# Design Movie Rental System Solution

## Design Movie Rental System Solution

You have a movie renting company consisting of n shops. You want to implement a renting system that supports searching for, booking, and returning movies. The system should also support generating a report of the currently rented movies.

Each movie is given as a 2D integer array entries where entries[i] = [shopi, moviei, pricei] indicates that there is a copy of movie moviei at shop shopi with a rental price of pricei. Each shop carries at most one copy of a movie moviei.

The system should support the following functions:

Search: Finds the cheapest 5 shops that have an unrented copy of a given movie. The shops should be sorted by price in ascending order, and in case of a tie, the one with the smaller shopi should appear first. If there are less than 5 matching shops, then all of them should be returned. If no shop has an unrented copy, then an empty list should be returned.

Rent: Rents an unrented copy of a given movie from a given shop.

Drop: Drops off a previously rented copy of a given movie at a given shop.

Report: Returns the cheapest 5 rented movies (possibly of the same movie ID) as a 2D list res where res[j] = [shopj, moviej] describes that the jth cheapest rented movie moviej was rented from the shop shopj. The movies in res should be sorted by price in ascending order, and in case of a tie, the one with the smaller shopj should appear first, and if there is still tie, the one with the smaller moviej should appear first. If there are fewer than 5 rented movies, then all of them should be returned. If no movies are currently being rented, then an empty list should be returned.

Implement the Movie Renting System class:

• MovieRentingSystem(int n, int[][] entries) Initializes the MovieRentingSystem object with n shops and the movies in entries.
• List search(int movie) Returns a list of shops that have an unrented copy of the given movie as described above.
• void rent(int shop, int movie) Rents the given movie from the given shop.
• void drop(int shop, int movie) Drops off a previously rented movie at the given shop.
• List> report() Returns a list of cheapest rented movies as described above.

Note: The test cases will be generated such that rent will only be called if the shop has an unrented copy of the movie, and drop will only be called if the shop had previously rented out the movie.

Example 1:

Input
[“MovieRentingSystem”, “search”, “rent”, “rent”, “report”, “drop”, “search”]
[[3, [[0, 1, 5], [0, 2, 6], [0, 3, 7], [1, 1, 4], [1, 2, 7], [2, 1, 5]]], [1], [0, 1], [1, 2], [], [1, 2], [2]]

Output
[null, [1, 0, 2], null, null, [[0, 1], [1, 2]], null, [0, 1]]

Explanation

• MovieRentingSystem movieRentingSystem = new MovieRentingSystem(3, [[0, 1, 5], [0, 2, 6], [0, 3, 7], [1, 1, 4], [1, 2, 7], [2, 1, 5]]);
• movieRentingSystem.search(1); // return [1, 0, 2], Movies of ID 1 are unrented at shops 1, 0, and 2. Shop 1 is cheapest; shop 0 and 2 are the same price, so order by shop number.
• movieRentingSystem.rent(0, 1); // Rent movie 1 from shop 0. Unrented movies at shop 0 are now [2,3].
• movieRentingSystem.rent(1, 2); // Rent movie 2 from shop 1. Unrented movies at shop 1 are now [1].
• movieRentingSystem.report(); // return [[0, 1], [1, 2]]. Movie 1 from shop 0 is cheapest, followed by movie 2 from shop 1.
• movieRentingSystem.drop(1, 2); // Drop off movie 2 at shop 1. Unrented movies at shop 1 are now [1,2].
• movieRentingSystem.search(2); // return [0, 1]. Movies of ID 2 are unrented at shops 0 and 1. Shop 0 is cheapest, followed by shop 1.

Constraints:

• 1 <= n <= 3 * 105
• 1 <= entries.length <= 105
• 0 <= shopi < n
• 1 <= moviei, pricei <= 104
• Each shop carries at most one copy of a movie moviei.
• At most 105 calls in total will be made to search, rent, drop and report.

### SOLUTION

Below is the explanation on the data structures:

``````Map<String, Entry> entryMap = new HashMap<>();
Map<Integer, TreeSet<Entry>> available = new HashMap<>();
Set<Entry> rented = new HashSet<>();``````
1. entryMap, is request to get the objects by a (shop & movie) combination in O(1). Because those are unique.
2. available, is required to get the available movies while serching, and Since those are already shorted, search will also be constant time. O(5) ~= O(1)
3. rented, is required to maintain the entries those are rented so that it can be used while the report call. This will be a O(N Log N), where N is items those are sold.
• By converting this to a TreeSet, we can reduce the report Time complexity to constant.

Program: Design Movie Rental System Solution in Java

``````class MovieRentingSystem {
int n, limit = 5;
Map<String, Entry> entryMap = new HashMap<>();
Map<Integer, TreeSet<Entry>> available = new HashMap<>();
Comparator<Entry> order = (e1, e2) -> e1.price == e2.price ? (e1.shop == e2.shop ? Integer.compare(e1.movie, e2.movie) : Integer.compare(e1.shop, e2.shop)) : Integer.compare(e1.price, e2.price);
//                TreeSet<Entry> rented = new TreeSet<>(order);
Set<Entry> rented = new HashSet<>();

public MovieRentingSystem(int n, int[][] entries) {
this.n = n;
for (int[] entry : entries) {
Entry entryObj = new Entry(entry[0], entry[1], entry[2]);
entryMap.put(getKey(entryObj), entryObj);
}
}

}

String getKey(Entry entry) {
return getKey(entry.shop, entry.movie);
}

String getKey(int shop, int movie) {
return shop + " -> " + movie;
}

public List<Integer> search(int movie) {
return available.getOrDefault(movie, new TreeSet<>(order)).stream().limit(limit).map(entry -> entry.shop).collect(Collectors.toList());
}

//            Time: Log m, where m is number of available Entries for the movie, because we are removing it from available sorted set.
public void rent(int shop, int movie) {
Entry entry = entryMap.get(getKey(shop, movie));
if (entry == null) return;
available.get(movie).remove(entry);
}

//            Time: Log m, where m is number of available Entries for the movie, because we are adding it to available sorted set.
public void drop(int shop, int movie) {
Entry entry = entryMap.get(getKey(shop, movie));
if (entry == null) return;
rented.remove(entry);
}

public List<List<Integer>> report() {
//            Loop though all the rented items and sort those by the order, and limit only 5.
return rented.stream().sorted(order).limit(limit).map(entry -> List.of(entry.shop, entry.movie)).collect(Collectors.toList());
}

class Entry {
int shop, movie, price;
Entry(int shop, int movie, int price) {
this.shop = shop;
this.movie = movie;
this.price = price;
}
}
}``````