[SQL문제풀기] Managers with at Least 5 Direct Reports

Oct 20, 2025

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문제

Managers with at Least 5 Direct Reports - LeetCode

Can you solve this real interview question? 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 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 | +------+

https://leetcode.com/problems/managers-with-at-least-5-direct-reports/description/?envType=study-plan-v2&envId=top-sql-50

MySQL

내가 작성한 정답

: where이 좀 더 빠르다


select name
from Employee 
where id in (select managerId 
             from Employee
             group by managerId 
             having count(id) >= 5);


select b.name
from Employee b
left join Employee e
on b.id = e.managerId
group by b.id
having count(e.id) >= 5;