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Commit e83943b

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feat: update solutions to lc problems (doocs#1799)
* No.1225.Report Contiguous Dates * No.1270.All People Report to the Given Manager * No.1412.Find the Quiet Students in All Exams * No.1596.The Most Frequently Ordered Products for Each Customer * No.1709.Biggest Window Between Visits * No.1767.Find the Subtasks That Did Not Execute
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‎.prettierignore‎

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/solution/1600-1699/1613.Find the Missing IDs/Solution.sql
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/solution/1600-1699/1635.Hopper Company Queries I/Solution.sql
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/solution/1600-1699/1651.Hopper Company Queries III/Solution.sql
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/solution/1700-1799/1767.Find the Subtasks That Did Not Execute/Solution.sql
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/solution/1800-1899/1873.Calculate Special Bonus/Solution.sql
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/solution/2100-2199/2118.Build the Equation/Solution.sql
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/solution/2100-2199/2153.The Number of Passengers in Each Bus II/Solution.sql

‎solution/1200-1299/1225.Report Contiguous Dates/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:合并 + 窗口函数 + 分组求最大最小值**
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我们可以将两个表合并,用一个字段 $st$ 表示状态,其中 `failed` 表示失败,`succeeded` 表示成功。然后我们可以使用窗口函数,将相同状态的记录分到一组,求出每个日期与其所在组排名的差值 $pt,ドル作为同一个连续状态的标识。最后我们可以按照 $st$ 和 $pt$ 分组,求出每组的最小日期和最大日期,然后按照最小日期排序即可。
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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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WITH
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T AS (
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SELECT fail_date AS dt, 'failed' AS st
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FROM Failed
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WHERE year(fail_date) = 2019
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UNION ALL
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SELECT success_date AS dt, 'succeeded' AS st
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FROM Succeeded
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WHERE year(success_date) = 2019
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)
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SELECT
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state AS period_state,
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st AS period_state,
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min(dt) AS start_date,
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max(dt) AS end_date
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FROM
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subdate(
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dt,
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rank() OVER (
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PARTITION BY state
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PARTITION BY st
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ORDER BY dt
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)
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) AS dif
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FROM
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(
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SELECT
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'failed' AS state,
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fail_date AS dt
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FROM failed
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WHERE fail_date BETWEEN '2019年01月01日' AND '2019年12月31日'
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UNION ALL
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SELECT
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'succeeded' AS state,
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success_date AS dt
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FROM succeeded
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WHERE success_date BETWEEN '2019年01月01日' AND '2019年12月31日'
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) AS t1
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) AS t2
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GROUP BY state, dif
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ORDER BY dt;
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) AS pt
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FROM T
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) AS t
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GROUP BY 1, pt
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ORDER BY 2;
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```
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<!-- tabs:end -->

‎solution/1200-1299/1225.Report Contiguous Dates/README_EN.md‎

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## Solutions
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**Solution 1: Union + Window Function + Group By**
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We can merge the two tables into one table with a field `st` representing the status, where `failed` indicates failure and `succeeded` indicates success. Then, we can use a window function to group the records with the same status into one group, and calculate the difference between each date and its rank within the group as `pt`, which serves as the identifier for the same continuous status. Finally, we can group by `st` and `pt`, and calculate the minimum and maximum dates for each group, and sort by the minimum date.
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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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WITH
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T AS (
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SELECT fail_date AS dt, 'failed' AS st
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FROM Failed
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WHERE year(fail_date) = 2019
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UNION ALL
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SELECT success_date AS dt, 'succeeded' AS st
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FROM Succeeded
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WHERE year(success_date) = 2019
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)
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SELECT
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state AS period_state,
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st AS period_state,
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min(dt) AS start_date,
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max(dt) AS end_date
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FROM
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subdate(
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dt,
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rank() OVER (
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PARTITION BY state
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PARTITION BY st
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ORDER BY dt
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)
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) AS dif
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FROM
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(
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SELECT
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'failed' AS state,
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fail_date AS dt
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FROM failed
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WHERE fail_date BETWEEN '2019年01月01日' AND '2019年12月31日'
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UNION ALL
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SELECT
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'succeeded' AS state,
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success_date AS dt
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FROM succeeded
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WHERE success_date BETWEEN '2019年01月01日' AND '2019年12月31日'
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) AS t1
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) AS t2
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GROUP BY state, dif
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ORDER BY dt;
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) AS pt
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FROM T
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) AS t
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GROUP BY 1, pt
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ORDER BY 2;
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```
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<!-- tabs:end -->
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# Write your MySQL query statement below
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WITH
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T AS (
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SELECT fail_date AS dt, 'failed' AS st
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FROM Failed
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WHERE year(fail_date) = 2019
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UNION ALL
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SELECT success_date AS dt, 'succeeded' AS st
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FROM Succeeded
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WHERE year(success_date) = 2019
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)
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SELECT
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state AS period_state,
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st AS period_state,
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min(dt) AS start_date,
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max(dt) AS end_date
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FROM
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subdate(
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dt,
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rank() OVER (
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PARTITION BY state
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PARTITION BY st
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ORDER BY dt
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)
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) AS dif
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FROM
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(
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SELECT
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'failed' AS state,
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fail_date AS dt
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FROM failed
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WHERE fail_date BETWEEN '2019年01月01日' AND '2019年12月31日'
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UNION ALL
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SELECT
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'succeeded' AS state,
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success_date AS dt
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FROM succeeded
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WHERE success_date BETWEEN '2019年01月01日' AND '2019年12月31日'
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) AS t1
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) AS t2
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GROUP BY state, dif
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ORDER BY dt;
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) AS pt
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FROM T
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) AS t
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GROUP BY 1, pt
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ORDER BY 2;

‎solution/1200-1299/1270.All People Report to the Given Manager/README.md‎

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@@ -71,6 +71,12 @@ employee_id 是 3, 8 ,9 的职员不会直接或间接的汇报给公司 CEO
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:两次连接**
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我们可以通过两次连接来找到所有直接或间接向公司 CEO 汇报工作的职工的 `employee_id`。
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具体地,我们首先通过一次连接,找到每个 `manager_id` 对应的上级经理的 `manager_id`,然后再通过一次连接,找到更上一级经理的 `manager_id`,最后,如果更上一级的 `manager_id` 为 1ドル,ドル且员工的 `employee_id` 不为 1ドル,ドル则说明该员工直接或间接向公司 CEO 汇报工作。
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<!-- tabs:start -->
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### **SQL**

‎solution/1200-1299/1270.All People Report to the Given Manager/README_EN.md‎

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## Solutions
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**Solution 1: Two Joins**
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We can use two joins to find all employees who report directly or indirectly to the company CEO.
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Specifically, we first use a join to find the `manager_id` of the superior manager for each `manager_id`, and then use another join to find the `manager_id` of the higher-level manager. Finally, if the `manager_id` of the higher-level manager is 1ドル$ and the `employee_id` of the employee is not 1ドル,ドル it means that the employee reports directly or indirectly to the company CEO.
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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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# Write your MySQL query statement below
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SELECT e1.employee_id
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FROM

‎solution/1400-1499/1412.Find the Quiet Students in All Exams/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:使用 RANK() 窗口函数 + 分组聚合**
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我们可以使用 `RANK()` 窗口函数来计算每个学生在每场考试中的正序排名 $rk1$ 和倒序排序 $rk2,ドル得到表 $T$。
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接下来,我们将表 $T$ 与表 $Student$ 进行内连接,然后按照学生编号进行分组聚合,得到每个学生在所有考试中的正序排名为 1ドル$ 的次数 $cnt1$ 和倒序排名为 1ドル$ 的次数 $cnt2$。如果 $cnt1$ 和 $cnt2$ 都为 0ドル,ドル则说明该学生在所有考试中都处于中游。
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### **SQL**
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```sql
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# Write your MySQL query statement below
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WITH
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t AS (
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T AS (
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SELECT
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*,
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student_id,
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rank() OVER (
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PARTITION BY exam_id
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ORDER BY scoreDESC
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ORDER BY score
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) AS rk1,
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rank() OVER (
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PARTITION BY exam_id
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ORDER BY score ASC
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ORDER BY score DESC
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) AS rk2
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FROM Exam
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)
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SELECT
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t.student_id,
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student_name
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SELECT student_id, student_name
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FROM
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t
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JOIN Student AS s ON t.student_id = s.student_id
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GROUP BY t.student_id
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HAVING sum(rk1 = 1) = 0 AND sum(rk2 = 1) = 0;
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T
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JOIN Student USING (student_id)
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GROUP BY 1
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HAVING sum(rk1 = 1) = 0 AND sum(rk2 = 1) = 0
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ORDER BY 1;
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```
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<!-- tabs:end -->

‎solution/1400-1499/1412.Find the Quiet Students in All Exams/README_EN.md‎

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## Solutions
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**Solution 1: Using RANK() Window Function + Group By**
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We can use the `RANK()` window function to calculate the ascending rank $rk1$ and descending rank $rk2$ of each student in each exam, and obtain the table $T$.
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Next, we can perform an inner join between the table $T$ and the table $Student,ドル and then group by student ID to obtain the number of times each student has a rank of 1ドル$ in ascending order $cnt1$ and descending order $cnt2$ in all exams. If both $cnt1$ and $cnt2$ are 0ドル,ドル it means that the student is in the middle of the pack in all exams.
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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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WITH
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t AS (
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T AS (
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SELECT
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*,
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student_id,
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rank() OVER (
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PARTITION BY exam_id
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ORDER BY scoreDESC
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ORDER BY score
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) AS rk1,
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rank() OVER (
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PARTITION BY exam_id
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ORDER BY score ASC
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ORDER BY score DESC
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) AS rk2
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FROM Exam
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)
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SELECT
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t.student_id,
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student_name
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SELECT student_id, student_name
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FROM
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t
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JOIN Student AS s ON t.student_id = s.student_id
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GROUP BY t.student_id
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HAVING sum(rk1 = 1) = 0 AND sum(rk2 = 1) = 0;
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T
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JOIN Student USING (student_id)
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GROUP BY 1
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HAVING sum(rk1 = 1) = 0 AND sum(rk2 = 1) = 0
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ORDER 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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WITH
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t AS (
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T AS (
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SELECT
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*,
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student_id,
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rank() OVER (
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PARTITION BY exam_id
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ORDER BY scoreDESC
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ORDER BY score
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) AS rk1,
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rank() OVER (
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PARTITION BY exam_id
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ORDER BY score ASC
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ORDER BY score DESC
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) AS rk2
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FROM Exam
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)
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SELECT
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t.student_id,
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student_name
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SELECT student_id, student_name
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FROM
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t
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JOIN Student AS s ON t.student_id = s.student_id
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GROUP BY t.student_id
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HAVING sum(rk1 = 1) = 0 AND sum(rk2 = 1) = 0;
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T
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JOIN Student USING (student_id)
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GROUP BY 1
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HAVING sum(rk1 = 1) = 0 AND sum(rk2 = 1) = 0
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ORDER BY 1;

‎solution/1500-1599/1596.The Most Frequently Ordered Products for Each Customer/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:分组 + 窗口函数**
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我们将 `Orders` 表按照 `customer_id` 和 `product_id` 进行分组,然后利用窗口函数 `rank()`,按照 `customer_id` 分区,并且按照 `count(1)` 降序排列,得到每个 `customer_id` 下对应的 `product_id` 的排名,排名为 1ドル$ 的就是该 `customer_id` 下最经常订购的商品。
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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
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customer_id,
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p.product_id,
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p.product_name
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FROM
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(
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WITH
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T AS (
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SELECT
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customer_id,
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product_id,
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ORDER BY count(1) DESC
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) AS rk
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GROUP BY customer_id, product_id
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) AS o
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JOIN Products AS p ON o.product_id = p.product_id
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GROUP BY 1, 2
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)
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SELECT customer_id, product_id, product_name
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FROM
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T
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JOIN Products USING (product_id)
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WHERE rk = 1;
153156
```
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