DEV Community

Cover image for 1912. Design Movie Rental System
MD ARIFUL HAQUE
MD ARIFUL HAQUE

Posted on

1912. Design Movie Rental System

1912. Design Movie Rental System

Difficulty: Hard

Topics: Array, Hash Table, Design, Heap (Priority Queue), Ordered Set, Biweekly Contest 55

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]]

Enter fullscreen mode Exit fullscreen mode
  • 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.

Enter fullscreen mode Exit fullscreen mode

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.

Hint:

  1. You need to maintain a sorted list for each movie and a sorted list for rented movies
  2. When renting a movie remove it from its movies sorted list and added it to the rented list and vice versa in the case of dropping a movie

Solution:

We need to design a movie rental system that supports searching for available movies, renting movies, returning rented movies, and generating a report of the currently rented movies. The system must efficiently handle these operations while ensuring that the results are sorted according to specific criteria.

Approach

  1. Initialization:

    • Store the price of each movie at each shop in a dictionary (prices).
    • Track the rental status of each movie at each shop using a dictionary (rented).
    • For each movie, maintain a min-heap (movieHeaps) that contains all copies (as [price, shop] pairs) available for that movie. This heap helps in efficiently retrieving the cheapest available shops during search operations.
    • Maintain a min-heap (rentedHeap) for all rented movies, stored as [price, shop, movie] tuples. This heap helps in generating the report of the cheapest rented movies.
    • Use a dictionary (inRentedHeap) to track whether a specific movie at a shop is currently in the rented heap to avoid duplicates.
  2. Search Operation:

    • For a given movie, retrieve its corresponding heap.
    • Pop elements from the heap until we find up to 5 available (unrented) movies. During popping, check the rental status of each movie-shop combination.
    • Temporarily store popped elements and push them back into the heap after processing to maintain the heap's integrity for future operations.
  3. Rent Operation:

    • Mark the movie at the given shop as rented.
    • If the movie-shop combination is not already in the rented heap, add it to the heap and mark it as present in the heap.
  4. Drop Operation:

    • Mark the movie at the given shop as available (unrented).
  5. Report Operation:

    • Pop elements from the rented heap until we find up to 5 currently rented movies. During popping, check the rental status of each movie-shop combination.
    • Temporarily store rented movies that are still rented and push them back into the heap after processing. Non-rented movies are not pushed back, effectively removing them from the heap.

Let's implement this solution in PHP: 1912. Design Movie Rental System

<?php
class MovieRentingSystem {
    private $prices;
    private $rented;
    private $movieHeaps;
    private $rentedHeap;
    private $inRentedHeap;

    /**
     * @param Integer $n
     * @param Integer[][] $entries
     */
    function __construct($n, $entries) {
        ...
        ...
        ...
        /**
         * go to ./solution.php
         */
    }

    /**
     * @param Integer $movie
     * @return Integer[]
     */
    function search($movie) {
        ...
        ...
        ...
        /**
         * go to ./solution.php
         */
    }

    /**
     * @param Integer $shop
     * @param Integer $movie
     * @return NULL
     */
    function rent($shop, $movie) {
        ...
        ...
        ...
        /**
         * go to ./solution.php
         */
    }

    /**
     * @param Integer $shop
     * @param Integer $movie
     * @return NULL
     */
    function drop($shop, $movie) {
        ...
        ...
        ...
        /**
         * go to ./solution.php
         */
    }

    /**
     * @return Integer[][]
     */
    function report() {
        ...
        ...
        ...
        /**
         * go to ./solution.php
         */
    }
}

/**
 * Your MovieRentingSystem object will be instantiated and called as such:
 * $obj = MovieRentingSystem($n, $entries);
 * $ret_1 = $obj->search($movie);
 * $obj->rent($shop, $movie);
 * $obj->drop($shop, $movie);
 * $ret_4 = $obj->report();
 */

// Test cases
$movieRentingSystem = new MovieRentingSystem(3, [
    [0, 1, 5], [0, 2, 6], [0, 3, 7],
    [1, 1, 4], [1, 2, 7], [2, 1, 5]
]);

print_r($movieRentingSystem->search(1)); 
// [1, 0, 2]

$movieRentingSystem->rent(0, 1);
$movieRentingSystem->rent(1, 2);

print_r($movieRentingSystem->report()); 
// [[0, 1], [1, 2]]

$movieRentingSystem->drop(1, 2);

print_r($movieRentingSystem->search(2)); 
// [0, 1]
?>
Enter fullscreen mode Exit fullscreen mode

Explanation:

  • Initialization: The constructor initializes data structures to store movie prices, rental status, and heaps for each movie and rented movies. It processes the input entries to populate these structures.
  • Search: The search function checks the heap of the specified movie, collecting up to 5 available shops by checking rental status. Popped elements are temporarily stored and pushed back to maintain the heap.
  • Rent: Marks a movie as rented and adds it to the rented heap if not already present.
  • Drop: Marks a movie as available (unrented).
  • Report: Collects up to 5 currently rented movies from the rented heap, skipping any that are no longer rented. Rented movies are pushed back into the heap, while non-rented ones are removed.

This approach efficiently manages the movie rental operations by leveraging min-heaps and lazy deletion, ensuring that the system meets the required performance constraints.

Contact Links

If you found this series helpful, please consider giving the repository a star on GitHub or sharing the post on your favorite social networks 😍. Your support would mean a lot to me!

If you want more helpful content like this, feel free to follow me:

Top comments (0)