SQL > LeetCode 1341. Movie Rating (MySQL)

문제

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.

 

두 문제이다. 하나는 가장 영화 리뷰를 많이 남긴 사람의 이름을 출력해야한다. 만약 리뷰 수가 같다면 사전순으로 더 작은 사람을 출력한다.

다른 하나는 2020년 2월에 평균 평점이 가장 높은 영화 이름을 찾야야한다. 동점인 경우에 사전순으로 더 작은 영화 이름을 출력한다.

 

테이블 및 예제

[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.

 

[Users]

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

 

[Examples]

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     |
+--------------+

 

풀이

임시테이블을 사용해서 조건에 맞는 테이블을 생성하고 limit을 걸어서 열을 뽑아낸다. 후에 UNION을 사용해서 합쳐주면 결과를 도출 가능하다.

WITH 
g_movie AS (
    SELECT r.movie_id,COUNT(r.user_id) AS cnt, m.title, u.name
    FROM MovieRating AS r
    JOIN Movies AS m ON r.movie_id = m.movie_id JOIN Users AS u ON u.user_id = r.user_id
    GROUP BY r.user_id
),
fin1 AS (
SELECT name AS results
FROM g_movie
ORDER BY cnt DESC, name
limit 1
),
FE20 AS(
    SELECT m.title AS results
    FROM MovieRating AS mr JOIN Movies AS m 
        ON mr.movie_id = m.movie_id
    WHERE mr.created_at BETWEEN '2020-02-01' AND '2020-02-28'
    GROUP BY mr.movie_id 
    ORDER BY AVG(mr.rating) DESC, m.title
    limit 1
)
SELECT * FROM FE20
UNION ALL
SELECT * FROM fin1

 

처음에는 각각 임시테이블을 만들고 조건에 맞는 열을 뽑아냈는데 UNION을 하는게 어려워졌다.

따라서 위와 같은 방식으로 열을 뽑아내는 임시테이블을 쭉 만들어낸 다음에 테이블을 가져와서 UNION 해주었다.