feat: add sql solutions to lc problems: No.2984~2991

solution/2900-2999/2984.Find Peak Calling Hours for Each City/README.md

@@ -69,7 +69,29 @@ Output table is ordered by peak_calling_hour and city in descending order.</pre>
 <!-- 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
+            *,
+            RANK() OVER (
+                PARTITION BY city
+                ORDER BY cnt DESC
+            ) AS rk
+        FROM
+            (
+                SELECT
+                    city,
+                    HOUR(call_time) AS h,
+                    COUNT(1) AS cnt
+                FROM Calls
+                GROUP BY 1, 2
+            ) AS t
+    )
+SELECT city, h AS peak_calling_hour, cnt AS number_of_calls
+FROM T
+WHERE rk = 1
+ORDER BY 2 DESC, 1 DESC;
 ```
 
 <!-- tabs:end -->

solution/2900-2999/2984.Find Peak Calling Hours for Each City/README_EN.md

@@ -63,7 +63,29 @@ Output table is ordered by peak_calling_hour and city in descending order.</pre>
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            *,
+            RANK() OVER (
+                PARTITION BY city
+                ORDER BY cnt DESC
+            ) AS rk
+        FROM
+            (
+                SELECT
+                    city,
+                    HOUR(call_time) AS h,
+                    COUNT(1) AS cnt
+                FROM Calls
+                GROUP BY 1, 2
+            ) AS t
+    )
+SELECT city, h AS peak_calling_hour, cnt AS number_of_calls
+FROM T
+WHERE rk = 1
+ORDER BY 2 DESC, 1 DESC;
 ```
 
 <!-- tabs:end -->

solution/2900-2999/2984.Find Peak Calling Hours for Each City/Solution.sql

@@ -0,0 +1,23 @@
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            *,
+            RANK() OVER (
+                PARTITION BY city
+                ORDER BY cnt DESC
+            ) AS rk
+        FROM
+            (
+                SELECT
+                    city,
+                    HOUR(call_time) AS h,
+                    COUNT(1) AS cnt
+                FROM Calls
+                GROUP BY 1, 2
+            ) AS t
+    )
+SELECT city, h AS peak_calling_hour, cnt AS number_of_calls
+FROM T
+WHERE rk = 1
+ORDER BY 2 DESC, 1 DESC;

solution/2900-2999/2985.Calculate Compressed Mean/README.md

@@ -63,7 +63,13 @@ The calculation is as follows:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    ROUND(
+        SUM(item_count * order_occurrences) / SUM(order_occurrences),
+        2
+    ) AS average_items_per_order
+FROM Orders;
 ```
 
 <!-- tabs:end -->

solution/2900-2999/2985.Calculate Compressed Mean/README_EN.md

@@ -57,7 +57,13 @@ The calculation is as follows:
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    ROUND(
+        SUM(item_count * order_occurrences) / SUM(order_occurrences),
+        2
+    ) AS average_items_per_order
+FROM Orders;
 ```
 
 <!-- tabs:end -->

solution/2900-2999/2985.Calculate Compressed Mean/Solution.sql

@@ -0,0 +1,7 @@
+# Write your MySQL query statement below
+SELECT
+    ROUND(
+        SUM(item_count * order_occurrences) / SUM(order_occurrences),
+        2
+    ) AS average_items_per_order
+FROM Orders;

solution/2900-2999/2986.Find Third Transaction/README.md

@@ -70,7 +70,32 @@ Output table is ordered by user_id in ascending order.
 <!-- 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
+            *,
+            RANK() OVER (
+                PARTITION BY user_id
+                ORDER BY transaction_date
+            ) AS rk,
+            spend > (
+                LAG(spend) OVER (
+                    PARTITION BY user_id
+                    ORDER BY transaction_date
+                )
+            )
+            AND spend > (
+                LAG(spend, 2) OVER (
+                    PARTITION BY user_id
+                    ORDER BY transaction_date
+                )
+            ) AS st
+        FROM Transactions
+    )
+SELECT user_id, spend AS third_transaction_spend, transaction_date AS third_transaction_date
+FROM T
+WHERE rk = 3 AND st = 1;
 ```
 
 <!-- tabs:end -->

solution/2900-2999/2986.Find Third Transaction/README_EN.md

@@ -64,7 +64,32 @@ Output table is ordered by user_id in ascending order.
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            *,
+            RANK() OVER (
+                PARTITION BY user_id
+                ORDER BY transaction_date
+            ) AS rk,
+            spend > (
+                LAG(spend) OVER (
+                    PARTITION BY user_id
+                    ORDER BY transaction_date
+                )
+            )
+            AND spend > (
+                LAG(spend, 2) OVER (
+                    PARTITION BY user_id
+                    ORDER BY transaction_date
+                )
+            ) AS st
+        FROM Transactions
+    )
+SELECT user_id, spend AS third_transaction_spend, transaction_date AS third_transaction_date
+FROM T
+WHERE rk = 3 AND st = 1;
 ```
 
 <!-- tabs:end -->

solution/2900-2999/2986.Find Third Transaction/Solution.sql

@@ -0,0 +1,26 @@
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            *,
+            RANK() OVER (
+                PARTITION BY user_id
+                ORDER BY transaction_date
+            ) AS rk,
+            spend > (
+                LAG(spend) OVER (
+                    PARTITION BY user_id
+                    ORDER BY transaction_date
+                )
+            )
+            AND spend > (
+                LAG(spend, 2) OVER (
+                    PARTITION BY user_id
+                    ORDER BY transaction_date
+                )
+            ) AS st
+        FROM Transactions
+    )
+SELECT user_id, spend AS third_transaction_spend, transaction_date AS third_transaction_date
+FROM T
+WHERE rk = 3 AND st = 1;

solution/2900-2999/2987.Find Expensive Cities/README.md

@@ -75,7 +75,12 @@ Only Chicago and Los Angeles have average home prices exceeding the national ave
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT city
+FROM Listings
+GROUP BY city
+HAVING AVG(price) > (SELECT AVG(price) FROM Listings)
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/2900-2999/2987.Find Expensive Cities/README_EN.md

@@ -69,7 +69,12 @@ Only Chicago and Los Angeles have average home prices exceeding the national ave
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT city
+FROM Listings
+GROUP BY city
+HAVING AVG(price) > (SELECT AVG(price) FROM Listings)
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/2900-2999/2987.Find Expensive Cities/Solution.sql

@@ -0,0 +1,6 @@
+# Write your MySQL query statement below
+SELECT city
+FROM Listings
+GROUP BY city
+HAVING AVG(price) > (SELECT AVG(price) FROM Listings)
+ORDER BY 1;

solution/2900-2999/2988.Manager of the Largest Department/README.md

@@ -72,7 +72,19 @@ Output table is ordered by dep_id in ascending order.
 <!-- 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 dep_id, COUNT(1) AS cnt
+        FROM Employees
+        GROUP BY 1
+    )
+SELECT emp_name AS manager_name, t.dep_id
+FROM
+    T AS t
+    JOIN Employees AS e ON t.dep_id = e.dep_id AND e.position = 'Manager'
+WHERE cnt = (SELECT MAX(cnt) FROM T)
+ORDER BY 2;
 ```
 
 <!-- tabs:end -->

solution/2900-2999/2988.Manager of the Largest Department/README_EN.md

@@ -66,7 +66,19 @@ Output table is ordered by dep_id in ascending order.
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT dep_id, COUNT(1) AS cnt
+        FROM Employees
+        GROUP BY 1
+    )
+SELECT emp_name AS manager_name, t.dep_id
+FROM
+    T AS t
+    JOIN Employees AS e ON t.dep_id = e.dep_id AND e.position = 'Manager'
+WHERE cnt = (SELECT MAX(cnt) FROM T)
+ORDER BY 2;
 ```
 
 <!-- tabs:end -->

solution/2900-2999/2988.Manager of the Largest Department/Solution.sql

@@ -0,0 +1,13 @@
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT dep_id, COUNT(1) AS cnt
+        FROM Employees
+        GROUP BY 1
+    )
+SELECT emp_name AS manager_name, t.dep_id
+FROM
+    T AS t
+    JOIN Employees AS e ON t.dep_id = e.dep_id AND e.position = 'Manager'
+WHERE cnt = (SELECT MAX(cnt) FROM T)
+ORDER BY 2;

solution/2900-2999/2989.Class Performance/README.md

@@ -64,14 +64,23 @@ student_id 321 has the highest score of 230, while student_id 896 has the lowest
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：最大值最小值**
+
+我们可以使用 `MAX` 和 `MIN` 函数来分别获取 `assignment1`、`assignment2`、`assignment3` 的和的最大值和最小值，然后相减即可。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    MAX(assignment1 + assignment2 + assignment3) - MIN(
+        assignment1 + assignment2 + assignment3
+    ) AS difference_in_score
+FROM Scores;
 ```
 
 <!-- tabs:end -->

solution/2900-2999/2989.Class Performance/README_EN.md

@@ -60,12 +60,21 @@ student_id 321 has the highest score of 230, while student_id 896 has the lowest
 
 ## Solutions
 
+**Solution 1: Maximum and Minimum**
+
+We can use the `MAX` and `MIN` functions to get the maximum and minimum sums of `assignment1`, `assignment2`, and `assignment3`, respectively. Then, subtract the minimum from the maximum.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT
+    MAX(assignment1 + assignment2 + assignment3) - MIN(
+        assignment1 + assignment2 + assignment3
+    ) AS difference_in_score
+FROM Scores;
 ```
 
 <!-- tabs:end -->

solution/2900-2999/2989.Class Performance/Solution.sql

@@ -0,0 +1,6 @@
+# Write your MySQL query statement below
+SELECT
+    MAX(assignment1 + assignment2 + assignment3) - MIN(
+        assignment1 + assignment2 + assignment3
+    ) AS difference_in_score
+FROM Scores;

solution/2900-2999/2990.Loan Types/README.md

@@ -62,14 +62,23 @@ Output table is ordered by user_id in ascending order.
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：分组求和**
+
+我们可以对 `Loans` 表按照 `user_id` 进行分组，找出既包含 `Refinance` 又包含 `Mortgage` 的用户，然后按照 `user_id` 进行排序。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+# Write your MySQL query statement below
+SELECT user_id
+FROM Loans
+GROUP BY 1
+HAVING SUM(loan_type = 'Refinance') > 0 AND SUM(loan_type = 'Mortgage') > 0
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->