LeetCode-in-Java
1341. Movie Rating
Medium
SQL Schema
Table: Movies
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| movie_id | int |
| title | varchar |
+---------------+---------+
movie_id is the primary key 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 for this table.
Table: MovieRating
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| movie_id | int |
| user_id | int |
| rating | int |
| created_at | date |
+---------------+---------+
(movie_id, user_id) is the primary key 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 an SQL query 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 query result format is in the following example.
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
# Write your MySQL query statement below
(SELECT name results
FROM Users as U, MovieRating as MR
WHERE U.user_id = MR.user_id
GROUP BY U.user_id
ORDER BY COUNT(MR.user_id) DESC, name ASC LIMIT 1)
UNION ALL
(SELECT title results
FROM Movies as M, MovieRating as MR
WHERE M.movie_id = MR.movie_id AND created_at BETWEEN '2020-02-01' AND '2020-02-29'
GROUP BY M.movie_id
ORDER BY AVG(rating) DESC, title ASC LIMIT 1)