[SQL_Q] 1341. Movie Rating

https://leetcode.com/problems/movie-rating/description/

문제

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.

 

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.

내 쿼리

with top_cnt_uer as (
SELECT u.name as results
FROM MovieRating m JOIN Users u
ON m.user_id = u.user_id
GROUP BY m.user_id
ORDER BY COUNT(*) DESC, results
LIMIT 1
),top_rate_mv as(
SELECT m.title
FROM (
    SELECT movie_id, user_id, rating 
    FROM MovieRating
    WHERE created_at >= '2020-02-01' and created_at < '2020-03-01'
) f join Movies m
on f.movie_id = m.movie_id
GROUP BY f.movie_id
ORDER BY avg(f.rating) DESC, m.title
LIMIT 1
)
SELECT * FROM top_cnt_uer
UNION ALL
SELECT * FROM top_rate_mv

다른 사람 쿼리

(
  SELECT u.name AS results
  FROM MovieRating mr
  JOIN Users u ON mr.user_id = u.user_id
  GROUP BY u.name
  ORDER BY COUNT(*) DESC, u.name
  LIMIT 1
)
UNION ALL
(
  SELECT m.title AS results
  FROM MovieRating mr
  JOIN Movies m ON mr.movie_id = m.movie_id
  WHERE DATE_FORMAT(mr.created_at, '%Y-%m') = '2020-02'
  GROUP BY m.title
  ORDER BY AVG(mr.rating) DESC, m.title
  LIMIT 1
);

UNION 에서는 LIMIT을 사용할 수 없다는 것은 알고 있었는데 ORDER BY 도 사용 못하는 건 까먹고 있었다.

UNION은 두 SELECT의 결과를 합친 후 하나의 테이블처럼 취급하므로, 개별 SELECT에 ORDER BY/LIMIT을 둘 수 없다.

with으로 만든 Common Table Expression 방법만 생각이 났었는데,
SELECT 문을 괄호로 묶으면 union이 되는지 이제 알았다.. 재사용성 & 성능 문제 없이 빠르게 테스트 해봐야 할 때 괜찮은 방법인 것 같다.

내 쿼리 수정

with top_cnt_usr as (
    SELECT u.name as results
    FROM Users u JOIN MovieRating m
    ON m.user_id = u.user_id
    GROUP BY m.user_id
    ORDER BY COUNT(*) DESC, results
    LIMIT 1
),
top_rate_mv as(
    SELECT m.title
    FROM Movies m JOIN MovieRating mr
    on mr.created_at >= '2020-02-01' AND mr.created_at < '2020-03-01'
    and mr.movie_id = m.movie_id
    GROUP BY mr.movie_id
    ORDER BY avg(mr.rating) DESC, m.title
    LIMIT 1
)
SELECT * FROM top_cnt_usr
UNION ALL 
SELECT * FROM top_rate_mv

1. 필요 없는 sub쿼리 지양하기
2. join 시 작은 테이블이 먼저 오도록 하기, where 필터링 조건 넣어도 됨