What is the time complexity of Keys and Rooms?

The optimal solution has a time complexity of O(N+E).

What data structures are used to solve Keys and Rooms?

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 Keys and Rooms?

Insight 1: The problem can be modeled as a graph, where rooms are nodes and keys represent edges between rooms. Insight 2: We can use Depth-First Search (DFS) or Breadth-First Search (BFS) to traverse the graph and find all reachable nodes (rooms).

Which companies ask LeetCode 841. Keys and Rooms in interviews?

This problem is frequently asked at Expedia, Graviton, Nvidia, Tinkoff, Walmart Labs.

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Keys and Rooms - LeetCode 841 Solution

Keys and Rooms - Complete Solution Guide

Keys and Rooms is LeetCode problem 841, 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 rooms labeled from 0 to n - 1 and all the rooms are locked except for room 0 . Your goal is to visit all the rooms. However, you cannot enter a locked room without having its key. When you visit a room, you may find a set of distinct keys in it. Each key has a number on it, denoting which room it unlocks, and you can take all of them with you to unlock the other rooms. Given an array rooms where rooms[i] is the set of keys that you can obtain if you visited room i , return true if yo

Detailed Explanation

The problem asks whether it's possible to visit all rooms in a building, given a list of rooms and the keys that can be found in each room. You start in room 0, which is unlocked. Each room contains a list of keys, where each key represents a room number that can be unlocked. The goal is to determine if you can visit all the rooms using the keys you find along the way. The input is a list of lists called `rooms`, where `rooms[i]` is the list of keys you find in room `i`. The output is a boolean value: `true` if you can visit all rooms, and `false` otherwise. The constraints specify the size of the input and valid ranges for the room and key numbers.

Solution Approach

The provided solutions use Depth-First Search (DFS) with a stack to traverse the rooms. We start with room 0 and mark it as visited. Then, we iterate through the keys found in that room. For each key, if the corresponding room hasn't been visited yet, we mark it as visited and add it to the stack. We continue this process until the stack is empty, indicating that we've explored all reachable rooms. Finally, we check if the number of visited rooms equals the total number of rooms. If it does, we return `true`; otherwise, we return `false`.

Step-by-Step Algorithm

Step 1: Initialize a `visited` set (or array) to keep track of the visited rooms, and add room 0 to the set (since we start there).
Step 2: Initialize a stack and push room 0 onto the stack.
Step 3: While the stack is not empty:
Step 4: Pop a room from the stack.
Step 5: Iterate through the keys found in the popped room.
Step 6: For each key, check if the corresponding room has been visited.
Step 7: If the room hasn't been visited, add it to the `visited` set and push it onto the stack.
Step 8: After the stack is empty, check if the size of the `visited` set equals the total number of rooms. If it does, return `true`; otherwise, return `false`.

Key Insights

Insight 1: The problem can be modeled as a graph, where rooms are nodes and keys represent edges between rooms.
Insight 2: We can use Depth-First Search (DFS) or Breadth-First Search (BFS) to traverse the graph and find all reachable nodes (rooms).
Insight 3: We need to keep track of which rooms have been visited to avoid infinite loops and ensure we explore each room only once. Using a `visited` set or array is crucial.

Complexity Analysis

Time Complexity: O(N+E)

Space Complexity: O(N)

Topics

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

Companies

Asked at: Expedia, Graviton, Nvidia, Tinkoff, Walmart Labs.

Keys and Rooms | LeetCode 841

Medium

Depth-First Search Breadth-First Search Graph

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

Solution Code

Available in:

import java.util.List;
import java.util.HashSet;
import java.util.Stack;
import java.util.Set;

class Solution {
    public boolean canVisitAllRooms(List<List<Integer>> rooms) {
        Set<Integer> visited = new HashSet<>();
        visited.add(0);
        Stack<Integer> stack = new Stack<>();
        stack.push(0);

        while (!stack.isEmpty()) {
            int room = stack.pop();

            for (int key : rooms.get(room)) {
                if (!visited.contains(key)) {
                    visited.add(key);
                    stack.push(key);
                }
            }
        }

        return visited.size() == rooms.size();
    }
}

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Problem Info

Description

Problem 841: Keys and Rooms - Detailed explanation and solution approaches.

Topics

Depth-First Search Breadth-First Search Graph

Companies

ExpediaGravitonNvidiaTinkoffWalmart Labs

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:841

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: Initialize a `visited` set (or array) to keep track of the visited rooms, and add room 0 to the set (since we start there).
2
Step 2: Initialize a stack and push room 0 onto the stack.
3
Step 3: While the stack is not empty:
4
Step 4: Pop a room from the stack.
5
Step 5: Iterate through the keys found in the popped room.
6
Step 6: For each key, check if the corresponding room has been visited.
7
Step 7: If the room hasn't been visited, add it to the `visited` set and push it onto the stack.
8
Step 8: After the stack is empty, check if the size of the `visited` set equals the total number of rooms. If it does, return `true`; otherwise, return `false`.

💡

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+E)

💾

Space Complexity

O(N)

📝

Why?

The time complexity is determined by the graph traversal. We process each node (room) and edge (key) at most once. The space complexity is due to the data structures used to store visited rooms and the recursion stack (or explicit stack in the iterative DFS implementation). In the worst-case scenario where all rooms are interconnected, both can grow linearly with the number of rooms.

📖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

Breadth-First Search (BFS)

BFS can also be used to solve this problem. Instead of using a stack, we use a queue to explore the rooms level by level.

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

Trade-offs: BFS and DFS have similar time and space complexities for this problem. BFS might be preferred if we wanted to find the shortest path to a specific room, although that's not the problem statement here. In general, the iterative DFS implementation might be slightly more memory-efficient than BFS in some cases, but in practice, the difference is often negligible.

✨

Key Insights

1Insight 1: The problem can be modeled as a graph, where rooms are nodes and keys represent edges between rooms.
2Insight 2: We can use Depth-First Search (DFS) or Breadth-First Search (BFS) to traverse the graph and find all reachable nodes (rooms).
3Insight 3: We need to keep track of which rooms have been visited to avoid infinite loops and ensure we explore each room only once. Using a `visited` set or array is crucial.

📚

Key Terms Explained

For Students

Graph

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

Edge Cases

Edge case 1: Empty list of rooms. The solution should handle this correctly by returning `true` because there are no rooms to visit besides the starting room.
Edge case 2: A room with no keys. The algorithm will still visit the room but won't be able to proceed to other rooms from that room. The solution should correctly handle this case.
Edge case 3: Disconnected graph where some rooms cannot be reached from room 0. The algorithm will only visit the connected component starting from room 0 and return `false`.
Edge case 4: All rooms are reachable directly from room 0. The algorithm will quickly visit all rooms and return `true`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to mark rooms as visited, which can lead to infinite loops and incorrect results.

✗Common mistake 2: Not handling the case where some rooms are unreachable from the starting room, which can lead to premature termination and an incorrect `true` result.

✗Common mistake 3: Not considering the possibility of duplicate keys in a single room (although the problem states that all the values of rooms[i] are unique, this is an important general consideration for graph problems).

✗Common mistake 4: Incorrectly initializing the `visited` set. Make sure to include the starting room (room 0) initially.

✅

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 connectivity analysis. Determining if all nodes in a network can be reached from a specific node, which can be used for network troubleshooting and security analysis.

🏦Real-world scenario 2: Dependency resolution in software packages. Determining if all required dependencies for a software package can be resolved and installed. The rooms represent the software packages and the keys represent the packages required by that software.

🎮Real-world scenario 3: Web crawling. Visiting all reachable web pages from a starting page. The rooms would be webpages and the keys would be links on the webpages.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: September 30, 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 13

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

@InterviewProJul 6

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

@CodeNinjaJun 29

❓

Frequently Asked Questions

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Keys and Rooms"?

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.

🤖 Need help with this problem?