feat: add sql solutions to lc problems: No.1789,1809 (#1111)

* No.1789.Primary Department for Each Employee
* No.1809.Ad-Free Sessions

Author: yanglbme

solution/1700-1799/1789.Primary Department for Each Employee/README.md

@@ -77,7 +77,15 @@ Result table:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT employee_id, department_id
+FROM Employee
+WHERE primary_flag = 'Y'
+UNION
+SELECT employee_id, department_id
+FROM Employee
+GROUP BY employee_id
+HAVING count(1) = 1;
 ```
 
 <!-- tabs:end -->

solution/1700-1799/1789.Primary Department for Each Employee/README_EN.md

@@ -70,7 +70,15 @@ Employee table:
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT employee_id, department_id
+FROM Employee
+WHERE primary_flag = 'Y'
+UNION
+SELECT employee_id, department_id
+FROM Employee
+GROUP BY employee_id
+HAVING count(1) = 1;
 ```
 
 <!-- tabs:end -->

solution/1700-1799/1789.Primary Department for Each Employee/Solution.sql

@@ -0,0 +1,9 @@
+# Write your MySQL query statement below
+SELECT employee_id, department_id
+FROM Employee
+WHERE primary_flag = 'Y'
+UNION
+SELECT employee_id, department_id
+FROM Employee
+GROUP BY employee_id
+HAVING count(1) = 1;

solution/1800-1899/1809.Ad-Free Sessions/README.md

@@ -93,7 +93,18 @@ Result table:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT session_id
+FROM Playback
+WHERE
+    session_id NOT IN (
+        SELECT session_id
+        FROM
+            Playback AS p
+            JOIN Ads AS a
+                ON p.customer_id = a.customer_id
+                AND a.timestamp BETWEEN p.start_time AND p.end_time
+    );
 ```
 
 <!-- tabs:end -->

solution/1800-1899/1809.Ad-Free Sessions/README_EN.md

@@ -90,7 +90,18 @@ We can see that sessions 1 and 4 had at least one ad. Sessions 2, 3, and 5 did n
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT session_id
+FROM Playback
+WHERE
+    session_id NOT IN (
+        SELECT session_id
+        FROM
+            Playback AS p
+            JOIN Ads AS a
+                ON p.customer_id = a.customer_id
+                AND a.timestamp BETWEEN p.start_time AND p.end_time
+    );
 ```
 
 <!-- tabs:end -->

solution/1800-1899/1809.Ad-Free Sessions/Solution.sql

@@ -0,0 +1,12 @@
+# Write your MySQL query statement below
+SELECT session_id
+FROM Playback
+WHERE
+    session_id NOT IN (
+        SELECT session_id
+        FROM
+            Playback AS p
+            JOIN Ads AS a
+                ON p.customer_id = a.customer_id
+                AND a.timestamp BETWEEN p.start_time AND p.end_time
+    );