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
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.
Pythonsql
import pandas as pd def find_managers(employee: pd.DataFrame) -> pd.DataFrame: # Group the employees by managerId and count the number of direct reports manager_report_count = ( employee.groupby("managerId").size().reset_index(name="directReports") ) # Filter managers with at least five direct reports result = manager_report_count[manager_report_count["directReports"] >= 5] # Merge with the Employee table to get the names of these managers result = result.merge( employee[["id", "name"]], left_on="managerId", right_on="id", how="inner" ) # Select only the 'name' column and drop the 'id' and 'directReports' columns result = result[["name"]] return result(code-box)
