hotelWithMethods.swift

hotelWithMethods.swift

// Again, same class as before.
class Hotel {

  // We now have the returned `roomID` optional.
  // It's possible, that there are no rooms available anymore,
  // so we need to make nil a valid option.
  func checkIn(_ dog: Dog) -> String? {
  
    // We want to return a roomID so each user 
    // is able to retrieve their dog using the ID.
  	var roomID: String?
    
    // We check for the dog size and we check if 
    // the desired room size is available
    if dog.size == .small, smallRooms.keys.count < smallRoomsCount {

      // Only then do we generate a PIN.
      // To safe memory. Mention that to your interviewer.
      // That's part of "scalablity".
      roomID = UUID()
      // We use the ID to assign a room for the dog.
      // Of course there are other ways to do this.
      // But this is an easy way and it makes sense.
      // No reason to overcomplicate it in an interview.
      smallRooms[roomID] = dog
    }

    // Same for medium as we did for small
    if dog.size == .medium, mediumRooms.keys.count < mediumRoomsCount {
      roomID = UUID()
      mediumRooms[roomID] = dog
    }

    // Same for large as we did for small and medium
    if dog.size == .large, largeRooms.keys.count < largeRoomsCount {
      roomID = UUID()
      largeRooms[roomID] = dog
    }

    // Lastly we return the ID that can be nil
    // in case of no availability.
    return roomID 
  }
  
  // For the checkOut we've created a more swifty syntax.
  // That shows that we care about writing well readable code.
  // We now require the dogID and the roomID to check out. 
  // That makes more sense.
  // If the IDs are invalid, then the return is nil.
  func checkOut(dogWithID dogID: String, outOfRoom roomID: String) -> Dog? {
  
    // To not overcomplicate it, we'll just check if 
    // there's a dog in the room, 
    // and then if the dogIDs are matching
    //
    // It's possible, that your interviewer asks you
    // about concerns regarding efficiency of this approach.
    // Let them know, that the Big O cost is constant 
    // and very efficient, even if checked in 3 dictionaries.
    if let dog = smallRooms[roomID], dog.uid == dogID {
      // If so, it's safe to return the dog
      return dog
    }

    // Same then for medium rooms.
    if let dog = mediumRooms[roomID], dog.uid == dogID {
      return dog
    }

    // And for large rooms.
    if let dog = largeRooms[roomID], dog.uid == dogID {
      return dog
    }

    // If there's no matches in the IDs, 
    // then we're not gonna return any pups.
    return nil
  }
}