570. Managers with at Least 5 Direct Reports
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 | +------+
Solutions
Solution 1: Grouping and Joining
Thinking
A manager is an employee with at least five direct reports. Count by managerId, then join back for names.
Group-count reports, keep those with \(\ge 5\), and join Employee on id. Aggregation and the name lookup stay separate.
We can first count the number of direct subordinates for each manager, and then join the Employee table to find the managers whose number of direct subordinates is greater than or equal to \(5\).
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