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‎problems/activity-participants/README.md‎

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## [1355. Activity Participants (Medium)](https://leetcode.com/problems/activity-participants "")
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<p>Table: <code>Friends</code></p>
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<pre>
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+---------------+---------+
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| Column Name | Type |
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+---------------+---------+
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| id | int |
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| name | varchar |
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| activity | varchar |
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+---------------+---------+
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id is the id of the friend and primary key for this table.
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name is the name of the friend.
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activity is the name of the activity which the friend takes part in.
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</pre>
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<p>Table: <code>Activities</code></p>
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<pre>
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+---------------+---------+
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| Column Name | Type |
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+---------------+---------+
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| id | int |
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| name | varchar |
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+---------------+---------+
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id is the primary key for this table.
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name is the name of the activity.
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</pre>
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Write an SQL query to find the names of all the activities with neither maximum, nor minimum number of participants.
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Return the result table in any order. Each activity in table Activities is performed by any person in the table Friends.
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The query result format is in the following example:
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Friends table:
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<pre>
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+------+--------------+---------------+
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| id | name | activity |
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+------+--------------+---------------+
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| 1 | Jonathan D. | Eating |
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| 2 | Jade W. | Singing |
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| 3 | Victor J. | Singing |
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| 4 | Elvis Q. | Eating |
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| 5 | Daniel A. | Eating |
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| 6 | Bob B. | Horse Riding |
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+------+--------------+---------------+
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Activities table:
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+------+--------------+
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| id | name |
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+------+--------------+
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| 1 | Eating |
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| 2 | Singing |
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| 3 | Horse Riding |
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+------+--------------+
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Result table:
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+--------------+
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| results |
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+--------------+
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| Singing |
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+--------------+
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Eating activity is performed by 3 friends, maximum number of participants, (Jonathan D. , Elvis Q. and Daniel A.)
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Horse Riding activity is performed by 1 friend, minimum number of participants, (Bob B.)
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Singing is performed by 2 friends (Victor J. and Jade W.)
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</pre>

‎problems/movie-rating/README.md‎

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## [1341. Movie Rating (Medium)](https://leetcode.com/problems/movie-rating "电影评分")
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<p>Table: <code>Movies</code></p>
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<pre>
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+---------------+---------+
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| Column Name | Type |
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+---------------+---------+
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| movie_id | int |
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| title | varchar |
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+---------------+---------+
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movie_id is the primary key for this table.
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title is the name of the movie.
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</pre>
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<p>Table: <code>Users</code></p>
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<pre>
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+---------------+---------+
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| Column Name | Type |
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+---------------+---------+
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| user_id | int |
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| name | varchar |
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+---------------+---------+
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user_id is the primary key for this table.
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</pre>
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<p>Table: <code>Movie_Rating</code></p>
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<pre>
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+---------------+---------+
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| Column Name | Type |
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+---------------+---------+
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| movie_id | int |
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| user_id | int |
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| rating | int |
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| created_at | date |
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+---------------+---------+
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(movie_id, user_id) is the primary key for this table.
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This table contains the rating of a movie by a user in their review.
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created_at is the user's review date.
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</pre>
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Write the following SQL query:
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- Find the name of the user who has rated the greatest number of the movies.
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In case of a tie, return lexicographically smaller user name.
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- Find the movie name with the highest average rating as of Feb 2020.
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In case of a tie, return lexicographically smaller movie name..
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Query is returned in 2 rows, the query result format is in the folowing example:
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<pre>
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Movie table:
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+-------------+--------------+
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| movie_id | title |
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+-------------+--------------+
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| 1 | Avengers |
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| 2 | Frozen 2 |
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| 3 | Joker |
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+-------------+--------------+
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Users table:
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+-------------+--------------+
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| user_id | name |
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+-------------+--------------+
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| 1 | Daniel |
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| 2 | Monica |
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| 3 | Maria |
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| 4 | James |
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+-------------+--------------+
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Movie_Rating table:
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+-------------+--------------+--------------+-------------+
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| movie_id | user_id | rating | created_at |
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+-------------+--------------+--------------+-------------+
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| 1 | 1 | 3 | 2020年01月12日 |
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| 1 | 2 | 4 | 2020年02月11日 |
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| 1 | 3 | 2 | 2020年02月12日 |
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| 1 | 4 | 1 | 2020年01月01日 |
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| 2 | 1 | 5 | 2020年02月17日 |
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| 2 | 2 | 2 | 2020年02月01日 |
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| 2 | 3 | 2 | 2020年03月01日 |
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| 3 | 1 | 3 | 2020年02月22日 |
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| 3 | 2 | 4 | 2020年02月25日 |
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+-------------+--------------+--------------+-------------+
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Result table:
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+--------------+
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| results |
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+--------------+
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| Daniel |
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| Frozen 2 |
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+--------------+
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Daniel and Maria have rated 3 movies ("Avengers", "Frozen 2" and "Joker") but Daniel is smaller lexicographically.
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Frozen 2 and Joker have a rating average of 3.5 in February but Frozen 2 is smaller lexicographically.
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</pre>

‎problems/students-with-invalid-departments/README.md‎

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## [1350. Students With Invalid Departments (Easy)](https://leetcode.com/problems/students-with-invalid-departments "院系无效的学生")
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<p>Table: <code>Departments</code></p>
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<pre>
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+---------------+---------+
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| Column Name | Type |
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+---------------+---------+
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| id | int |
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| name | varchar |
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+---------------+---------+
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id is the primary key of this table.
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The table has information about the id of each department of a university.
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</pre>
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<p>Table: <code>Students</code></p>
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<pre>
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+---------------+---------+
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| Column Name | Type |
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+---------------+---------+
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| id | int |
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| name | varchar |
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| department_id | int |
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+---------------+---------+
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id is the primary key of this table.
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The table has information about the id of each student at a university and the id of the department he/she studies at.
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</pre>
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Write an SQL query to find the id and the name of all students who are enrolled in departments that no longer exists.
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Return the result table in any order.
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The query result format is in the following example:
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<pre>
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Departments table:
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+------+--------------------------+
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| id | name |
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+------+--------------------------+
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| 1 | Electrical Engineering |
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| 7 | Computer Engineering |
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| 13 | Bussiness Administration |
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+------+--------------------------+
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Students table:
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+------+----------+---------------+
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| id | name | department_id |
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+------+----------+---------------+
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| 23 | Alice | 1 |
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| 1 | Bob | 7 |
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| 5 | Jennifer | 13 |
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| 2 | John | 14 |
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| 4 | Jasmine | 77 |
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| 3 | Steve | 74 |
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| 6 | Luis | 1 |
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| 8 | Jonathan | 7 |
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| 7 | Daiana | 33 |
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| 11 | Madelynn | 1 |
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+------+----------+---------------+
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Result table:
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+------+----------+
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| id | name |
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+------+----------+
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| 2 | John |
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| 7 | Daiana |
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| 4 | Jasmine |
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| 3 | Steve |
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+------+----------+
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John, Daiana, Steve and Jasmine are enrolled in departments 14, 33, 74 and 77 respectively. department 14, 33, 74 and 77 doesn't exist in the Departments table.
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</pre>

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