How do you solve Movie Rating on LeetCode?

Movie Rating (LeetCode #1341) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using hash maps can significantly reduce time complexity.

What is the brute force solution for Movie Rating?

The brute force approach for Movie Rating is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will iterate through each user and count how many movies they have rated. Then, we will check the average ratings of all movies in February 2020 to find the highest-rated movie. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Movie Rating?

Movie Rating uses the following concepts: Database. The recommended patterns to study are: Hash Map, Aggregation Functions.

What are the interview tips for Movie Rating?

Always clarify the requirements and constraints before jumping into coding. Think about edge cases, such as no ratings or all ratings being the same. Practice writing SQL queries on paper or a whiteboard to get comfortable with syntax.

What are common mistakes when solving Movie Rating?

Not considering ties when selecting the user or movie. Overcomplicating the SQL query instead of breaking it down into manageable parts.

#1341

Movie Rating

Medium

DatabaseHash MapAggregation Functions

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

In the optimal approach, we will use hash maps to efficiently count ratings and calculate averages in a single pass through the data. This reduces the time complexity significantly.

⚙️

Algorithm

4 steps

1Step 1: Use a hash map to count the number of ratings for each user.
2Step 2: Use another hash map to accumulate ratings and count for each movie rated in February 2020.
3Step 3: Determine the user with the maximum ratings count and resolve ties lexicographically.
4Step 4: Calculate the average rating for each movie and find the one with the highest average, resolving ties lexicographically.

solution.py13 lines

1WITH UserCount AS (
2    SELECT user_id, COUNT(movie_id) AS rating_count
3    FROM MovieRating
4    GROUP BY user_id
5),
6MovieAverage AS (
7    SELECT movie_id, AVG(rating) AS avg_rating, COUNT(rating) AS rating_count
8    FROM MovieRating
9    WHERE created_at BETWEEN '2020-02-01' AND '2020-02-29'
10    GROUP BY movie_id
11)
12SELECT (SELECT name FROM Users u JOIN UserCount uc ON u.user_id = uc.user_id ORDER BY uc.rating_count DESC, u.name ASC LIMIT 1) AS user_name,
13       (SELECT title FROM Movies m JOIN MovieAverage ma ON m.movie_id = ma.movie_id ORDER BY ma.avg_rating DESC, m.title ASC LIMIT 1) AS movie_name;

ℹ

Complexity note: The time complexity is O(n) because we traverse the MovieRating table a limited number of times, and space complexity is O(n) due to the hash maps storing user counts and movie averages.

1Using hash maps can significantly reduce time complexity.
2Understanding how to group and aggregate data is crucial for SQL queries.

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