841. Keys and Rooms

1. Description

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.

2. Example

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.

3. 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.

4. Solutions

Breadth-First Search

n = rooms.size()
Time complexity: O(n)
Space complexity: O(n)

class Solution {
public:
    bool canVisitAllRooms(vector<vector<int>> &rooms) {
        vector<bool> visited(rooms.size(), false);
        queue<int> to_visit({0});
        int visited_count = 0;
        while (!to_visit.empty()) {
            int index = to_visit.front();
            to_visit.pop();

            if (!visited[index]) {
                ++visited_count;
                if (visited_count == rooms.size()) {
                    return true;
                }
                visited[index] = true;

                for (int i : rooms[index]) {
                    if (!visited[i]) {
                        to_visit.push(i);
                    }
                }
            }
        }

        return false;
    }
};