# 841. Keys and Rooms

#### Medium

***

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**.

#### Time Complexity  : O(N+E)

Why is it O(N+E), but not O(N \* E), that's because E (as per the solution author) is the total number of keys from all rooms combined. It's not the total number of unique rooms that can be opened using the available keys.

```python
class Solution:
    def canVisitAllRooms(self, rooms: List[List[int]]) -> bool:
        d = defaultdict(list)
        for room, keys in enumerate(rooms):
            d[room].extend(keys)
        unlocked = set()
        unlocked.add(0) # Since Room 0 in unlocked by default
        nodes = 0
        dq = deque([(0, d[0])])
        while dq:
            room, keys = dq.popleft()
            for adj in d[room]:
                if adj not in unlocked:
                    unlocked.add(adj)
                    dq.append((adj, d[adj]))
                    if len(unlocked) == len(rooms):
                        return True
        return False
        
```