Introduction

In this blog post, we will explore how to solve the LeetCode problem "841" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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 you can visit all the rooms, or false otherwise. Example 1: Input: rooms = [[1],[2],[3],[]] Output: true Explanation: We visit room 0 and pick up key 1. We then visit room 1 and pick up key 2. We then visit room 2 and pick up key 3. We then visit room 3. Since we were able to visit every room, we return true. Example 2: Input: rooms = [[1,3],[3,0,1],[2],[0]] Output: false Explanation: We can not enter room number 2 since the only key that unlocks it is in that room. Constraints: n == rooms.length 2 <= n <= 1000 0 <= rooms[i].length <= 1000 1 <= sum(rooms[i].length) <= 3000 0 <= rooms[i][j] < n All the values of rooms[i] are unique.

Explanation

Here's a breakdown of the solution:

• Core Idea: Use a Breadth-First Search (BFS) approach to explore the rooms. Start from room 0 and use the keys found in each room to unlock and visit other rooms.
• Tracking Visited Rooms: Maintain a set of visited rooms to avoid cycles and redundant exploration.

• Termination Condition: If the number of visited rooms equals the total number of rooms, it means we can visit all rooms.

• Complexity: Time: O(N + E), Space: O(N), where N is the number of rooms and E is the total number of keys.

Code

    from collections import deque

def canVisitAllRooms(rooms):
    """
    Determines if all rooms can be visited starting from room 0.

    Args:
        rooms: A list of lists representing the keys in each room.

    Returns:
        True if all rooms can be visited, False otherwise.
    """

    n = len(rooms)
    visited = {0}  # Start with room 0 visited
    queue = deque([0])  # Start BFS from room 0

    while queue:
        room_num = queue.popleft()
        for key in rooms[room_num]:
            if key not in visited:
                visited.add(key)
                queue.append(key)

    return len(visited) == n