Added tasks 1327, 1341, 1378, 1517, 1633

Author: ThanhNIT

src/main/java/g1301_1400/s1327_list_the_products_ordered_in_a_period/readme.md

@@ -0,0 +1,85 @@
+1327\. List the Products Ordered in a Period
+
+Easy
+
+SQL Schema
+
+Table: `Products`
+
+    +------------------+---------+ 
+    | Column Name      | Type    | 
+    +------------------+---------+ 
+    | product_id       | int     | 
+    | product_name     | varchar | 
+    | product_category | varchar | 
+    +------------------+---------+ 
+    
+product_id is the primary key for this table. This table contains data about the company's products.
+
+Table: `Orders`
+
+    +---------------+---------+ 
+    | Column Name   | Type    | 
+    +---------------+---------+ 
+    | product_id    | int     | 
+    | order_date    | date    | 
+    | unit          | int     | 
+    +---------------+---------+ 
+
+There is no primary key for this table. It may have duplicate rows. product_id is a foreign key to the Products table. unit is the number of products ordered in order_date.
+
+Write an SQL query to get the names of products that have at least `100` units ordered in **February 2020** and their amount.
+
+Return result table in **any order**.
+
+The query result format is in the following example.
+
+**Example 1:**
+
+**Input:** Products table: 
+
+    +-------------+-----------------------+------------------+ 
+    | product_id  | product_name          | product_category | 
+    +-------------+-----------------------+------------------+ 
+    | 1           | Leetcode Solutions    | Book             | 
+    | 2           | Jewels of Stringology | Book             | 
+    | 3           | HP                    | Laptop           | 
+    | 4           | Lenovo                | Laptop           | 
+    | 5           | Leetcode Kit          | T-shirt          | 
+    +-------------+-----------------------+------------------+ 
+
+Orders table: 
+
+    +--------------+--------------+----------+ 
+    | product_id   | order_date   | unit     | 
+    +--------------+--------------+----------+ 
+    | 1            | 2020-02-05   | 60       | 
+    | 1            | 2020-02-10   | 70       | 
+    | 2            | 2020-01-18   | 30       | 
+    | 2            | 2020-02-11   | 80       | 
+    | 3            | 2020-02-17   | 2        | 
+    | 3            | 2020-02-24   | 3        | 
+    | 4            | 2020-03-01   | 20       | 
+    | 4            | 2020-03-04   | 30       | 
+    | 4            | 2020-03-04   | 60       | 
+    | 5            | 2020-02-25   | 50       | 
+    | 5            | 2020-02-27   | 50       | 
+    | 5            | 2020-03-01   | 50       | 
+    +--------------+--------------+----------+
+
+**Output:** 
+
+    +--------------------+---------+ 
+    | product_name       | unit    | 
+    +--------------------+---------+ 
+    | Leetcode Solutions | 130     | 
+    | Leetcode Kit       | 100     | 
+    +--------------------+---------+
+
+**Explanation:**
+
+Products with product_id = 1 is ordered in February a total of (60 + 70) = 130. 
+
+Products with product_id = 2 is ordered in February a total of 80. Products with product_id = 3 is ordered in February a total of (2 + 3) = 5. 
+
+Products with product_id = 4 was not ordered in February 2020. Products with product_id = 5 is ordered in February a total of (50 + 50) = 100. 

src/main/java/g1301_1400/s1327_list_the_products_ordered_in_a_period/script.sql

@@ -0,0 +1,14 @@
+# Write your MySQL query statement below
+# #Easy #Database #2023_08_16_Time_1374_ms_(80.18%)_Space_0B_(100.00%)
+SELECT * FROM (
+  SELECT
+    a.product_name,
+    SUM(b.unit) as unit
+  FROM Products a
+  LEFT JOIN Orders b
+  ON a.product_id = b.product_id
+  WHERE b.order_date BETWEEN '2020-02-01' AND '2020-02-29'
+  GROUP BY a.product_name
+) AS d
+GROUP BY d.product_name
+HAVING d.unit >= 100

src/main/java/g1301_1400/s1341_movie_rating/readme.md

@@ -0,0 +1,101 @@
+1341\. Movie Rating
+
+Medium
+
+SQL Schema
+
+Table: `Movies`
+
+    +---------------+---------+ 
+    | Column Name   | Type    | 
+    +---------------+---------+ 
+    | movie_id      | int     | 
+    | title         | varchar | 
+    +---------------+---------+ 
+
+movie_id is the primary key 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 for this table.
+
+Table: `MovieRating`
+
+    +---------------+---------+ 
+    | Column Name   | Type    | 
+    +---------------+---------+ 
+    | movie_id      | int     | 
+    | user_id       | int     | 
+    | rating        | int     | 
+    | created_at    | date    | 
+    +---------------+---------+ 
+
+(movie_id, user_id) is the primary key 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 an SQL query 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 query 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. 

src/main/java/g1301_1400/s1341_movie_rating/script.sql

@@ -0,0 +1,13 @@
+# Write your MySQL query statement below
+# #Medium #Database #2023_08_16_Time_2515_ms_(78.94%)_Space_0B_(100.00%)
+(SELECT name results
+FROM Users as U, MovieRating as  MR
+WHERE U.user_id = MR.user_id
+GROUP BY U.user_id
+ORDER BY COUNT(MR.user_id) DESC, name ASC LIMIT 1)
+UNION ALL
+(SELECT title results
+FROM Movies as M, MovieRating as MR
+WHERE M.movie_id = MR.movie_id AND created_at BETWEEN '2020-02-01' AND '2020-02-29'
+GROUP BY M.movie_id
+ORDER BY AVG(rating) DESC, title ASC LIMIT 1)

src/main/java/g1301_1400/s1378_replace_employee_id_with_the_unique_identifier/readme.md

@@ -0,0 +1,79 @@
+1378\. Replace Employee ID With The Unique Identifier
+
+Easy
+
+SQL Schema
+
+Table: `Employees`
+
+    +---------------+---------+
+    | Column Name   | Type    |
+    +---------------+---------+
+    | id            | int     |
+    | name          | varchar |
+    +---------------+---------+
+    id is the primary key for this table.
+    Each row of this table contains the id and the name of an employee in a company. 
+
+Table: `EmployeeUNI`
+
+    +---------------+---------+
+    | Column Name   | Type    |
+    +---------------+---------+
+    | id            | int     |
+    | unique_id     | int     |
+    +---------------+---------+ 
+    (id, unique_id) is the primary key for this table.
+    Each row of this table contains the id and the corresponding unique id of an employee in the company. 
+
+Write an SQL query to show the **unique ID** of each user, If a user does not have a unique ID replace just show `null`.
+
+Return the result table in **any** order.
+
+The query result format is in the following example.
+
+**Example 1:**
+
+**Input:**,
+
+    Employees table:
+    +----+----------+
+    | id | name     |
+    +----+----------+
+    | 1  | Alice    |
+    | 7  | Bob      |
+    | 11 | Meir     |
+    | 90 | Winston  |
+    | 3  | Jonathan |
+    +----+----------+
+    
+    EmployeeUNI table:
+    +----+-----------+
+    | id | unique_id |
+    +----+-----------+
+    | 3  | 1         |
+    | 11 | 2         |
+    | 90 | 3         |
+    +----+-----------+
+
+**Output:**
+
+    +-----------+----------+
+    | unique_id | name     |
+    +-----------+----------+
+    | null      | Alice    |
+    | null      | Bob      |
+    | 2         | Meir     |
+    | 3         | Winston  |
+    | 1         | Jonathan |
+    +-----------+----------+
+
+**Explanation:**
+
+Alice and Bob do not have a unique ID, We will show null instead.
+
+The unique ID of Meir is 2.
+
+The unique ID of Winston is 3.
+
+The unique ID of Jonathan is 1.

src/main/java/g1301_1400/s1378_replace_employee_id_with_the_unique_identifier/script.sql

@@ -0,0 +1,6 @@
+# Write your MySQL query statement below
+# #Easy #Database #2023_08_16_Time_2498_ms_(63.60%)_Space_0B_(100.00%)
+select u.unique_id, e.name
+from Employees e
+left join EmployeeUNI u
+on e.id = u.id;

src/main/java/g1501_1600/s1517_find_users_with_valid_e_mails/readme.md

@@ -0,0 +1,65 @@
+1517\. Find Users With Valid E-Mails
+
+Easy
+
+SQL Schema
+
+Table: `Users`
+
+    +---------------+---------+
+    | Column Name   | Type    |
+    +---------------+---------+
+    | user_id       | int     |
+    | name          | varchar |
+    | mail          | varchar |
+    +---------------+---------+
+    user_id is the primary key for this table.
+    This table contains information of the users signed up in a website. Some e-mails are invalid. 
+
+Write an SQL query to find the users who have **valid emails**.
+
+A valid e-mail has a prefix name and a domain where:
+
+*   **The prefix name** is a string that may contain letters (upper or lower case), digits, underscore `'_'`, period `'.'`, and/or dash `'-'`. The prefix name **must** start with a letter.
+*   **The domain** is `'@leetcode.com'`.
+
+Return the result table in **any order**.
+
+The query result format is in the following example.
+
+**Example 1:**
+
+**Input:**
+
+    Users table:
+    +---------+-----------+-------------------------+
+    | user_id | name      | mail                    |
+    +---------+-----------+-------------------------+
+    | 1       | Winston   | winston@leetcode.com    |
+    | 2       | Jonathan  | jonathanisgreat         |
+    | 3       | Annabelle | bella-@leetcode.com     |
+    | 4       | Sally     | sally.come@leetcode.com |
+    | 5       | Marwan    | quarz#2020@leetcode.com |
+    | 6       | David     | david69@gmail.com       |
+    | 7       | Shapiro   | .shapo@leetcode.com     |
+    +---------+-----------+-------------------------+
+
+**Output:**
+
+    +---------+-----------+-------------------------+
+    | user_id | name      | mail                    |
+    +---------+-----------+-------------------------+
+    | 1       | Winston   | winston@leetcode.com    |
+    | 3       | Annabelle | bella-@leetcode.com     |
+    | 4       | Sally     | sally.come@leetcode.com |
+    +---------+-----------+-------------------------+
+
+**Explanation:**
+
+The mail of user 2 does not have a domain.
+
+The mail of user 5 has the # sign which is not allowed.
+
+The mail of user 6 does not have the leetcode domain.
+
+The mail of user 7 starts with a period.

src/main/java/g1501_1600/s1517_find_users_with_valid_e_mails/script.sql

@@ -0,0 +1,5 @@
+# Write your MySQL query statement below
+# #Easy #Database #2023_08_16_Time_1356_ms_(75.95%)_Space_0B_(100.00%)
+SELECT *
+FROM Users
+WHERE mail REGEXP '^[a-zA-Z][a-zA-Z0-9_.-]*@leetcode[.]com'