2988. Manager of the Largest Department 🔒

Description

Table: Employees

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| emp_id      | int     |
| emp_name    | varchar |
| dep_id      | int     |
| position    | varchar |
+-------------+---------+
emp_id is column of unique values for this table.
This table contains emp_id, emp_name, dep_id, and position.

Write a solution to find the name of the manager from the largest department. There may be multiple largest departments when the number of employees in those departments is the same.

Return the result table sorted by dep_id in ascending order.

The result format is in the following example.

 

Example 1:

Input: 
Employees table:
+--------+----------+--------+---------------+
| emp_id | emp_name | dep_id | position      | 
+--------+----------+--------+---------------+
| 156    | Michael  | 107    | Manager       |
| 112    | Lucas    | 107    | Consultant    |    
| 8      | Isabella | 101    | Manager       | 
| 160    | Joseph   | 100    | Manager       | 
| 80     | Aiden    | 100    | Engineer      | 
| 190    | Skylar   | 100    | Freelancer    | 
| 196    | Stella   | 101    | Coordinator   |
| 167    | Audrey   | 100    | Consultant    |
| 97     | Nathan   | 101    | Supervisor    |
| 128    | Ian      | 101    | Administrator |
| 81     | Ethan    | 107    | Administrator |
+--------+----------+--------+---------------+
Output
+--------------+--------+
| manager_name | dep_id | 
+--------------+--------+
| Joseph       | 100    | 
| Isabella     | 101    | 
+--------------+--------+
Explanation
- Departments with IDs 100 and 101 each has a total of 4 employees, while department 107 has 3 employees. Since both departments 100 and 101 have an equal number of employees, their respective managers will be included.
Output table is ordered by dep_id in ascending order.

Solutions

Solution 1: Grouping + Equi-Join + Subquery

We can first count the number of employees in each department, denoted as table T. Then we join T with the Employees table, with the join condition being T.dep_id = Employees.dep_id and Employees.position = 'Manager'. This way, we can get the manager of each department. Finally, we filter out the department with the most employees.

MySQL

# Write your MySQL query statement below
WITH
    T AS (
        SELECT dep_id, COUNT(1) AS cnt
        FROM Employees
        GROUP BY 1
    )
SELECT emp_name AS manager_name, t.dep_id
FROM
    T AS t
    JOIN Employees AS e ON t.dep_id = e.dep_id AND e.position = 'Manager'
WHERE cnt = (SELECT MAX(cnt) FROM T)
ORDER BY 2;

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