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yanglbmeyanglbme
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feat: add solutions to lc problems
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+150
-6
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12 files changed

+150
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‎solution/1000-1099/1075.Project Employees I/README.md

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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 project_id
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,round(AVG(experience_years),2) AS average_years
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FROM Project p
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LEFT JOIN Employee e
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ON p.employee_id = e.employee_id
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GROUP BY project_id
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```
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<!-- tabs:end -->

‎solution/1000-1099/1075.Project Employees I/README_EN.md

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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 project_id
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,round(AVG(experience_years),2) AS average_years
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FROM Project p
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LEFT JOIN Employee e
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ON p.employee_id = e.employee_id
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GROUP BY project_id
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```
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<!-- tabs:end -->

‎solution/1000-1099/1075.Project Employees I/Solution.sql

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# Write your MySQL query statement below
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SELECT project_id
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,round(AVG(experience_years),2) AS average_years
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FROM Project p
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LEFT JOIN Employee e
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ON p.employee_id = e.employee_id
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GROUP BY project_id

‎solution/1000-1099/1076.Project Employees II/README.md

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@@ -73,7 +73,27 @@ Result table:
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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 project_id
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FROM Project p
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GROUP BY project_id
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HAVING COUNT(employee_id) >= all(
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SELECT COUNT(employee_id)
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FROM Project
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GROUP BY project_id )
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```
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```sql
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# Write your MySQL query statement below
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SELECT project_id
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FROM
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(
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SELECT project_id
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,dense_rank() over(order by COUNT(employee_id) desc) AS rk
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FROM Project
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GROUP BY project_id
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) t
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WHERE rk = 1
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```
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<!-- tabs:end -->

‎solution/1000-1099/1076.Project Employees II/README_EN.md

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@@ -82,7 +82,27 @@ Employee table:
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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 project_id
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FROM Project p
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GROUP BY project_id
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HAVING COUNT(employee_id) >= all(
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SELECT COUNT(employee_id)
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FROM Project
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GROUP BY project_id )
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```
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```sql
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# Write your MySQL query statement below
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SELECT project_id
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FROM
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(
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SELECT project_id
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,dense_rank() over(order by COUNT(employee_id) desc) AS rk
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FROM Project
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GROUP BY project_id
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) t
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WHERE rk = 1
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```
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<!-- tabs:end -->

‎solution/1000-1099/1076.Project Employees II/Solution.sql

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# Write your MySQL query statement below
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SELECT project_id
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FROM Project p
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GROUP BY project_id
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HAVING COUNT(employee_id) >= all(
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SELECT COUNT(employee_id)
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FROM Project
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GROUP BY project_id )

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

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@@ -79,7 +79,19 @@ employee_id 为 1 和 3 的员工在 project_id 为 1 的项目中拥有最丰
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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 project_id
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,employee_id
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FROM
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(
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SELECT p.project_id
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,p.employee_id
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,rank() over(partition by p.project_id ORDER BY e.experience_years desc) AS rk
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FROM Project p
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LEFT JOIN Employee e
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ON p.employee_id = e.employee_id
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) t
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WHERE rk = 1
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```
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<!-- tabs:end -->

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

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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 project_id
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,employee_id
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FROM
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(
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SELECT p.project_id
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,p.employee_id
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,rank() over(partition by p.project_id ORDER BY e.experience_years desc) AS rk
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FROM Project p
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LEFT JOIN Employee e
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ON p.employee_id = e.employee_id
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) t
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WHERE rk = 1
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```
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<!-- tabs:end -->

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

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# Write your MySQL query statement below
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SELECT project_id
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,employee_id
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FROM
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(
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SELECT p.project_id
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,p.employee_id
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,rank() over(partition by p.project_id ORDER BY e.experience_years desc) AS rk
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FROM Project p
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LEFT JOIN Employee e
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ON p.employee_id = e.employee_id
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) t
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WHERE rk = 1

‎solution/1100-1199/1126.Active Businesses/README.md

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@@ -60,7 +60,20 @@ id 为 1 的业务有 7 个 &#39;reviews&#39; 事件（大于 5）和 11 个 &#3
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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 business_id
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FROM EVENTS AS t1
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JOIN
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(
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SELECT event_type
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,AVG( occurences ) AS occurences
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FROM EVENTS
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GROUP BY event_type
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) AS t2
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ON t1.event_type = t2.event_type
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WHERE t1.occurences > t2.occurences
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GROUP BY business_id
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HAVING COUNT( 1 ) > 1
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```
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<!-- tabs:end -->

‎solution/1100-1199/1126.Active Businesses/README_EN.md

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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 business_id
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FROM EVENTS AS t1
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JOIN
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(
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SELECT event_type
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,AVG( occurences ) AS occurences
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FROM EVENTS
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GROUP BY event_type
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) AS t2
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ON t1.event_type = t2.event_type
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WHERE t1.occurences > t2.occurences
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GROUP BY business_id
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HAVING COUNT( 1 ) > 1
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```
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<!-- tabs:end -->

‎solution/1100-1199/1126.Active Businesses/Solution.sql

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# Write your MySQL query statement below
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SELECT business_id
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FROM EVENTS AS t1
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JOIN
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(
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SELECT event_type
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,AVG( occurences ) AS occurences
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FROM EVENTS
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GROUP BY event_type
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) AS t2
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ON t1.event_type = t2.event_type
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WHERE t1.occurences > t2.occurences
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GROUP BY business_id
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HAVING COUNT( 1 ) > 1

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