LeetCode 1912. Design Movie Rental System Solution in Java, C++, Python & Go | Explanation + Code

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1912. Design Movie Rental System

Description

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 MovieRentingSystem class:

  • MovieRentingSystem(int n, int[][] entries) Initializes the MovieRentingSystem object with n shops and the movies in entries.
  • List<Integer> 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<List<Integer>> 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.

Solutions

Solution 1: Ordered Set

We define an ordered set available, where available[movie] stores a list of all shops that have not rented out the movie movie. Each element in the list is (price, shop), sorted in ascending order by price, and if prices are equal, by shop in ascending order.

Additionally, we define a hash map pricemap, where pricemap[f(shop, movie)] stores the rental price of movie movie in shop shop.

We also define an ordered set rented, which stores all rented movies as (price, shop, movie), sorted in ascending order by price, then by shop, and if both are equal, by movie.

For the MovieRentingSystem(n, entries) operation, we iterate through entries and add each shop's movie information to both available and pricemap. The time complexity is O(m log m), where m is the length of entries.

For the search(movie) operation, we return the shop IDs of the first 5 shops in available[movie]. The time complexity is O(1).

For the rent(shop, movie) operation, we remove (price, shop) from available[movie] and add (price, shop, movie) to rented. The time complexity is O(log m).

For the drop(shop, movie) operation, we remove (price, shop, movie) from rented and add (price, shop) back to available[movie]. The time complexity is O(log m).

For the report() operation, we return the shop and movie IDs of the first 5 rented movies in rented. The time complexity is O(1).

The space complexity is O(m), where m is the length of entries.

PythonJavaC++Go
class MovieRentingSystem: def __init__(self, n: int, entries: List[List[int]]): self.available = defaultdict(lambda: SortedList()) self.price_map = {} for shop, movie, price in entries: self.available[movie].add((price, shop)) self.price_map[self.f(shop, movie)] = price self.rented = SortedList() def search(self, movie: int) -> List[int]: return [shop for _, shop in self.available[movie][:5]] def rent(self, shop: int, movie: int) -> None: price = self.price_map[self.f(shop, movie)] self.available[movie].remove((price, shop)) self.rented.add((price, shop, movie)) def drop(self, shop: int, movie: int) -> None: price = self.price_map[self.f(shop, movie)] self.rented.remove((price, shop, movie)) self.available[movie].add((price, shop)) def report(self) -> List[List[int]]: return [[shop, movie] for _, shop, movie in self.rented[:5]] def f(self, shop: int, movie: int) -> int: return shop << 30 | movie # Your MovieRentingSystem object will be instantiated and called as such: # obj = MovieRentingSystem(n, entries) # param_1 = obj.search(movie) # obj.rent(shop,movie) # obj.drop(shop,movie) # param_4 = obj.report()(code-box)

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