feat: update solutions to lc problems: No.0570,0620,1251,1280,1934 (doocs#1834)

* No.0570.Managers with at Least 5 Direct Reports
* No.0620.Not Boring Movies
* No.1251.Average Selling Price
* No.1280.Students and Examinations
* No.1934.Confirmation Rate

Author: yanglbme

solution/0500-0599/0570.Managers with at Least 5 Direct Reports/README.md

@@ -59,41 +59,26 @@ Employee 表:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：分组统计 + 连接**
+
+我们可以先统计每个经理的直接下属人数，然后再连接 `Employee` 表，找出直接下属人数大于等于 $5$ 的经理。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    name
+SELECT name
 FROM
-    Employee AS e1
+    Employee
     JOIN (
-        SELECT
-            managerId
-        FROM Employee
-        WHERE managerId IS NOT NULL
-        GROUP BY managerId
-        HAVING COUNT(1) >= 5
-    ) AS e2
-        ON e1.id = e2.managerId;
-```
-
-```sql
-# Write your MySQL query statement below
-WITH
-    T AS (
-        SELECT
-            managerId,
-            COUNT(1) OVER (PARTITION BY managerId) AS cnt
+        SELECT managerId AS id, COUNT(1) AS cnt
         FROM Employee
-    )
-SELECT DISTINCT name
-FROM
-    Employee AS e
-    JOIN T AS t ON e.id = t.managerId
-WHERE cnt >= 5;
+        GROUP BY 1
+        HAVING cnt >= 5
+    ) AS t
+        USING (id);
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0570.Managers with at Least 5 Direct Reports/README_EN.md

@@ -55,41 +55,26 @@ Employee table:
 
 ## Solutions
 
+**Solution 1: Grouping and Joining**
+
+We can first count the number of direct subordinates for each manager, and then join the `Employee` table to find the managers whose number of direct subordinates is greater than or equal to $5$.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    name
+SELECT name
 FROM
-    Employee AS e1
+    Employee
     JOIN (
-        SELECT
-            managerId
-        FROM Employee
-        WHERE managerId IS NOT NULL
-        GROUP BY managerId
-        HAVING COUNT(1) >= 5
-    ) AS e2
-        ON e1.id = e2.managerId;
-```
-
-```sql
-# Write your MySQL query statement below
-WITH
-    T AS (
-        SELECT
-            managerId,
-            COUNT(1) OVER (PARTITION BY managerId) AS cnt
+        SELECT managerId AS id, COUNT(1) AS cnt
         FROM Employee
-    )
-SELECT DISTINCT name
-FROM
-    Employee AS e
-    JOIN T AS t ON e.id = t.managerId
-WHERE cnt >= 5;
+        GROUP BY 1
+        HAVING cnt >= 5
+    ) AS t
+        USING (id);
 ```
 
 <!-- tabs:end -->

solution/0500-0599/0570.Managers with at Least 5 Direct Reports/Solution.sql

@@ -1,13 +1,11 @@
 # Write your MySQL query statement below
-WITH
-    T AS (
-        SELECT
-            managerId,
-            COUNT(1) OVER (PARTITION BY managerId) AS cnt
-        FROM Employee
-    )
-SELECT DISTINCT name
+SELECT name
 FROM
-    Employee AS e
-    JOIN T AS t ON e.id = t.managerId
-WHERE cnt >= 5;
+    Employee
+    JOIN (
+        SELECT managerId AS id, COUNT(1) AS cnt
+        FROM Employee
+        GROUP BY 1
+        HAVING cnt >= 5
+    ) AS t
+        USING (id);

solution/0600-0699/0620.Not Boring Movies/README.md

@@ -60,6 +60,10 @@ id 是该表的主键(具有唯一值的列)。
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：条件筛选 + 排序**
+
+我们可以使用 `WHERE` 子句筛选出 `description` 不为 `boring`，并且 `id` 为奇数的记录，然后使用 `ORDER BY` 子句对结果按照 `rating` 降序排序。
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -68,8 +72,8 @@ id 是该表的主键(具有唯一值的列)。
 # Write your MySQL query statement below
 SELECT *
 FROM Cinema
-WHERE description != 'boring' AND id % 2 = 1
-ORDER BY rating DESC;
+WHERE description != 'boring' AND id & 1 = 1
+ORDER BY 4 DESC;
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0620.Not Boring Movies/README_EN.md

@@ -56,6 +56,10 @@ We have three movies with odd-numbered IDs: 1, 3, and 5. The movie with ID = 3 i
 
 ## Solutions
 
+**Solution 1: Conditional Filtering + Sorting**
+
+We can use the `WHERE` clause to filter out the records where `description` is not `boring` and `id` is odd, and then use the `ORDER BY` clause to sort the result in descending order by `rating`.
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -64,8 +68,8 @@ We have three movies with odd-numbered IDs: 1, 3, and 5. The movie with ID = 3 i
 # Write your MySQL query statement below
 SELECT *
 FROM Cinema
-WHERE description != 'boring' AND id % 2 = 1
-ORDER BY rating DESC;
+WHERE description != 'boring' AND id & 1 = 1
+ORDER BY 4 DESC;
 ```
 
 <!-- tabs:end -->

solution/0600-0699/0620.Not Boring Movies/Solution.sql

@@ -1,5 +1,5 @@
 # Write your MySQL query statement below
 SELECT *
 FROM Cinema
-WHERE description != 'boring' AND id % 2 = 1
-ORDER BY rating DESC;
+WHERE description != 'boring' AND id & 1 = 1
+ORDER BY 4 DESC;

solution/1200-1299/1251.Average Selling Price/README.md

@@ -84,19 +84,24 @@ UnitsSold table:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：左连接 + 分组统计**
+
+我们可以使用左连接，将 `Prices` 表和 `UnitsSold` 表连接起来，连接条件为 `product_id` 相等，并且 `purchase_date` 在 `start_date` 和 `end_date` 之间。然后使用 `GROUP BY` 子句对 `product_id` 进行分组，使用 `AVG` 函数计算平均价格。注意，如果某个产品没有销售记录，那么 `AVG` 函数会返回 `NULL`，因此我们可以使用 `IFNULL` 函数将其转换为 $0$。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
+# Write your MySQL query statement below
 SELECT
     p.product_id,
-    IFNULL(ROUND(SUM(units * price) / SUM(units), 2), 0) AS average_price
+    IFNULL(ROUND(SUM(price * units) / SUM(units), 2), 0) AS average_price
 FROM
     Prices AS p
     LEFT JOIN UnitsSold AS u
         ON p.product_id = u.product_id AND purchase_date BETWEEN start_date AND end_date
-GROUP BY product_id;
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

solution/1200-1299/1251.Average Selling Price/README_EN.md

@@ -82,19 +82,24 @@ Average selling price for product 2 = ((200 * 15) + (30 * 30)) / 230 = 16.96
 
 ## Solutions
 
+**Solution 1: Left Join + Grouping**
+
+We can use a left join to join the `Prices` table and the `UnitsSold` table on `product_id`, and the condition that `purchase_date` is between `start_date` and `end_date`. Then, we can use `GROUP BY` to group by `product_id` for aggregation, and use the `AVG` function to calculate the average price. Note that if a product has no sales records, the `AVG` function will return `NULL`, so we can use the `IFNULL` function to convert it to $0$.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
+# Write your MySQL query statement below
 SELECT
     p.product_id,
-    IFNULL(ROUND(SUM(units * price) / SUM(units), 2), 0) AS average_price
+    IFNULL(ROUND(SUM(price * units) / SUM(units), 2), 0) AS average_price
 FROM
     Prices AS p
     LEFT JOIN UnitsSold AS u
         ON p.product_id = u.product_id AND purchase_date BETWEEN start_date AND end_date
-GROUP BY product_id;
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

solution/1200-1299/1251.Average Selling Price/Solution.sql

@@ -1,8 +1,9 @@
+# Write your MySQL query statement below
 SELECT
     p.product_id,
-    IFNULL(ROUND(SUM(units * price) / SUM(units), 2), 0) AS average_price
+    IFNULL(ROUND(SUM(price * units) / SUM(units), 2), 0) AS average_price
 FROM
     Prices AS p
     LEFT JOIN UnitsSold AS u
         ON p.product_id = u.product_id AND purchase_date BETWEEN start_date AND end_date
-GROUP BY product_id;
+GROUP BY 1;

solution/1200-1299/1280.Students and Examinations/README.md

@@ -123,23 +123,23 @@ John  参加了数学、物理、编程测试各 1 次。
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：两次连接 + 分组统计**
+
+我们可以先连接 `Students` 表和 `Subjects` 表，得到所有学生和所有科目的组合，然后再连接 `Examinations` 表，连接条件为 `student_id` 和 `subject_name`，这样就得到了每个学生参加每一门科目测试的次数，最后按 `student_id` 和 `subject_name` 分组统计即可。
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    a.student_id,
-    student_name,
-    b.subject_name,
-    COUNT(c.subject_name) AS attended_exams
+SELECT student_id, student_name, subject_name, COUNT(e.student_id) AS attended_exams
 FROM
-    Students AS a
-    CROSS JOIN Subjects AS b
-    LEFT JOIN Examinations AS c ON a.student_id = c.student_id AND b.subject_name = c.subject_name
-GROUP BY a.student_id, b.subject_name
-ORDER BY a.student_id, b.subject_name;
+    Students
+    JOIN Subjects
+    LEFT JOIN Examinations AS e USING (student_id, subject_name)
+GROUP BY 1, 3
+ORDER BY 1, 3;
 ```
 
 <!-- tabs:end -->

solution/1200-1299/1280.Students and Examinations/README_EN.md

@@ -120,23 +120,23 @@ John attended the Math exam 1 time, the Physics exam 1 time, and the Programming
 
 ## Solutions
 
+**Solution 1: Two Joins + Grouping**
+
+We can first join the `Students` table and the `Subjects` table to obtain all combinations of students and subjects, and then join the `Examinations` table with the condition of `student_id` and `subject_name`. This way, we can get the number of times each student has taken each subject's test. Finally, we can group by `student_id` and `subject_name` to count the number of times each student has taken each subject's test.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-SELECT
-    a.student_id,
-    student_name,
-    b.subject_name,
-    COUNT(c.subject_name) AS attended_exams
+SELECT student_id, student_name, subject_name, COUNT(e.student_id) AS attended_exams
 FROM
-    Students AS a
-    CROSS JOIN Subjects AS b
-    LEFT JOIN Examinations AS c ON a.student_id = c.student_id AND b.subject_name = c.subject_name
-GROUP BY a.student_id, b.subject_name
-ORDER BY a.student_id, b.subject_name;
+    Students
+    JOIN Subjects
+    LEFT JOIN Examinations AS e USING (student_id, subject_name)
+GROUP BY 1, 3
+ORDER BY 1, 3;
 ```
 
 <!-- tabs:end -->

solution/1200-1299/1280.Students and Examinations/Solution.sql

@@ -1,12 +1,8 @@
 # Write your MySQL query statement below
-SELECT
-    a.student_id,
-    student_name,
-    b.subject_name,
-    COUNT(c.subject_name) AS attended_exams
+SELECT student_id, student_name, subject_name, COUNT(e.student_id) AS attended_exams
 FROM
-    Students AS a
-    CROSS JOIN Subjects AS b
-    LEFT JOIN Examinations AS c ON a.student_id = c.student_id AND b.subject_name = c.subject_name
-GROUP BY a.student_id, b.subject_name
-ORDER BY a.student_id, b.subject_name;
+    Students
+    JOIN Subjects
+    LEFT JOIN Examinations AS e USING (student_id, subject_name)
+GROUP BY 1, 3
+ORDER BY 1, 3;

solution/1900-1999/1934.Confirmation Rate/README.md

@@ -91,6 +91,10 @@ Confirmations 表:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：左连接 + 分组统计**
+
+我们可以使用左连接，将 `Signups` 表和 `Confirmations` 表按照 `user_id` 进行连接，然后使用 `GROUP BY` 对 `user_id` 进行分组统计。
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -101,14 +105,11 @@ Confirmations 表:
 # Write your MySQL query statement below
 SELECT
     user_id,
-    ROUND(
-        SUM(IF(action = 'confirmed', 1, 0)) / COUNT(1),
-        2
-    ) AS confirmation_rate
+    ROUND(IFNULL(SUM(action = 'confirmed') / COUNT(1), 0), 2) AS confirmation_rate
 FROM
-    Signups
+    SignUps
     LEFT JOIN Confirmations USING (user_id)
-GROUP BY user_id;
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->

solution/1900-1999/1934.Confirmation Rate/README_EN.md

@@ -89,6 +89,10 @@ User 2 made 2 requests where one was confirmed and the other timed out. The conf
 
 ## Solutions
 
+**Solution 1: Left Join + Grouping**
+
+We can use a left join to join the `Signups` table and the `Confirmations` table on `user_id`, and then use `GROUP BY` to group by `user_id` for aggregation.
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -97,14 +101,11 @@ User 2 made 2 requests where one was confirmed and the other timed out. The conf
 # Write your MySQL query statement below
 SELECT
     user_id,
-    ROUND(
-        SUM(IF(action = 'confirmed', 1, 0)) / COUNT(1),
-        2
-    ) AS confirmation_rate
+    ROUND(IFNULL(SUM(action = 'confirmed') / COUNT(1), 0), 2) AS confirmation_rate
 FROM
-    Signups
+    SignUps
     LEFT JOIN Confirmations USING (user_id)
-GROUP BY user_id;
+GROUP BY 1;
 ```
 
 <!-- tabs:end -->