0570-managers-with-at-least-5-direct-reports.sql

-- 570. Managers with at Least 5 Direct Reports
-- Medium
-- https://leetcode.com/problems/managers-with-at-least-5-direct-reports

/*
Table: Employee
+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| name        | varchar |
| department  | varchar |
| managerId   | int     |
+-------------+---------+
id is the primary key column 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.

Find the managers with at least five direct reports.
Return the result table in any order.
The result format is in the following example.

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 |
+------+
*/

SELECT e1.name
FROM Employee e1
JOIN (
  SELECT managerId
  FROM Employee
  GROUP BY managerId
  HAVING COUNT(id) >= 5
) e2
ON e1.id = e2.managerId;