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Commit 5591807

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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
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13 files changed

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

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@@ -79,7 +79,7 @@ WHERE
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player_id,
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min(event_date) AS event_date
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FROM Activity
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GROUP BY player_id
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GROUP BY 1
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);
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```
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‎solution/0500-0599/0512.Game Play Analysis II/README_EN.md‎

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@@ -55,6 +55,14 @@ Activity table:
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## Solutions
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**Solution 1: Subquery**
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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.
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**Solution 2: Window Function**
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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ドル$.
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<!-- tabs:start -->
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### **SQL**
@@ -71,7 +79,7 @@ WHERE
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player_id,
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min(event_date) AS event_date
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FROM Activity
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GROUP BY player_id
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GROUP BY 1
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);
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```
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‎solution/0500-0599/0512.Game Play Analysis II/Solution.sql‎

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

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

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@@ -65,40 +65,39 @@ ID 为 5、6 的员工是 team_id 为 9 的团队的成员。
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:分组统计 + 等值连接**
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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。
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**方法二:左连接**
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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。
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<!-- tabs:start -->
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### **SQL**
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解法 1:
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```sql
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# Write your MySQL query statement below
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SELECT
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e.employee_id,
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t.team_size
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FROM
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Employee AS e
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LEFT JOIN (
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SELECT
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team_id,
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count(1) AS team_size
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WITH
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T AS (
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SELECT team_id, count(1) AS team_size
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FROM Employee
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GROUP BY team_id
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) AS t
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ON e.team_id = t.team_id;
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GROUP BY 1
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)
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SELECT employee_id, team_size
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FROM
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Employee
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JOIN T USING (team_id);
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```
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解法 2:
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```sql
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# Write your MySQL query statement below
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SELECT
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e1.employee_id,
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count(*) AS team_size
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SELECT e1.employee_id, count(1) AS team_size
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FROM
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Employee AS e1
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LEFT JOIN Employee AS e2 ONe1.team_id=e2.team_id
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GROUP BY e1.employee_id;
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LEFT JOIN Employee AS e2 USING (team_id)
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GROUP BY 1;
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```
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<!-- tabs:end -->

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

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@@ -60,40 +60,39 @@ Employees with Id 5,6 are part of a team with team_id = 9.
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## Solutions
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**Solution 1: Group By + Equi-Join**
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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.
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**Solution 2: Left Join**
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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.
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<!-- tabs:start -->
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### **SQL**
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Solution 1:
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```sql
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# Write your MySQL query statement below
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SELECT
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e.employee_id,
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t.team_size
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FROM
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Employee AS e
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LEFT JOIN (
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SELECT
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team_id,
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count(1) AS team_size
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WITH
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T AS (
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SELECT team_id, count(1) AS team_size
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FROM Employee
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GROUP BY team_id
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) AS t
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ON e.team_id = t.team_id;
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GROUP BY 1
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)
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SELECT employee_id, team_size
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FROM
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Employee
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JOIN T USING (team_id);
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```
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Solution 2:
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```sql
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# Write your MySQL query statement below
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SELECT
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e1.employee_id,
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count(*) AS team_size
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SELECT e1.employee_id, count(1) AS team_size
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FROM
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Employee AS e1
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LEFT JOIN Employee AS e2 ONe1.team_id=e2.team_id
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GROUP BY e1.employee_id;
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LEFT JOIN Employee AS e2 USING (team_id)
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GROUP BY 1;
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```
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<!-- tabs:end -->
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# Write your MySQL query statement below
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SELECT
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e1.employee_id,
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count(*) AS team_size
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SELECT e1.employee_id, count(1) AS team_size
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FROM
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Employee AS e1
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LEFT JOIN Employee AS e2 ONe1.team_id=e2.team_id
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GROUP BY e1.employee_id;
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LEFT JOIN Employee AS e2 USING (team_id)
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GROUP BY 1;

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

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@@ -88,14 +88,25 @@ John, Daiana, Steve 和 Jasmine 所在的院系分别是 14, 33, 74 和 77,
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:左连接**
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**方法一:子查询**
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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。
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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。
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**方法二:左连接**
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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。
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<!-- tabs:start -->
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### **SQL**
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```sql
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# Write your MySQL query statement below
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SELECT id, name
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FROM Students
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WHERE department_id NOT IN (SELECT id FROM Departments);
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```
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```sql
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# Write your MySQL query statement below
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SELECT s.id, s.name

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

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@@ -84,10 +84,25 @@ John, Daiana, Steve, and Jasmine are enrolled in departments 14, 33, 74, and 77
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## Solutions
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**Solution 1: Subquery**
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We can directly use a subquery to find all students who are not in the `Departments` table.
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**Solution 2: Left Join**
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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`.
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### **SQL**
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```sql
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# Write your MySQL query statement below
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SELECT id, name
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FROM Students
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WHERE department_id NOT IN (SELECT id FROM Departments);
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```
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```sql
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# Write your MySQL query statement below
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SELECT s.id, s.name

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

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@@ -66,6 +66,14 @@ Account ID 4 --&gt; 该账户从 "2021年02月01日 17:00:00" 到 "2021年02月01日 17:00:00"
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:自连接**
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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:
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- 账户编号相同
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- 网络地址不同
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- 一次登录的时间在另一次"登录-退出"的时间范围内
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### **SQL**

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

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@@ -62,6 +62,14 @@ Account ID 4 --&gt; The account was active from &quot;2021年02月01日 17:00:00&quot;
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## Solutions
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**Solution 1: Self-Join**
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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:
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- The account numbers are the same.
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- The IP addresses are different.
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- The login time of one record is within the login-logout time range of another record.
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### **SQL**

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