feat: update solutions to lc problems: No.1907,1978 (doocs#1878)

* No.1907.Count Salary Categories
* No.1978.Employees Whose Manager Left the Company

Author: yanglbme

solution/1900-1999/1907.Count Salary Categories/README.md

@@ -69,6 +69,14 @@ Accounts 表:
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：构建临时表 + 分组统计 + 左连接**
+
+我们可以先构建一个临时表，包含所有工资类别，然后再统计每个工资类别的银行账户数量。最后我们使用左连接，将临时表和统计结果表连接起来，这样就可以保证结果表中包含所有工资类别。
+
+**方法二：筛选 + 合并**
+
+我们可以分别筛选出每个工资类别的银行账户数量，然后再将结果合并起来。这里我们使用 `UNION` 来合并结果。
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -94,12 +102,24 @@ WITH
             END AS category,
             COUNT(1) AS accounts_count
         FROM Accounts
-        GROUP BY category
+        GROUP BY 1
     )
-SELECT s.category, IFNULL(accounts_count, 0) AS accounts_count
+SELECT category, IFNULL(accounts_count, 0) AS accounts_count
 FROM
-    S AS s
-    LEFT JOIN T AS t ON s.category = t.category;
+    S
+    LEFT JOIN T USING (category);
+```
+
+```sql
+# Write your MySQL query statement below
+SELECT 'Low Salary' AS category, IFNULL(SUM(income < 20000), 0) AS accounts_count FROM Accounts
+UNION
+SELECT
+    'Average Salary' AS category,
+    IFNULL(SUM(income BETWEEN 20000 AND 50000), 0) AS accounts_count
+FROM Accounts
+UNION
+SELECT 'High Salary' AS category, IFNULL(SUM(income > 50000), 0) AS accounts_count FROM Accounts;
 ```
 
 <!-- tabs:end -->

solution/1900-1999/1907.Count Salary Categories/README_EN.md

@@ -63,6 +63,14 @@ High Salary: Accounts 3, 6, and 8.
 
 ## Solutions
 
+**Solution 1: Temporary Table + Grouping + Left Join**
+
+We can first create a temporary table containing all salary categories, and then count the number of bank accounts for each salary category. Finally, we use a left join to connect the temporary table with the result table to ensure that the result table contains all salary categories.
+
+**Solution 2: Filtering + Merging**
+
+We can filter out the number of bank accounts for each salary category separately, and then merge the results. Here, we use `UNION` to merge the results.
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -86,12 +94,24 @@ WITH
             END AS category,
             COUNT(1) AS accounts_count
         FROM Accounts
-        GROUP BY category
+        GROUP BY 1
     )
-SELECT s.category, IFNULL(accounts_count, 0) AS accounts_count
+SELECT category, IFNULL(accounts_count, 0) AS accounts_count
 FROM
-    S AS s
-    LEFT JOIN T AS t ON s.category = t.category;
+    S
+    LEFT JOIN T USING (category);
+```
+
+```sql
+# Write your MySQL query statement below
+SELECT 'Low Salary' AS category, IFNULL(SUM(income < 20000), 0) AS accounts_count FROM Accounts
+UNION
+SELECT
+    'Average Salary' AS category,
+    IFNULL(SUM(income BETWEEN 20000 AND 50000), 0) AS accounts_count
+FROM Accounts
+UNION
+SELECT 'High Salary' AS category, IFNULL(SUM(income > 50000), 0) AS accounts_count FROM Accounts;
 ```
 
 <!-- tabs:end -->

solution/1900-1999/1907.Count Salary Categories/Solution.sql

@@ -16,9 +16,9 @@ WITH
             END AS category,
             COUNT(1) AS accounts_count
         FROM Accounts
-        GROUP BY category
+        GROUP BY 1
     )
-SELECT s.category, IFNULL(accounts_count, 0) AS accounts_count
+SELECT category, IFNULL(accounts_count, 0) AS accounts_count
 FROM
-    S AS s
-    LEFT JOIN T AS t ON s.category = t.category;
+    S
+    LEFT JOIN T USING (category);

solution/1900-1999/1978.Employees Whose Manager Left the Company/README.md

@@ -64,6 +64,14 @@ Joziah 的上级经理是 6 号员工，他已经离职，因为员工表里面
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：左连接**
+
+我们可以使用左连接，将员工表自身连接一次，然后筛选出薪水小于 30000 的员工，且有上级经理，但是上级经理已经离职的员工。
+
+**方法二：子查询**
+
+我们也可以使用子查询，先找出所有已经离职的经理，然后再找出薪水小于 30000 的员工，且他们的上级经理不在已经离职的经理列表中。
+
 <!-- tabs:start -->
 
 ### **SQL**
@@ -72,20 +80,20 @@ Joziah 的上级经理是 6 号员工，他已经离职，因为员工表里面
 
 ```sql
 # Write your MySQL query statement below
-SELECT a.employee_id
+SELECT e1.employee_id
 FROM
-    Employees AS a
-    LEFT JOIN Employees AS b ON a.manager_id = b.employee_id
-WHERE b.employee_id IS NULL AND a.salary < 30000 AND a.manager_id IS NOT NULL
-ORDER BY a.employee_id;
+    Employees AS e1
+    LEFT JOIN Employees AS e2 ON e1.manager_id = e2.employee_id
+WHERE e1.salary < 30000 AND e1.manager_id IS NOT NULL AND e2.employee_id IS NULL
+ORDER BY 1;
 ```
 
 ```sql
 # Write your MySQL query statement below
 SELECT employee_id
 FROM Employees
 WHERE salary < 30000 AND manager_id NOT IN (SELECT employee_id FROM Employees)
-ORDER BY employee_id;
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/1900-1999/1978.Employees Whose Manager Left the Company/README_EN.md

@@ -58,26 +58,34 @@ Joziah's manager is employee 6, who left the company because there is no row
 
 ## Solutions
 
+**Solution 1: Left Join**
+
+We can use a left join to connect the employee table with itself, and then filter out the employees whose salary is less than $30000$ and have a superior manager who has left the company.
+
+**Solution 2: Subquery**
+
+We can also use a subquery to first find all the managers who have left the company, and then find the employees whose salary is less than $30000$ and whose superior manager is not in the list of managers who have left the company.
+
 <!-- tabs:start -->
 
 ### **SQL**
 
 ```sql
 # Write your MySQL query statement below
-SELECT a.employee_id
+SELECT e1.employee_id
 FROM
-    Employees AS a
-    LEFT JOIN Employees AS b ON a.manager_id = b.employee_id
-WHERE b.employee_id IS NULL AND a.salary < 30000 AND a.manager_id IS NOT NULL
-ORDER BY a.employee_id;
+    Employees AS e1
+    LEFT JOIN Employees AS e2 ON e1.manager_id = e2.employee_id
+WHERE e1.salary < 30000 AND e1.manager_id IS NOT NULL AND e2.employee_id IS NULL
+ORDER BY 1;
 ```
 
 ```sql
 # Write your MySQL query statement below
 SELECT employee_id
 FROM Employees
 WHERE salary < 30000 AND manager_id NOT IN (SELECT employee_id FROM Employees)
-ORDER BY employee_id;
+ORDER BY 1;
 ```
 
 <!-- tabs:end -->

solution/1900-1999/1978.Employees Whose Manager Left the Company/Solution.sql

@@ -1,7 +1,7 @@
 # Write your MySQL query statement below
-SELECT a.employee_id
+SELECT e1.employee_id
 FROM
-    Employees AS a
-    LEFT JOIN Employees AS b ON a.manager_id = b.employee_id
-WHERE b.employee_id IS NULL AND a.salary < 30000 AND a.manager_id IS NOT NULL
-ORDER BY a.employee_id;
+    Employees AS e1
+    LEFT JOIN Employees AS e2 ON e1.manager_id = e2.employee_id
+WHERE e1.salary < 30000 AND e1.manager_id IS NOT NULL AND e2.employee_id IS NULL
+ORDER BY 1;