Problem
Table: Employee
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| id | int |
| name | varchar |
| department | varchar |
| managerId | int |
+-------------+---------+
id
is the primary key (column with unique values) for this table.
Each row of this table indicates the name of an employee, their department, and the id of their manager.
If managerId is null, then the employee does not have a manager.
No employee will be the manager of themself.
Write a solution to find managers with at least five direct reports.
Return the result table in any order.
The result format is in the following example.
Examples
Example 1:
Input: Employee table:
+-----+-------+------------+-----------+
| id | name | department | managerId |
+-----+-------+------------+-----------+
| 101 | John | A | None |
| 102 | Dan | A | 101 |
| 103 | James | A | 101 |
| 104 | Amy | A | 101 |
| 105 | Anne | A | 101 |
| 106 | Ron | B | 101 |
+-----+-------+------------+-----------+
Output:
+------+
| name |
+------+
| John |
+------+
Solution
Method 1 - CTE and Join
Code
SQL
WITH m5 AS
(
select managerId, count(id) as cnt FROM employee
GROUP BY managerId
HAVING cnt >= 5
)
SELECT name FROM employee e
JOIN m5 ON m5.managerId = e.id
Method 2 - Self Join
Code
SQL
SELECT e1.name
FROM Employee e1
JOIN Employee e2
ON e1.id = e2.managerId
GROUP BY e1.id
HAVING count(e2.managerId) >= 5;
Python
import pandas as pd
def find_managers(employee: pd.DataFrame) -> pd.DataFrame:
managers = employee.groupby(
'managerId', as_index=False
).agg(
reporting=('id', 'count'),
).query(
'5 <= reporting'
)['managerId']
return employee[
employee['id'].isin(managers)
][['name']]