Managers with at Least 5 Direct Reports - Leetcode 570 | Week 4 | Qs 16

Leetcode - 570. Managers with at Least 5 Direct Reports
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SQL Question
Write an SQL query to report the managers with at least five direct reports.
SQL Schema
Create table If Not Exists Employee (id int, name varchar(255), department varchar(255), managerId int)
Truncate table Employee
insert into Employee (id, name, department, managerId) values ('101', 'John', 'A', 'None')
insert into Employee (id, name, department, managerId) values ('102', 'Dan', 'A', '101')
insert into Employee (id, name, department, managerId) values ('103', 'James', 'A', '101')
insert into Employee (id, name, department, managerId) values ('104', 'Amy', 'A', '101')
insert into Employee (id, name, department, managerId) values ('105', 'Anne', 'A', '101')
insert into Employee (id, name, department, managerId) values ('106', 'Ron', 'B', '101')
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.
Write an SQL query to report the managers with at least five direct reports.
Return the result table in any order.
The query 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 |
+------+

[@vishalbhandari8875]

WHAT AN EXPLANATION. KEEP UP THE GOOD WORK

[@adityasaini8437]

I am have a doubt , how inner join working here ? as it defines inner join tables on matching column values , now according to your on cond. e. managerID = m.id , we have only one matching column then why we are getting all the value as output