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Maddie & Anibel - VideoStoreAPI - Edges #12
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Video Store API
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Comprehension Questions
O(n)where n is the number of movie records in the database since we are retrieving all movies. The time complexity of/customersis at best-case scenarioO(n)if no order is required. Otherwise, the time complexity will be based on active record's.ordermethod and will depend on the database's implementationO(n)or worse.n- is this the number of customers/movies, the number of rentals, or something else?Second, since
.ordersorts the data, we know it is at bestO(c log(c)), wherecis the number of customers.Third, when you generate JSON for a customer, you call the
movies_checked_out_countmethod, which means the time depends on the time to find the rentals for that customer - probablyO(log(r))whereris the total number of rentals, depending on how the database index is built. Then the total time would beO(c log(r) + c log(c))POST /rentals/check-inendpoint? What does the time complexity depend on? Explain your reasoning.POST /rentals/check-inendpoint is restricted by the finding the record in the db and saving it. Assuming that find queries to active record databases are modeled like a hash table, this endpoint should run inO(1)time.find_byis checking a number of properties. Usually databases use trees for indexed fields (movie_id,customer_id), and have to resort to linear search for unindexed fields (checked_out?). The database will do the faster filters first, which means you end up with something slightly worse thanO(log(r)), whereris the number of rentals.available_inventoryMovie method that returns the available inventory for an instance of a movie. We wrapped this functionality in a method since it made our controller code more readable and could also be used in our .rabl views.