Keys and Rooms


There are N rooms and you start in room 0. Each room has a distinct number in 0, 1, 2, ..., N-1, and each room may have some keys to access the next room.

Formally, each room i has a list of keys rooms[i], and each key rooms[i][j] is an integer in [0, 1, ..., N-1] where N = rooms.length. A key rooms[i][j] = v opens the room with number v.

Initially, all the rooms start locked (except for room 0). You can walk back and forth between rooms freely.

Return true if and only if you can enter every room.


class Solution:
    def canVisitAllRooms(self, rooms):
        if len(rooms) <= 1: #guard for short or empty list
            return True
        key_inventory = set([0]) # append only
        rooms_visited = set() #append

        while key_inventory - rooms_visited: # We have keys to rooms we have not seen
            chosen_room = (key_inventory - rooms_visited).pop()  #choose the room
            keys = set(rooms[chosen_room])  # visit chosen room, get keys
            key_inventory |= keys  # add keys found in  room to inventory
            rooms_visited.add(chosen_room)  # room has been visited
            if key_inventory >= set(range(len(rooms))):  # Stop once we have a key for each room
                return True

        return False # We have no keys to rooms we have not seen, and we didnt have a key for each room

Solution: Recursive

class Solution:

    def __init__(self):
      self.visited = set()
      self.number_of_rooms = 0

    def dfs(self, rms, rm):
        if rm not in self.visited:

            for room_key in rms[rm]:
              if room_key < self.number_of_rooms:
                self.dfs(rms, room_key)

    def canVisitAllRooms(self, rooms) -> bool:

        if rooms:
          self.number_of_rooms = len(rooms)
          self.dfs(rooms, 0)

          if len(self.visited) != self.number_of_rooms:
            return False

            return True

          print("No Rooms to Enter")
          return False


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