task: #1978

Author: Ivanbyone

README.md

@@ -66,6 +66,7 @@ Have a good contributing!
     - [1729. Find Followers Count](./leetcode/easy/1729.%20Find%20Followers%20Count.sql)
     - [1741. Find Total Time Spent by Each Employee](./leetcode/easy/1741.%20Find%20Total%20Time%20Spent%20by%20Each%20Employee.sql)
     - [1965. Employees With Missing Information](./leetcode/easy/1965.%20Employees%20With%20Missing%20Information.sql)
+    - [1978. Employees Whose Manager Left the Company](./leetcode/easy/1978.%20Employees%20Whose%20Manager%20Left%20the%20Company.sql)
     - [2356. Number of Unique Subjects Taught by Each Teacher](./leetcode/easy/2356.%20Number%20of%20Unique%20Subjects%20Taught%20by%20Each%20Teacher.sql)
 2. [Medium](./leetcode/medium/)
     - [176. Second Highest Salary](./leetcode/medium/176.%20Second%20Highest%20Salary.sql)

leetcode/easy/1978. Employees Whose Manager Left the Company.sql

@@ -0,0 +1,31 @@
+/*
+Question 1978. Employees Whose Manager Left the Company
+Link: https://leetcode.com/problems/employees-whose-manager-left-the-company/description/?envType=study-plan-v2&envId=top-sql-50
+
+Table: Employees
+
++-------------+----------+
+| Column Name | Type     |
++-------------+----------+
+| employee_id | int      |
+| name        | varchar  |
+| manager_id  | int      |
+| salary      | int      |
++-------------+----------+
+In SQL, employee_id is the primary key for this table.
+This table contains information about the employees, their salary, and the ID of their manager. Some employees do not have a manager (manager_id is null). 
+ 
+
+Find the IDs of the employees whose salary is strictly less than $30000 and whose manager left the company. When a manager leaves the company, their information is deleted from the Employees table, but the reports still have their manager_id set to the manager that left.
+
+Return the result table ordered by employee_id.
+*/
+
+SELECT e1.employee_id
+FROM Employees AS e1
+WHERE
+    e1.salary < 30000 AND e1.manager_id NOT IN (
+        SELECT e2.employee_id
+        FROM Employees AS e2
+    )
+ORDER BY e1.employee_id ASC