problemmediumdatabaseleetcode-1341leetcode 1341leetcode1341

Movie Rating

MediumUpdated: Oct 13, 2025
Practice on:
LeetCode

Problem

Table: Movies

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| movie_id      | int     |
| title         | varchar |
+---------------+---------+
movie_id is the primary key (column with unique values) for this table.
title is the name of the movie.

Table: `Users`

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| user_id       | int     |
| name          | varchar |
+---------------+---------+
user_id is the primary key (column with unique values) for this table.
The column 'name' has unique values.

Table: `MovieRating`

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| movie_id      | int     |
| user_id       | int     |
| rating        | int     |
| created_at    | date    |
+---------------+---------+
(movie_id, user_id) is the primary key (column with unique values) for this table.
This table contains the rating of a movie by a user in their review.
created_at is the user's review date.

Write a solution to:

  • Find the name of the user who has rated the greatest number of movies. In case of a tie, return the lexicographically smaller user name.
  • Find the movie name with the highest average rating in February 2020. In case of a tie, return the lexicographically smaller movie name.

The result format is in the following example.

Examples

Example 1

Input: 
Movies table:
+-------------+--------------+
| movie_id    |  title       |
+-------------+--------------+
| 1           | Avengers     |
| 2           | Frozen 2     |
| 3           | Joker        |
+-------------+--------------+
Users table:
+-------------+--------------+
| user_id     |  name        |
+-------------+--------------+
| 1           | Daniel       |
| 2           | Monica       |
| 3           | Maria        |
| 4           | James        |
+-------------+--------------+
MovieRating table:
+-------------+--------------+--------------+-------------+
| movie_id    | user_id      | rating       | created_at  |
+-------------+--------------+--------------+-------------+
| 1           | 1            | 3            | 2020-01-12  |
| 1           | 2            | 4            | 2020-02-11  |
| 1           | 3            | 2            | 2020-02-12  |
| 1           | 4            | 1            | 2020-01-01  |
| 2           | 1            | 5            | 2020-02-17  | 
| 2           | 2            | 2            | 2020-02-01  | 
| 2           | 3            | 2            | 2020-03-01  |
| 3           | 1            | 3            | 2020-02-22  | 
| 3           | 2            | 4            | 2020-02-25  | 
+-------------+--------------+--------------+-------------+
Output: 
+--------------+
| results      |
+--------------+
| Daniel       |
| Frozen 2     |
+--------------+
Explanation: 
Daniel and Monica have rated 3 movies ("Avengers", "Frozen 2" and "Joker") but Daniel is smaller lexicographically.
Frozen 2 and Joker have a rating average of 3.5 in February but Frozen 2 is smaller lexicographically.

Solution

Method 1 - Aggregate + Join with Tie-Break Ordering

Intuition

We need two queries: one to find the user who rated the most movies (with lexicographical tie-break), and one to find the movie with the highest average rating in February 2020 (with lexicographical tie-break).

Approach

  1. For the user: Group MovieRating by user_id, count ratings, join with Users, order by count desc and name asc, limit 1.
  2. For the movie: Filter MovieRating for February 2020, group by movie_id, calculate average, join with Movies, order by average desc and title asc, limit 1.

Code

SQL
-- User with most ratings
SELECT name AS results
FROM Users u
JOIN (
  SELECT user_id, COUNT(*) AS cnt
  FROM MovieRating
  GROUP BY user_id
) r ON u.user_id = r.user_id
ORDER BY cnt DESC, name ASC
LIMIT 1
UNION ALL
-- Movie with highest average rating in Feb 2020
SELECT title AS results
FROM Movies m
JOIN (
  SELECT movie_id, AVG(rating) AS avg_rating
  FROM MovieRating
  WHERE created_at BETWEEN '2020-02-01' AND '2020-02-29'
  GROUP BY movie_id
) r ON m.movie_id = r.movie_id
ORDER BY avg_rating DESC, title ASC
LIMIT 1;
Python (using pandas)
# User with most ratings
user_counts = movierating.groupby('user_id').size()
max_count = user_counts.max()
user_id = user_counts[user_counts == max_count].index.min()
user_name = users.loc[users['user_id'] == user_id, 'name'].values[0]
# Movie with highest average rating in Feb 2020
feb = movierating[(movierating['created_at'] >= '2020-02-01') & (movierating['created_at'] <= '2020-02-29')]
movie_avg = feb.groupby('movie_id')['rating'].mean()
max_avg = movie_avg.max()
movie_id = movie_avg[movie_avg == max_avg].index.min()
movie_title = movies.loc[movies['movie_id'] == movie_id, 'title'].values[0]

Complexity

  • ⏰ Time complexity: O(N) for each query (group by, join, sort).
  • 🧺 Space complexity: O(U + M) for intermediate results.

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