Last updated on January 22nd, 2025 at 11:16 pm
Here, we see the All People Report to the Given Manager LeetCode Solution. This Leetcode problem is solved using MySQL and Pandas.
List of all LeetCode Solution
Level of Question
Medium

All People Report to the Given Manager LeetCode Solution
Table of Contents
1. Problem Statement
Column Name | Type |
employee_id | int |
employee_name | varchar |
manager_id | int |
employee_id is the primary key for this table.
Each row of this table indicates that the employee with ID employee_id and name employee_name reports his work to his/her direct manager with manager_id
The head of the company is the employee with employee_id = 1.
Write an SQL query to find employee_id of all employees that directly or indirectly report their work to the head of the company. The indirect relation between managers will not exceed 3 managers as the company is small. Return result table in any order without duplicates.
The result format is in the following example.
Example 1:
Input:
employee_id | employee_name | manager_id |
1 | Boss | 1 |
3 | Alice | 3 |
2 | Bob | 1 |
4 | Daniel | 2 |
7 | Luis | 4 |
8 | Jhon | 3 |
9 | Angela | 8 |
77 | Robert | 1 |
Output:
employee_id |
2 |
77 |
4 |
7 |
Explanation:
The head of the company is the employee with employee_id 1.
The employees with employee_id 2 and 77 report their work directly to the head of the company.
The employee with employee_id 4 report his work indirectly to the head of the company 4 –> 2 –> 1.
The employee with employee_id 7 report his work indirectly to the head of the company 7 –> 4 –> 2 –> 1.
The employees with employee_id 3, 8 and 9 don’t report their work to head of company directly or indirectly.
2. Code Implementation in Different Languages
2.1 All People Report to the Given Manager MySQL
select distinct e1.employee_id from Employees as e1 inner join Employees as e2 inner join Employees as e3 on e1.manager_id = e2.employee_id and e2.manager_id = e3.employee_id where e1.employee_id <> 1 and ( e1.manager_id = 1 or e2.manager_id = 1 or e3.manager_id = 1 );