Leecode_SQL50_570. Managers with at Least 5 Direct Reports

Leecode

570. Managers with at Least 5 Direct Reports

Problem description

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.

Example 1:

Input:
Employee table:

id	name	department	managerId
101	John	A	null
102	Dan	A	101
103	James	A	101
104	Amy	A	101
105	Anne	A	101
106	Ron	B	101

Output
:

name
John

My solution

WITH a AS(
    SELECT m.name, COUNT(e.managerId) AS coun
    FROM Employee e
        JOIN Employee m
            ON e.managerId = m.id
    GROUP BY e.managerId
)
SELECT a.name
FROM a
WHERE coun >= 5

My failed solution, test case not passed when name=null:

# Write your MySQL query statement below
WITH a AS (
    SELECT COUNT(id) AS count_connected, managerId
    FROM Employee e
    GROUP BY managerId
    HAVING COUNT(name) >= 5
)
SELECT e.name
FROM a
    JOIN Employee e
        ON a.managerId = e.id