graph where bus stops are nodes, and an edge exists between two stops if they are on the same bus route. The number of buses taken corresponds to the number of edges traversed.
2Insight 2: Breadth-First Search (BFS) is suitable for finding the shortest path (in terms of bus rides) in an unweighted graph.
3Insight 3: To optimize the search, we should avoid revisiting the same bus route multiple times. We should also create a mapping from bus stop to the list of buses that stop there for efficient searching.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: `source == target`. In this case, the solution returns 0 immediately, as no buses are needed.
Edge case 2: No path exists between `source` and `target`. In this case, the BFS queue becomes empty before finding the target, and the solution returns -1.
Edge case 3: The source or target stop does not exist in any routes. The solution will still return -1 as the BFS will not find a path.
Edge case 4: Empty `routes` array. If there are no routes, and source != target, the program should return -1. If source == target, it should return 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the `source == target` edge case, leading to incorrect output or errors.

✗Common mistake 2: Not marking visited routes, leading to infinite loops and incorrect results due to revisiting the same bus routes repeatedly.

✗Common mistake 3: Failing to construct the `stop_to_routes` map efficiently, resulting in a large time complexity for the search.

✗Common mistake 4: Using DFS instead of BFS. DFS might find *a* path, but it doesn't guarantee the *shortest* path in terms of the number of buses taken.

✅

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: Public transportation route planning. Finding the minimum number of transfers to get from one location to another using public transport systems (buses, trains, subways).

🏦Real-world scenario 2: Network routing. Determining the fewest number of hops (routers) to send data from a source to a destination in a computer network.

💡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

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

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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: September 4, 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

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

@DevExpert92Jul 13

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

@AlgoMasterJul 6

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

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

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 "Bus Routes"?

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

🤖 Need help with this problem?