feat: add sql solutions to lc problems: No.1731,1867,1988 (#1063)

* No.1731.The Number of Employees Which Report to Each Employee
* No.1867.Orders With Maximum Quantity Above Average
* No.1988.Find Cutoff Score for Each School

Author: yanglbme

solution/1700-1799/1731.The Number of Employees Which Report to Each Employee/README.md

@@ -69,7 +69,6 @@ SELECT
 FROM
     Employees AS e1
     JOIN Employees AS e2 ON e1.reports_to = e2.employee_id
-WHERE e2.employee_id IS NOT NULL
 GROUP BY e2.employee_id
 ORDER BY e2.employee_id;
 ```

solution/1700-1799/1731.The Number of Employees Which Report to Each Employee/README_EN.md

@@ -68,7 +68,6 @@ SELECT
 FROM
     Employees AS e1
     JOIN Employees AS e2 ON e1.reports_to = e2.employee_id
-WHERE e2.employee_id IS NOT NULL
 GROUP BY e2.employee_id
 ORDER BY e2.employee_id;
 ```

solution/1700-1799/1731.The Number of Employees Which Report to Each Employee/Solution.sql

@@ -7,6 +7,5 @@ SELECT
 FROM
     Employees AS e1
     JOIN Employees AS e2 ON e1.reports_to = e2.employee_id
-WHERE e2.employee_id IS NOT NULL
 GROUP BY e2.employee_id
 ORDER BY e2.employee_id;

solution/1800-1899/1867.Orders With Maximum Quantity Above Average/README.md

@@ -94,7 +94,19 @@ OrdersDetails 表:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    t AS (
+        SELECT
+            order_id,
+            max(quantity) AS max_quantity,
+            sum(quantity) / count(1) AS avg_quantity
+        FROM OrdersDetails
+        GROUP BY order_id
+    )
+SELECT order_id
+FROM t
+WHERE max_quantity > (SELECT max(avg_quantity) FROM t);
 ```
 
 <!-- tabs:end -->

solution/1800-1899/1867.Orders With Maximum Quantity Above Average/README_EN.md

@@ -87,7 +87,19 @@ Orders 1 and 3 are imbalanced because they have a maximum quantity that exceeds
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    t AS (
+        SELECT
+            order_id,
+            max(quantity) AS max_quantity,
+            sum(quantity) / count(1) AS avg_quantity
+        FROM OrdersDetails
+        GROUP BY order_id
+    )
+SELECT order_id
+FROM t
+WHERE max_quantity > (SELECT max(avg_quantity) FROM t);
 ```
 
 <!-- tabs:end -->

solution/1800-1899/1867.Orders With Maximum Quantity Above Average/Solution.sql

@@ -0,0 +1,13 @@
+# Write your MySQL query statement below
+WITH
+    t AS (
+        SELECT
+            order_id,
+            max(quantity) AS max_quantity,
+            sum(quantity) / count(1) AS avg_quantity
+        FROM OrdersDetails
+        GROUP BY order_id
+    )
+SELECT order_id
+FROM t
+WHERE max_quantity > (SELECT max(avg_quantity) FROM t);

solution/1900-1999/1988.Find Cutoff Score for Each School/README.md

@@ -99,7 +99,12 @@ Exam 表:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT school_id, min(ifnull(score, -1)) AS score
+FROM
+    Schools AS s
+    LEFT JOIN Exam AS e ON s.capacity >= e.student_count
+GROUP BY school_id;
 ```
 
 <!-- tabs:end -->

solution/1900-1999/1988.Find Cutoff Score for Each School/README_EN.md

@@ -96,7 +96,12 @@ Exam table:
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT school_id, min(ifnull(score, -1)) AS score
+FROM
+    Schools AS s
+    LEFT JOIN Exam AS e ON s.capacity >= e.student_count
+GROUP BY school_id;
 ```
 
 <!-- tabs:end -->

solution/1900-1999/1988.Find Cutoff Score for Each School/Solution.sql

@@ -0,0 +1,6 @@
+# Write your MySQL query statement below
+SELECT school_id, min(ifnull(score, -1)) AS score
+FROM
+    Schools AS s
+    LEFT JOIN Exam AS e ON s.capacity >= e.student_count
+GROUP BY school_id;