841. Keys and Rooms

Medium

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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.

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