LeetCode 코딩 테스트 - Movie Rating(LV.Medium)

 

 

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.

 

 

# 쿼리를 작성하는 목표, 확인할 지표 : 가장 많이 평가를 남긴 유저 이름 & 가장 높은 평가를 받은 영화 이름 출력 / user_id, rating
# 쿼리 계산 방법 : 1. group by user_id -> 2. order by로 count 내림차순, name오름차순 limit 1로 첫 행만 추출 -> 3. 2020년 2월에 한해서 group by movie_id -> 4. order by로 avg 내림차순, title오름차순 limit 1 -> 5. union all
# 데이터의 기간 : avg 구할때 한해서 2020년 2월
# 사용할 테이블 : movies, users, movierating
# JOIN KEY : movie_id, user_id
# 데이터 특징 : x
WITH name AS (
    SELECT
        u.name
    FROM movierating AS mr
    INNER JOIN users AS u
    ON mr.user_id = u.user_id
    # 1
    GROUP BY
        mr.user_id
    # 2
    ORDER BY
        COUNT(*) DESC, u.name ASC
    LIMIT 1
), title AS (
    SELECT
        m.title
    FROM movierating AS mr
    INNER JOIN movies AS m
    ON mr.movie_id = m.movie_id
    # 3
    WHERE
        DATE_FORMAT(created_at, '%Y-%m') = '2020-02'
    GROUP BY
        mr.movie_id
    # 4
    ORDER BY
        AVG(mr.rating) DESC, m.title ASC
    LIMIT 1
)
SELECT
    name AS results
FROM name
# 5
UNION ALL
SELECT
    title
FROM title