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Commit 0e4c841

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feat: update solutions to lc problems (doocs#1796)
* No.0184.Department Highest Salary * No.1077.Project Employees III * No.1488.Avoid Flood in The City * No.1532.The Most Recent Three Orders * No.1549.The Most Recent Orders for Each Product * No.1831.Maximum Transaction Each Day
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‎solution/0100-0199/0184.Department Highest Salary/README.md‎

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@@ -80,42 +80,29 @@ Department 表:
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:等值连接 + 子查询**
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我们可以使用等值连接,将 `Employee` 表和 `Department` 表连接起来,连接条件为 `Employee.departmentId = Department.id`,然后使用子查询来找到每个部门的最高工资,最后使用 `WHERE` 子句来筛选出每个部门中薪资最高的员工。
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**方法二:等值连接 + 窗口函数**
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我们可以使用等值连接,将 `Employee` 表和 `Department` 表连接起来,连接条件为 `Employee.departmentId = Department.id`,然后使用窗口函数 `rank()`,它可以为每个部门的每个员工分配一个排名,然后我们可以选择排名为 1ドル$ 的行即可。
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<!-- tabs:start -->
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### **SQL**
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```sql
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SELECT
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Department.NAME AS Department,
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Employee.NAME AS Employee,
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Salary
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FROM
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Employee,
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Department
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WHERE
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Employee.DepartmentId = Department.Id
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AND (Employee.DepartmentId, Salary) IN (
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SELECT DepartmentId, max(Salary)
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FROM Employee
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GROUP BY DepartmentId
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);
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```
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```sql
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# Write your MySQL query statement below
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SELECT
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d.NAME AS Department,
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e1.NAME AS Employee,
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e1.salary AS Salary
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SELECT d.name AS department, e.name AS employee, salary
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FROM
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Employee AS e1
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JOIN Department AS d ON e1.departmentId = d.id
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Employee AS e
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JOIN Department AS d ON e.departmentId = d.id
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WHERE
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e1.salary = (
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SELECT
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MAX(Salary)
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FROM Employee AS e2
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WHERE e2.departmentId = d.id
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(d.id, salary) IN (
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SELECT departmentId, max(salary)
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FROM Employee
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GROUP BY 1
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);
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```
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WITH
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T AS (
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SELECT
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*,
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d.name AS department,
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e.name AS employee,
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salary,
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rank() OVER (
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PARTITION BY departmentId
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PARTITION BY d.name
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ORDER BY salary DESC
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) AS rk
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FROM Employee
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FROM
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Employee AS e
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JOIN Department AS d ON e.departmentId = d.id
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)
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SELECT d.name AS Department, t.name AS Employee, salary AS Salary
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FROM
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T AS t
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JOIN Department AS d ON t.departmentId = d.id
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SELECT department, employee, salary
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FROM T
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WHERE rk = 1;
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```
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‎solution/0100-0199/0184.Department Highest Salary/README_EN.md‎

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## Solutions
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**Solution 1: Equi-Join + Subquery**
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We can use an equi-join to join the `Employee` table and the `Department` table based on `Employee.departmentId = Department.id`, and then use a subquery to find the highest salary for each department. Finally, we can use a `WHERE` clause to filter out the employees with the highest salary in each department.
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**Solution 2: Equi-Join + Window Function**
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We can use an equi-join to join the `Employee` table and the `Department` table based on `Employee.departmentId = Department.id`, and then use the window function `rank()`, which assigns a rank to each employee in each department based on their salary. Finally, we can select the rows with a rank of 1ドル$ for each department.
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<!-- tabs:start -->
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### **SQL**
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```sql
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SELECT
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Department.NAME AS Department,
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Employee.NAME AS Employee,
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Salary
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FROM
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Employee,
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Department
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WHERE
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Employee.DepartmentId = Department.Id
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AND (Employee.DepartmentId, Salary) IN (
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SELECT DepartmentId, max(Salary)
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FROM Employee
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GROUP BY DepartmentId
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);
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```
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```sql
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# Write your MySQL query statement below
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SELECT
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d.NAME AS Department,
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e1.NAME AS Employee,
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e1.salary AS Salary
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SELECT d.name AS department, e.name AS employee, salary
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FROM
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Employee AS e1
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JOIN Department AS d ON e1.departmentId = d.id
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Employee AS e
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JOIN Department AS d ON e.departmentId = d.id
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WHERE
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e1.salary = (
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SELECT
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MAX(Salary)
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FROM Employee AS e2
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WHERE e2.departmentId = d.id
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(d.id, salary) IN (
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SELECT departmentId, max(salary)
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FROM Employee
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GROUP BY 1
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);
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```
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WITH
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T AS (
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SELECT
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*,
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d.name AS department,
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e.name AS employee,
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salary,
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rank() OVER (
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PARTITION BY departmentId
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PARTITION BY d.name
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ORDER BY salary DESC
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) AS rk
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FROM Employee
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FROM
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Employee AS e
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JOIN Department AS d ON e.departmentId = d.id
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)
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SELECT d.name AS Department, t.name AS Employee, salary AS Salary
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FROM
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T AS t
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JOIN Department AS d ON t.departmentId = d.id
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SELECT department, employee, salary
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FROM T
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WHERE rk = 1;
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```
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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
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*,
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rank() OVER (
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PARTITION BY departmentId
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ORDER BY salary DESC
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) AS rk
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FROM Employee
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)
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SELECT d.name AS Department, t.name AS Employee, salary AS Salary
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SELECT d.name AS department, e.name AS employee, salary
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FROM
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T AS t
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JOIN Department AS d ON t.departmentId = d.id
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WHERE rk = 1;
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Employee AS e
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JOIN Department AS d ON e.departmentId = d.id
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WHERE
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(d.id, salary) IN (
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SELECT departmentId, max(salary)
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FROM Employee
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GROUP BY 1
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);

‎solution/1000-1099/1077.Project Employees III/README.md‎

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WITH
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T AS (
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SELECT
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project_id,
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employee_id,
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*,
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rank() OVER (
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PARTITION BY project_id
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ORDER BY experience_years DESC

‎solution/1000-1099/1077.Project Employees III/README_EN.md‎

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@@ -79,6 +79,10 @@ Employee table:
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## Solutions
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**Solution 1: Inner Join + Window Function**
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We can first perform an inner join between the `Project` table and the `Employee` table, and then use the window function `rank()` to group the `Project` table, sort it in descending order by `experience_years`, and finally select the most experienced employee for each project.
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<!-- tabs:start -->
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### **SQL**
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WITH
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T AS (
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SELECT
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project_id,
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employee_id,
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*,
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rank() OVER (
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PARTITION BY project_id
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ORDER BY experience_years DESC

‎solution/1000-1099/1077.Project Employees III/Solution.sql‎

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WITH
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T AS (
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SELECT
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project_id,
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employee_id,
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*,
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rank() OVER (
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PARTITION BY project_id
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ORDER BY experience_years DESC

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