Formatted question description: https://leetcode.ca/all/1912.html

1912. Design Movie Rental System (Hard)

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.

Companies:
Flipkart

Related Topics:
Heap, Design, Ordered Map

Solution 1.

Intuition: Since search is reporting unrented movies while report is reporting rented movies, we can use two data structures to store unrented and rented movies, and rent and drop just moves the entry between these two data structures.

Algorithm:

search: for a specific movie, we need the cheapest 5 {price, shop} pairs. So we can use unordered_map<Movie, set<pair<Price, Shop>>> for unrented (movie -> set of {price, shop}) so that we can simply return the first 5 shops in the set of unrented[movie].

report: we need the cheapest 5 {price, shop, movie} tuple. So we can use set<tuple<Price, Shop, Movie>> for rented so that we can simply return the first 5 {shop, movie} pairs in the rented set.

rent: Given shop, movie, we need to move this entry from unrented to rented. We need a separate map map<pair<Shop, Movie>, Price> price ({shop, movie} -> price) to query the price given pair {shop, movie}. Then we just need to unrented[movie].erase({price, shop}) and rented.emplace(price, shop, movie).

drop: Similar to rent, we first get the corresponding price then move the entry from rented to unrented.

// OJ: https://leetcode.com/problems/design-movie-rental-system/
// Time:
//     MovieRentingSystem: O(ElogE)
//     search: O(1)
//     rent: O(logE)
//     drop: O(logE)
//     report: O(1)
// Space: O(E)
class MovieRentingSystem {
    map<pair<int, int>, int> price; // {shop, movie} -> price
    unordered_map<int, set<pair<int, int>>> unrented; // movie -> set of {price, shop}
    set<tuple<int, int, int>> rented; // set of {price, shop, movie}
public:
    MovieRentingSystem(int n, vector<vector<int>>& entries) {
        for (auto &e : entries) {// shop, movie, price
            int shop = e[0], movie = e[1], p = e[2];
            price[{shop, movie}] = p; 
            unrented[movie].emplace(p, shop);
        }
    }
    vector<int> search(int movie) {
        auto &s = unrented[movie];
        vector<int> ans;
        int i = 0;
        for (auto it = s.begin(); i < 5 && it != s.end(); ++it, ++i) {
            ans.push_back(it->second);
        }
        return ans;
    }
    void rent(int shop, int movie) {
        int p = price[{shop, movie}];
        unrented[movie].erase({p, shop});
        rented.emplace(p, shop, movie);
    }
    void drop(int shop, int movie) {
        int p = price[{shop, movie}];
        rented.erase({p, shop, movie});
        unrented[movie].emplace(p, shop);
    }
    vector<vector<int>> report() {// shop, movie
        vector<vector<int>> ans;
        int i = 0;
        for (auto it = rented.begin(); it != rented.end() && i < 5; ++i, ++it) {
            auto [p, s, m] = *it;
            ans.push_back({s, m});
        }
        return ans;
    }
};

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