task: #3482 CHALLENGE COMPLETED (all SQL freemium tasks have solutions)

Author: Ivanbyone

README.md

@@ -199,6 +199,7 @@ Useful for preparing for technical interviews and improving your SQL skills.
     - [601. Human Traffic of Stadium](./leetcode/hard/601.%20Human%20Traffic%20of%20Stadium.sql)
     - [3374. First Letter Capitalization II](./leetcode/hard/3374.%20First%20Letter%20Capitalization%20II.sql)
     - [3451. Find Invalid IP Addresses](./leetcode/hard/3451.%20Find%20Invalid%20IP%20Addresses.sql)
+    - [3482. Analyze Organization Hierarchy](./leetcode/hard/3482.%20Analyze%20Organization%20Hierarchy.sql)
     - [3554. Find Category Recommendation Pairs](./leetcode/hard/3554.%20Find%20Category%20Recommendation%20Pairs.sql)
 
 ## Contributing

leetcode/hard/3482. Analyze Organization Hierarchy.sql

@@ -0,0 +1,81 @@
+/*
+Question 3482. Analyze Organization Hierarchy
+Link: https://leetcode.com/problems/analyze-organization-hierarchy/description/?envType=problem-list-v2&envId=database
+
+Table: Employees
+
++----------------+---------+
+| Column Name    | Type    | 
++----------------+---------+
+| employee_id    | int     |
+| employee_name  | varchar |
+| manager_id     | int     |
+| salary         | int     |
+| department     | varchar |
++----------------+----------+
+employee_id is the unique key for this table.
+Each row contains information about an employee, including their ID, name, their manager's ID, salary, and department.
+manager_id is null for the top-level manager (CEO).
+Write a solution to analyze the organizational hierarchy and answer the following:
+
+Hierarchy Levels: For each employee, determine their level in the organization (CEO is level 1, employees reporting directly to the CEO are level 2, and so on).
+Team Size: For each employee who is a manager, count the total number of employees under them (direct and indirect reports).
+Salary Budget: For each manager, calculate the total salary budget they control (sum of salaries of all employees under them, including indirect reports, plus their own salary).
+Return the result table ordered by the result ordered by level in ascending order, then by budget in descending order, and finally by employee_name in ascending order.
+*/
+
+WITH RECURSIVE cte AS (
+    SELECT
+        e1.employee_id,
+        e1.employee_name,
+        e1.manager_id,
+        e1.salary,
+        1 AS rank
+    FROM Employees AS e1
+    WHERE e1.manager_id IS NULL
+
+    UNION ALL 
+
+    SELECT
+        e2.employee_id,
+        e2.employee_name,
+        e2.manager_id,
+        e2.salary,
+        c.rank + 1 AS rank
+    FROM Employees AS e2
+    INNER JOIN 
+        cte AS c
+        ON e2.manager_id = c.employee_id
+),
+
+teamsize AS (
+    SELECT
+        e3.employee_id AS manager_id,
+        e3.employee_id AS member_id,
+        e3.salary AS member_salary
+    FROM Employees AS e3
+
+    UNION ALL
+
+    SELECT 
+        ts.manager_id,
+        e4.employee_id AS member_id,
+        e4.salary AS member_salary
+    FROM Employees AS e4
+    JOIN --noqa: AM05
+        teamsize AS ts
+        ON e4.manager_id = ts.member_id
+)
+
+SELECT
+    c.employee_id,
+    c.employee_name,
+    c.rank AS level, --noqa: RF04
+    COUNT(*) - 1 AS team_size,
+    SUM(ts.member_salary) AS budget
+FROM cte AS c
+INNER JOIN 
+    teamsize AS ts
+    ON c.employee_id = ts.manager_id
+GROUP BY c.employee_id, c.employee_name, c.rank
+ORDER BY level ASC, budget DESC, c.employee_name ASC