feat: update sql solutions to lc problems (doocs#1793)

* No.0512.Game Play Analysis II
* No.1303.Find the Team Size
* No.1350.Students With Invalid Departments
* No.1747.Leetflex Banned Accounts
* No.1783.Grand Slam Titles

Author: yanglbme

solution/0500-0599/0512.Game Play Analysis II/README.md

@@ -79,7 +79,7 @@ WHERE
             player_id,
             min(event_date) AS event_date
         FROM Activity
-        GROUP BY player_id
+        GROUP BY 1
     );
 ```
 

solution/0500-0599/0512.Game Play Analysis II/README_EN.md

@@ -55,6 +55,14 @@ Activity table:
 
 ## Solutions
 
+**Solution 1: Subquery**
+
+We can use `GROUP BY` and `MIN` functions to find the first login date for each player, and then use a subquery with a composite key to find the first login device for each player.
+
+**Solution 2: Window Function**
+
+We can use the window function `rank()`, which assigns a rank to each login date for each player, and then select the rows with a rank of $1$.
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -71,7 +79,7 @@ WHERE
             player_id,
             min(event_date) AS event_date
         FROM Activity
-        GROUP BY player_id
+        GROUP BY 1
     );
 ```
 

solution/0500-0599/0512.Game Play Analysis II/Solution.sql

@@ -9,5 +9,5 @@ WHERE
             player_id,
             min(event_date) AS event_date
         FROM Activity
-        GROUP BY player_id
+        GROUP BY 1
     );

solution/1300-1399/1303.Find the Team Size/README.md

@@ -65,40 +65,39 @@ ID 为 5、6 的员工是 team_id 为 9 的团队的成员。
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：分组统计 + 等值连接**
+
+我们可以先统计出每个团队的人数，记录在 `T` 表中，然后我们将 `Employee` 表与 `T` 表按照 `team_id` 进行等值连接，即可得到每个员工所在团队的总人数。
+
+**方法二：左连接**
+
+我们也可以使用左连接，将 `Employee` 表按照 `team_id` 进行自连接，然后按照 `employee_id` 进行分组，统计每个员工所在团队的总人数。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
-解法 1：
-
 ```sql
 # Write your MySQL query statement below
-SELECT
-    e.employee_id,
-    t.team_size
-FROM
-    Employee AS e
-    LEFT JOIN (
-        SELECT
-            team_id,
-            count(1) AS team_size
+WITH
+    T AS (
+        SELECT team_id, count(1) AS team_size
         FROM Employee
-        GROUP BY team_id
-    ) AS t
-        ON e.team_id = t.team_id;
+        GROUP BY 1
+    )
+SELECT employee_id, team_size
+FROM
+    Employee
+    JOIN T USING (team_id);
 ```
 
-解法 2：
-
 ```sql
 # Write your MySQL query statement below
-SELECT
-    e1.employee_id,
-    count(*) AS team_size
+SELECT e1.employee_id, count(1) AS team_size
 FROM
     Employee AS e1
-    LEFT JOIN Employee AS e2 ON e1.team_id = e2.team_id
-GROUP BY e1.employee_id;
+    LEFT JOIN Employee AS e2 USING (team_id)
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

solution/1300-1399/1303.Find the Team Size/README_EN.md

@@ -60,40 +60,39 @@ Employees with Id 5,6 are part of a team with team_id = 9.
 
 ## Solutions
 
+**Solution 1: Group By + Equi-Join**
+
+We can first count the number of people in each team and record it in the `T` table. Then, we can use an equi-join to join the `Employee` table and the `T` table based on `team_id`, and obtain the total number of people in each team.
+
+**Solution 2: Left Join**
+
+We can also use a left join to join the `Employee` table with itself based on `team_id`, and then group by `employee_id` to count the total number of people in each team that the employee belongs to.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
-Solution 1:
-
 ```sql
 # Write your MySQL query statement below
-SELECT
-    e.employee_id,
-    t.team_size
-FROM
-    Employee AS e
-    LEFT JOIN (
-        SELECT
-            team_id,
-            count(1) AS team_size
+WITH
+    T AS (
+        SELECT team_id, count(1) AS team_size
         FROM Employee
-        GROUP BY team_id
-    ) AS t
-        ON e.team_id = t.team_id;
+        GROUP BY 1
+    )
+SELECT employee_id, team_size
+FROM
+    Employee
+    JOIN T USING (team_id);
 ```
 
-Solution 2:
-
 ```sql
 # Write your MySQL query statement below
-SELECT
-    e1.employee_id,
-    count(*) AS team_size
+SELECT e1.employee_id, count(1) AS team_size
 FROM
     Employee AS e1
-    LEFT JOIN Employee AS e2 ON e1.team_id = e2.team_id
-GROUP BY e1.employee_id;
+    LEFT JOIN Employee AS e2 USING (team_id)
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

solution/1300-1399/1303.Find the Team Size/Solution.sql

@@ -1,8 +1,6 @@
 # Write your MySQL query statement below
-SELECT
-    e1.employee_id,
-    count(*) AS team_size
+SELECT e1.employee_id, count(1) AS team_size
 FROM
     Employee AS e1
-    LEFT JOIN Employee AS e2 ON e1.team_id = e2.team_id
-GROUP BY e1.employee_id;
+    LEFT JOIN Employee AS e2 USING (team_id)
+GROUP BY 1;

solution/1300-1399/1350.Students With Invalid Departments/README.md

@@ -88,14 +88,25 @@ John, Daiana, Steve 和 Jasmine 所在的院系分别是 14, 33, 74 和 77， 
 
 <!-- 这里可写通用的实现逻辑 -->
 
-**方法一：左连接**
+**方法一：子查询**
 
-我们将 `Students` 表左连接 `Departments` 表，然后筛选出所有 `Departments` 表中 `id` 为 `NULL` 的记录即可。
+我们直接使用子查询的方式，找出所有不在院系表中的学生即可。
+
+**方法二：左连接**
+
+我们也可以使用左连接，将 `Students` 表和 `Departments` 连接，连接条件为 `Students.department_id = Departments.id`，然后筛选出 `Departments.id` 为空的学生即可。
 
 <!-- tabs:start -->
 
 ### **SQL**
 
+```sql
+# Write your MySQL query statement below
+SELECT id, name
+FROM Students
+WHERE department_id NOT IN (SELECT id FROM Departments);
+```
+
 ```sql
 # Write your MySQL query statement below
 SELECT s.id, s.name

solution/1300-1399/1350.Students With Invalid Departments/README_EN.md

@@ -84,10 +84,25 @@ John, Daiana, Steve, and Jasmine are enrolled in departments 14, 33, 74, and 77
 
 ## Solutions
 
+**Solution 1: Subquery**
+
+We can directly use a subquery to find all students who are not in the `Departments` table.
+
+**Solution 2: Left Join**
+
+We can also use a left join to join the `Students` table with the `Departments` table on the condition of `Students.department_id = Departments.id`, and then filter out the students whose `Departments.id` is `NULL`.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
+```sql
+# Write your MySQL query statement below
+SELECT id, name
+FROM Students
+WHERE department_id NOT IN (SELECT id FROM Departments);
+```
+
 ```sql
 # Write your MySQL query statement below
 SELECT s.id, s.name

solution/1700-1799/1747.Leetflex Banned Accounts/README.md

@@ -66,6 +66,14 @@ Account ID 4 --> 该账户从 "2021-02-01 17:00:00" 到 "2021-02-01 17:00:00"
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：自连接**
+
+我们可以通过自连接的方式，找出每个账户在同一天内，从不同的网络地址登录的情况。连接的条件是：
+
+-   账户编号相同
+-   网络地址不同
+-   一次登录的时间在另一次“登录-退出”的时间范围内
+
 <!-- tabs:start -->
 
 ### **SQL**

solution/1700-1799/1747.Leetflex Banned Accounts/README_EN.md

@@ -62,6 +62,14 @@ Account ID 4 --> The account was active from "2021-02-01 17:00:00"
 
 ## Solutions
 
+**Solution 1: Self-Join**
+
+We can use a self-join to find out the cases where each account logs in from different IP addresses on the same day. The conditions for joining are:
+
+-   The account numbers are the same.
+-   The IP addresses are different.
+-   The login time of one record is within the login-logout time range of another record.
+
 <!-- tabs:start -->
 
 ### **SQL**