Leetcode #1912: Design Movie Rental System
In this guide, we solve Leetcode #1912 Design Movie Rental System in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Design, Array, Hash Table, Ordered Set, Heap (Priority Queue)
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
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()
Complexity
The time complexity is , where is the length of . The space complexity is , where is the length of .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.