How do you solve Design Movie Rental System on LeetCode?

Design Movie Rental System (LeetCode #1912) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Using priority queues can significantly improve the efficiency of searching and managing rented movies.

What is the brute force solution for Design Movie Rental System?

The brute force approach for Design Movie Rental System is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we simply iterate through all the shops to find the cheapest options for renting a movie. This is straightforward but inefficient, especially as the number of shops increases. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Design Movie Rental System?

Clearly explain your thought process and the data structures you choose to use. Consider edge cases and how your solution handles them. Practice implementing priority queues and HashMaps, as they are commonly used in design problems.

What are common mistakes when solving Design Movie Rental System?

Not properly managing the state of rented and available movies, leading to incorrect results. Forgetting to handle edge cases, such as renting from a shop with no available copies.

#1912

Design Movie Rental System

Hard

ArrayHash TableDesignHeap (Priority Queue)Ordered SetHash MapPriority Queue

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

By using a combination of data structures like a HashMap for storing available movies and a priority queue for managing rented movies, we can efficiently handle searching, renting, and reporting with better performance.

⚙️

Algorithm

5 steps

1Step 1: Use a HashMap to store available movies with their prices and shops.
2Step 2: Use a priority queue (min-heap) to manage rented movies sorted by price.
3Step 3: For searching, retrieve from the HashMap and maintain a sorted list of available movies.
4Step 4: For renting, remove the movie from the HashMap and add it to the priority queue.
5Step 5: For dropping, move the movie back to the HashMap and adjust the priority queue.

solution.py33 lines

1# Full working Python code
2import heapq
3class MovieRentalSystem:
4    def __init__(self, entries):
5        self.available = {}
6        self.rented = []
7        for shop, movie, price in entries:
8            if movie not in self.available:
9                self.available[movie] = []
10            heapq.heappush(self.available[movie], (price, shop))
11
12    def search(self, movie):
13        if movie not in self.available:
14            return []
15        return sorted(self.available[movie])[:5]
16
17    def rent(self, shop, movie):
18        if movie in self.available and self.available[movie]:
19            price, available_shop = heapq.heappop(self.available[movie])
20            if available_shop == shop:
21                heapq.heappush(self.rented, (price, shop, movie))
22            else:
23                heapq.heappush(self.available[movie], (price, available_shop))
24
25    def drop(self, shop, movie):
26        for i in range(len(self.rented)):
27            if self.rented[i][1] == shop and self.rented[i][2] == movie:
28                price, _, _ = self.rented.pop(i)
29                heapq.heappush(self.available[movie], (price, shop))
30                return
31
32    def report(self):
33        return sorted(self.rented)[:5]

ℹ

Complexity note: The time complexity is O(n log n) due to the use of priority queues for managing the rented movies and the sorting of available movies. The space complexity is O(n) because we store all the movies and rented entries.

1Using priority queues can significantly improve the efficiency of searching and managing rented movies.
2Maintaining a HashMap allows for quick access to available movies, reducing the time complexity of searches.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.