1341. Movie Rating

Description

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.

Solutions

Solution 1

MySQL

# Write your MySQL query statement below
(
    SELECT name AS results
    FROM
        Users
        JOIN MovieRating USING (user_id)
    GROUP BY user_id
    ORDER BY COUNT(1) DESC, name
    LIMIT 1
)
UNION ALL
(
    SELECT title
    FROM
        MovieRating
        JOIN Movies USING (movie_id)
    WHERE DATE_FORMAT(created_at, '%Y-%m') = '2020-02'
    GROUP BY movie_id
    ORDER BY AVG(rating) DESC, title
    LIMIT 1
);

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