feat: add sql solutions to lc problems (#1228)

* No.1341.Movie Rating
* No.1350.Students With Invalid Departments
* No.1378.Replace Employee ID With The Unique Identifier
* No.1393.Capital GainLoss
* No.1398.Customers Who Bought Products A and B but Not C
* No.1407.Top Travellers

Author: yanglbme

.prettierrc

@@ -27,6 +27,7 @@
                 "solution/1200-1299/1205.Monthly Transactions II/Solution.sql",
                 "solution/1300-1399/1384.Total Sales Amount by Year/Solution.sql",
                 "solution/1300-1399/1322.Ads Performance/Solution.sql",
+                "solution/1300-1399/1393.Capital GainLoss/Solution.sql",
                 "solution/1300-1399/1398.Customers Who Bought Products A and B but Not C/Solution.sql",
                 "solution/1400-1499/1479.Sales by Day of the Week/Solution.sql",
                 "solution/1500-1599/1501.Countries You Can Safely Invest In/Solution.sql",

solution/1300-1399/1341.Movie Rating/README.md

@@ -125,24 +125,22 @@ Frozen 2 和 Joker 在 2 月的评分都是 3.5，但是 Frozen 2 的字典序
 ```sql
 # Write your MySQL query statement below
 (
-    SELECT
-        name AS results
+    SELECT name AS results
     FROM
-        MovieRating AS r
-        LEFT JOIN Users AS u ON u.user_id = r.user_id
-    GROUP BY r.user_id
+        Users
+        JOIN MovieRating USING (user_id)
+    GROUP BY user_id
     ORDER BY count(1) DESC, name
     LIMIT 1
 )
 UNION ALL
 (
-    SELECT
-        title AS results
+    SELECT title
     FROM
-        MovieRating AS r
-        LEFT JOIN Movies AS m ON m.movie_id = r.movie_id
+        MovieRating
+        JOIN Movies USING (movie_id)
     WHERE date_format(created_at, '%Y-%m') = '2020-02'
-    GROUP BY m.movie_id
+    GROUP BY movie_id
     ORDER BY avg(rating) DESC, title
     LIMIT 1
 );

solution/1300-1399/1341.Movie Rating/README_EN.md

@@ -117,24 +117,22 @@ Frozen 2 and Joker have a rating average of 3.5 in February but Frozen 2 is smal
 ```sql
 # Write your MySQL query statement below
 (
-    SELECT
-        name AS results
+    SELECT name AS results
     FROM
-        MovieRating AS r
-        LEFT JOIN Users AS u ON u.user_id = r.user_id
-    GROUP BY r.user_id
+        Users
+        JOIN MovieRating USING (user_id)
+    GROUP BY user_id
     ORDER BY count(1) DESC, name
     LIMIT 1
 )
 UNION ALL
 (
-    SELECT
-        title AS results
+    SELECT title
     FROM
-        MovieRating AS r
-        LEFT JOIN Movies AS m ON m.movie_id = r.movie_id
+        MovieRating
+        JOIN Movies USING (movie_id)
     WHERE date_format(created_at, '%Y-%m') = '2020-02'
-    GROUP BY m.movie_id
+    GROUP BY movie_id
     ORDER BY avg(rating) DESC, title
     LIMIT 1
 );

solution/1300-1399/1341.Movie Rating/Solution.sql

@@ -1,23 +1,21 @@
 # Write your MySQL query statement below
 (
-    SELECT
-        name AS results
+    SELECT name AS results
     FROM
-        MovieRating AS r
-        LEFT JOIN Users AS u ON u.user_id = r.user_id
-    GROUP BY r.user_id
+        Users
+        JOIN MovieRating USING (user_id)
+    GROUP BY user_id
     ORDER BY count(1) DESC, name
     LIMIT 1
 )
 UNION ALL
 (
-    SELECT
-        title AS results
+    SELECT title
     FROM
-        MovieRating AS r
-        LEFT JOIN Movies AS m ON m.movie_id = r.movie_id
+        MovieRating
+        JOIN Movies USING (movie_id)
     WHERE date_format(created_at, '%Y-%m') = '2020-02'
-    GROUP BY m.movie_id
+    GROUP BY movie_id
     ORDER BY avg(rating) DESC, title
     LIMIT 1
 );

solution/1300-1399/1350.Students With Invalid Departments/README.md

@@ -88,17 +88,17 @@ John, Daiana, Steve 和 Jasmine 所在的院系分别是 14, 33, 74 和 77， 
 
 <!-- 这里可写通用的实现逻辑 -->
 
-外连接查询。
+**方法一：左连接**
+
+我们将 `Students` 表左连接 `Departments` 表，然后筛选出所有 `Departments` 表中 `id` 为 `NULL` 的记录即可。
 
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    s.id,
-    s.name
+SELECT s.id, s.name
 FROM
     Students AS s
     LEFT JOIN Departments AS d ON s.department_id = d.id

solution/1300-1399/1350.Students With Invalid Departments/README_EN.md

@@ -90,9 +90,7 @@ John, Daiana, Steve, and Jasmine are enrolled in departments 14, 33, 74, and 77
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    s.id,
-    s.name
+SELECT s.id, s.name
 FROM
     Students AS s
     LEFT JOIN Departments AS d ON s.department_id = d.id

solution/1300-1399/1350.Students With Invalid Departments/Solution.sql

@@ -1,7 +1,5 @@
 # Write your MySQL query statement below
-SELECT
-    s.id,
-    s.name
+SELECT s.id, s.name
 FROM
     Students AS s
     LEFT JOIN Departments AS d ON s.department_id = d.id

solution/1300-1399/1378.Replace Employee ID With The Unique Identifier/README.md

@@ -86,18 +86,20 @@ Jonathan 唯一标识码是 1 。</pre>
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：LEFT JOIN**
+
+我们可以使用 `LEFT JOIN` 来连接 `Employees` 和 `EmployeeUNI` 表，然后使用 `SELECT` 语句来选择 `unique_id` 和 `name` 列。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    b.unique_id AS unique_id,
-    a.name AS name
+SELECT unique_id, name
 FROM
-    Employees AS a
-    LEFT JOIN EmployeeUNI AS b ON a.id = b.id;
+    Employees
+    LEFT JOIN EmployeeUNI USING (id);
 ```
 
 <!-- tabs:end -->

solution/1300-1399/1378.Replace Employee ID With The Unique Identifier/README_EN.md

@@ -88,12 +88,10 @@ The unique ID of Jonathan is 1.
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    b.unique_id AS unique_id,
-    a.name AS name
+SELECT unique_id, name
 FROM
-    Employees AS a
-    LEFT JOIN EmployeeUNI AS b ON a.id = b.id;
+    Employees
+    LEFT JOIN EmployeeUNI USING (id);
 ```
 
 <!-- tabs:end -->

solution/1300-1399/1378.Replace Employee ID With The Unique Identifier/Solution.sql

@@ -1,7 +1,5 @@
 # Write your MySQL query statement below
-SELECT
-    b.unique_id AS unique_id,
-    a.name AS name
+SELECT unique_id, name
 FROM
-    Employees AS a
-    LEFT JOIN EmployeeUNI AS b ON a.id = b.id;
+    Employees
+    LEFT JOIN EmployeeUNI USING (id);

solution/1300-1399/1393.Capital GainLoss/README.md

@@ -65,23 +65,21 @@ Corona Masks 股票在第1天以10美元的价格买入，在第3天以1010美
 
 <!-- 这里可写通用的实现逻辑 -->
 
-`CASE WHEN` + `GROUP BY`。
+**方法一：GROUP BY + SUM(IF())**
+
+我们使用 `GROUP BY` 将同一支股票的买卖操作分组，然后使用 `SUM(IF())` 计算每支股票的资本损益。
 
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
+# Write your MySQL query statement below
 SELECT
     stock_name,
-    SUM(
-        CASE
-            WHEN operation = 'Buy' THEN -price
-            ELSE price
-        END
-    ) AS capital_gain_loss
+    sum(if(operation = 'Buy', -price, price)) AS capital_gain_loss
 FROM Stocks
-GROUP BY stock_name;
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

solution/1300-1399/1393.Capital GainLoss/README_EN.md

@@ -72,16 +72,12 @@ Corona Masks stock was bought at day 1 for 10$ and was sold at day 3 for 1010$.
 ### **SQL**
 
 ```sql
+# Write your MySQL query statement below
 SELECT
     stock_name,
-    SUM(
-        CASE
-            WHEN operation = 'Buy' THEN -price
-            ELSE price
-        END
-    ) AS capital_gain_loss
+    sum(if(operation = 'Buy', -price, price)) AS capital_gain_loss
 FROM Stocks
-GROUP BY stock_name;
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

solution/1300-1399/1393.Capital GainLoss/Solution.sql

@@ -1,10 +1,6 @@
+# Write your MySQL query statement below
 SELECT
     stock_name,
-    SUM(
-        CASE
-            WHEN operation = 'Buy' THEN -price
-            ELSE price
-        END
-    ) AS capital_gain_loss
+    sum(if(operation = 'Buy', -price, price)) AS capital_gain_loss
 FROM Stocks
-GROUP BY stock_name;
+GROUP BY 1;

solution/1300-1399/1398.Customers Who Bought Products A and B but Not C/README.md

@@ -87,18 +87,17 @@ Result table:
 
 ```sql
 # Write your MySQL query statement below
-select
-    o.customer_id,
-    c.customer_name
-from
-    orders o
-    left join customers c on o.customer_id = c.customer_id
-group by
-    customer_id
-having
-    sum(if(product_name = 'A', 1, 0)) > 0
-    and sum(if(product_name = 'B', 1, 0)) > 0
-    and sum(if(product_name = 'C', 1, 0)) = 0
+SELECT
+    customer_id,
+    customer_name
+FROM
+    Orders
+    JOIN Customers USING (customer_id)
+GROUP BY 1
+HAVING
+    sum(product_name = 'A') > 0
+    AND sum(product_name = 'B') > 0
+    AND sum(product_name = 'C') = 0;
 ```
 
 <!-- tabs:end -->

solution/1300-1399/1398.Customers Who Bought Products A and B but Not C/README_EN.md

@@ -85,18 +85,17 @@ Orders table:
 
 ```sql
 # Write your MySQL query statement below
-select
-    o.customer_id,
-    c.customer_name
-from
-    orders o
-    left join customers c on o.customer_id = c.customer_id
-group by
-    customer_id
-having
-    sum(if(product_name = 'A', 1, 0)) > 0
-    and sum(if(product_name = 'B', 1, 0)) > 0
-    and sum(if(product_name = 'C', 1, 0)) = 0
+SELECT
+    customer_id,
+    customer_name
+FROM
+    Orders
+    JOIN Customers USING (customer_id)
+GROUP BY 1
+HAVING
+    sum(product_name = 'A') > 0
+    AND sum(product_name = 'B') > 0
+    AND sum(product_name = 'C') = 0;
 ```
 
 <!-- tabs:end -->

solution/1300-1399/1398.Customers Who Bought Products A and B but Not C/Solution.sql

@@ -1,12 +1,12 @@
 # Write your MySQL query statement below
 SELECT
-    o.customer_id,
-    c.customer_name
+    customer_id,
+    customer_name
 FROM
-    orders AS o
-    LEFT JOIN customers AS c ON o.customer_id = c.customer_id
-GROUP BY customer_id
+    Orders
+    JOIN Customers USING (customer_id)
+GROUP BY 1
 HAVING
-    sum(if(product_name = 'A', 1, 0)) > 0
-    AND sum(if(product_name = 'B', 1, 0)) > 0
-    AND sum(if(product_name = 'C', 1, 0)) = 0;
+    sum(product_name = 'A') > 0
+    AND sum(product_name = 'B') > 0
+    AND sum(product_name = 'C') = 0;

solution/1400-1499/1407.Top Travellers/README.md

@@ -96,27 +96,14 @@ Donald 没有任何行程, 他的旅行距离为 0。
 
 ### **SQL**
 
-```sql
-SELECT
-    name,
-    COALESCE(SUM(distance), 0) AS travelled_distance
-FROM
-    Users AS u
-    LEFT JOIN Rides AS r ON u.id = r.user_id
-GROUP BY name
-ORDER BY travelled_distance DESC, name;
-```
-
 ```sql
 # Write your MySQL query statement below
-SELECT
-    name,
-    sum(ifnull(distance, 0)) AS travelled_distance
+SELECT name, ifnull(sum(distance), 0) AS travelled_distance
 FROM
     Users AS u
     LEFT JOIN Rides AS r ON u.id = r.user_id
-GROUP BY (u.id)
-ORDER BY travelled_distance DESC, name ASC;
+GROUP BY u.id
+ORDER BY 2 DESC, 1;
 ```
 
 <!-- tabs:end -->