How to count bookings if allowing for efficient overlap - ruby-on-rails

Question

I've looked over some SO discussions here and, well at least I haven't seen this perspective. I'm trying to write code to count bookings of a given resource, where I want to find the MINIMUM number of resources I need to fulfill all bookings.
Let's use an example of hotel rooms. Given that I have the following bookings
Chris: July 4-July 17
Pat: July 15-July 19
Taylor: July 10-July 11
Chris calls and would like to add some room(s) to their reservation for friends, and wonders how many rooms I have available.
Rooms_available = Rooms_in_hotel - Rooms_booked
The Rooms_booked is where I'm having trouble. It seems like most questions (and indeed my code) just looks at overlapping dates. So it would do something like this:
Booking.where("booking_end >= :start_check AND booking_start <= :end_check", { start_check: "July 4, 2021".to_date, end_check: "July 7, 2021".to_date})
This code would return 3. Which means that if the hotel theoretically had 5 rooms, I would tell Chris that there were 2 more rooms left available.
However, while this method of counting is technically accurate, it misses the possibility of an efficient overlap. Namely that since Taylor checks out 4 days before Pat, they can both be "assigned" the same room. So technically, I can offer 3 more rooms to Chris.
So my question is how do I more accurately calculate Rooms_booked allowing for efficient overlap (i.e., efficient resource allocation)? Is there a query using ActiveRecord or what calculation do I impose on top of the existing query?

Answer 1

i don't think just only a query could solve your problem (or very very complex query).
my idea is group (and count) by (booking_start, booking_end) in order booking_start asc then reassign, e.g. if there're 2 bookings July 15-July 19 and 3 bookings July 10-July 11 then we only could re-assign for 2 pairs, and we need 3 rooms (2 rooms for July 15-July 19-July 10-July 11 and 1 for July 10-July 11.
and re-assign in code not query (we can optimize by pick a narrow range of time)
# when Chris calls and would like to add some room(s) to their reservation for friends,
# and wonders how many available rooms.
# pick (start_time, end_time) so that the range time long enough that
# including those booking {n} month ago
# but no need too long or all the time
scope :booking_available_count, -> (start_time, end_time) {
group_by_time = \
Booking.where("booking_start >= ? AND booking_end <= ?", start_time, end_time)
.group(:booking_start, :booking_end).order(:booking_start).count
# result: {[booking_start, booking_end] => 1, [booking_start, booking_end] => 2, ... }
# in order booking_start ASC
# then we could re-assign from left to right as below
booked = 0
group_by_time.each do |(start_time, end_time), count|
group_by_time.each do |(assign_start_time, assign_end_time), assign_count|
next if end_time > assign_start_time
count -= assign_count # re-assign
break if count <= 0
end
booked += count if count > 0
end
# return number of available rooms
# allow negative number
Room.count - booked
}