feat: add sql solutions to lc problems (#1133)

* No.2205.The Number of Users That Are Eligible for Discount
* No.2292.Products With Three or More Orders in Two Consecutive Years
* No.2314.The First Day of the Maximum Recorded Degree in Each City
* No.2324.Product Sales Analysis IV
* No.2329.Product Sales Analysis V
* No.2339.All the Matches of the League
* No.2377.Sort the Olympic Table

Author: yanglbme

.prettierignore

@@ -24,4 +24,5 @@ node_modules/
 /solution/1600-1699/1613.Find the Missing IDs/Solution.sql
 /solution/1600-1699/1635.Hopper Company Queries I/Solution.sql
 /solution/1600-1699/1651.Hopper Company Queries III/Solution.sql
-/solution/1700-1799/1767.Find the Subtasks That Did Not Execute/Solution.sql
+/solution/1700-1799/1767.Find the Subtasks That Did Not Execute/Solution.sql
+/solution/2200-2299/2205.The Number of Users That Are Eligible for Discount/Solution.sql

solution/2200-2299/2205.The Number of Users That Are Eligible for Discount/README.md

@@ -60,14 +60,27 @@ startDate = 2022-03-08, endDate = 2022-03-20, minAmount = 1000
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：使用 count(distinct) 函数**
+
+注意需要判断的是单次购买金额是否大于等于 `minAmount`，而不是累计购买金额是否大于等于 `minAmount`。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
-
+CREATE FUNCTION getUserIDs(startDate DATE, endDate DATE, minAmount INT) RETURNS INT
+BEGIN
+  RETURN (
+      # Write your MySQL query statement below.
+      # Write your MySQL query statement below.
+      SELECT count(DISTINCT user_id) AS user_cnt
+      FROM Purchases
+      WHERE time_stamp BETWEEN startDate AND endDate AND amount >= minAmount;
+  );
+END
 ```
 
 <!-- tabs:end -->

solution/2200-2299/2205.The Number of Users That Are Eligible for Discount/README_EN.md

@@ -61,7 +61,16 @@ Out of the three users, only User 3 is eligible for a discount.
 ### **SQL**
 
 ```sql
-
+CREATE FUNCTION getUserIDs(startDate DATE, endDate DATE, minAmount INT) RETURNS INT
+BEGIN
+  RETURN (
+      # Write your MySQL query statement below.
+      # Write your MySQL query statement below.
+      SELECT count(DISTINCT user_id) AS user_cnt
+      FROM Purchases
+      WHERE time_stamp BETWEEN startDate AND endDate AND amount >= minAmount;
+  );
+END
 ```
 
 <!-- tabs:end -->

solution/2200-2299/2205.The Number of Users That Are Eligible for Discount/Solution.sql

@@ -0,0 +1,10 @@
+CREATE FUNCTION getUserIDs(startDate DATE, endDate DATE, minAmount INT) RETURNS INT
+BEGIN
+  RETURN (
+      # Write your MySQL query statement below.
+      # Write your MySQL query statement below.
+      SELECT count(DISTINCT user_id) AS user_cnt
+      FROM Purchases
+      WHERE time_stamp BETWEEN startDate AND endDate AND amount >= minAmount;
+  );
+END

solution/2200-2299/2292.Products With Three or More Orders in Two Consecutive Years/README.md

@@ -69,7 +69,33 @@ Orders 表:
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```sql
+# Write your MySQL query statement below
+WITH
+    P AS (
+        SELECT product_id, year(purchase_date) AS y, count(1) >= 3 AS mark
+        FROM Orders
+        GROUP BY 1, 2
+    )
+SELECT DISTINCT p1.product_id
+FROM
+    P AS p1
+    JOIN P AS p2 ON p1.y = p2.y - 1 AND p1.product_id = p2.product_id
+WHERE p1.mark AND p2.mark;
+```
 
+```sql
+# Write your MySQL query statement below
+WITH
+    P AS (
+        SELECT product_id, year(purchase_date) AS y
+        FROM Orders
+        GROUP BY 1, 2
+        HAVING count(1) >= 3
+    )
+SELECT DISTINCT p1.product_id
+FROM
+    P AS p1
+    JOIN P AS p2 ON p1.y = p2.y - 1 AND p1.product_id = p2.product_id;
 ```
 
 <!-- tabs:end -->

solution/2200-2299/2292.Products With Three or More Orders in Two Consecutive Years/README_EN.md

@@ -62,7 +62,33 @@ Product 2 was ordered one time in 2022. We do not include it in the answer.
 ### **SQL**
 
 ```sql
+# Write your MySQL query statement below
+WITH
+    P AS (
+        SELECT product_id, year(purchase_date) AS y, count(1) >= 3 AS mark
+        FROM Orders
+        GROUP BY 1, 2
+    )
+SELECT DISTINCT p1.product_id
+FROM
+    P AS p1
+    JOIN P AS p2 ON p1.y = p2.y - 1 AND p1.product_id = p2.product_id
+WHERE p1.mark AND p2.mark;
+```
 
+```sql
+# Write your MySQL query statement below
+WITH
+    P AS (
+        SELECT product_id, year(purchase_date) AS y
+        FROM Orders
+        GROUP BY 1, 2
+        HAVING count(1) >= 3
+    )
+SELECT DISTINCT p1.product_id
+FROM
+    P AS p1
+    JOIN P AS p2 ON p1.y = p2.y - 1 AND p1.product_id = p2.product_id;
 ```
 
 <!-- tabs:end -->

solution/2200-2299/2292.Products With Three or More Orders in Two Consecutive Years/Solution.sql

@@ -0,0 +1,12 @@
+# Write your MySQL query statement below
+WITH
+    P AS (
+        SELECT product_id, year(purchase_date) AS y
+        FROM Orders
+        GROUP BY 1, 2
+        HAVING count(1) >= 3
+    )
+SELECT DISTINCT p1.product_id
+FROM
+    P AS p1
+    JOIN P AS p2 ON p1.y = p2.y - 1 AND p1.product_id = p2.product_id;

solution/2300-2399/2314.The First Day of the Maximum Recorded Degree in Each City/README.md

@@ -72,7 +72,21 @@ Weather 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
+            *,
+            rank() OVER (
+                PARTITION BY city_id
+                ORDER BY degree DESC, day
+            ) AS rk
+        FROM Weather
+    )
+SELECT city_id, day, degree
+FROM T
+WHERE rk = 1
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/2300-2399/2314.The First Day of the Maximum Recorded Degree in Each City/README_EN.md

@@ -65,7 +65,21 @@ For city 3, the maximum degree was recorded on 2022-12-07 with -6 degrees.
 ### **SQL**
 
 ```sql
-
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            *,
+            rank() OVER (
+                PARTITION BY city_id
+                ORDER BY degree DESC, day
+            ) AS rk
+        FROM Weather
+    )
+SELECT city_id, day, degree
+FROM T
+WHERE rk = 1
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/2300-2399/2314.The First Day of the Maximum Recorded Degree in Each City/Solution.sql

@@ -0,0 +1,15 @@
+# Write your MySQL query statement below
+WITH
+    T AS (
+        SELECT
+            *,
+            rank() OVER (
+                PARTITION BY city_id
+                ORDER BY degree DESC, day
+            ) AS rk
+        FROM Weather
+    )
+SELECT city_id, day, degree
+FROM T
+WHERE rk = 1
+ORDER BY 1;