코딩테스트 SQL 코드카타 (Movie Rating)

문제

 

Movie Rating - LeetCode

 

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.

 

---

풀이

results 는 union으로 값을 합쳐야 한다.

1. 사용자의 rating을 카운트 한다.
2. max(count(raiting)) 이 사용자의 count와 동일한 사용자를 필터
3. order by로 정렬 해주고 limit 1 으로 한명만 값을 뽑는다.
4. 영화의 average 점수를 뽑는다.
5. max(average) 가 영화의  average와 동일한 영화를 필터
6. order by 로 정렬하고  limit 1 으로 1개만.

---

코드

with Userrating as (
    select u.name as name, count(*) as count
    from Users u
    join MovieRating m
    on u.user_id = m.user_id
    group by u.name
),
Movierating as (
    select m.title as movie, avg(mr.rating) as average
    from Movies m
    join MovieRating mr
    on m.movie_id = mr.movie_id
    where date_format(created_at,'%Y-%m') = "2020-02"
    group by title
),

Userresult as(
select name  as results
from Userrating
where count = (select max(count) from Userrating)
order by name
limit 1
),


Movieresult as(
select movie as results
from Movierating
where average = (select max(average) from Movierating)
order by movie
limit 1
)

select results from Userresult
union all
select results from Movieresult

 

Userresult와 Movieresult는 Union all 이 안되서 같이 서브쿼리화 시켰다.

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링크

LeetCode/1341-movie-rating at main · K-MarkLee/LeetCode

LeetCode/1341-movie-rating at main · K-MarkLee/LeetCode