Leetcode #570: Managers with at Least 5 Direct Reports
In this guide, we solve Leetcode #570 Managers with at Least 5 Direct Reports in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Database
Intuition
The task is relational in nature, which maps cleanly to DataFrame operations in Python.
By treating tables as DataFrames, joins and group-bys become concise and readable.
Approach
Load the inputs as DataFrames and apply the appropriate merge, filter, or group-by.
Select or rename the columns to match the required output.
Steps:
- Load inputs as DataFrames.
- Apply merge/groupby/filter operations.
- Select the output columns.
Example
+-------------+---------+
| 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.
Python Solution
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
Complexity
The time complexity is O(n log n) (typical). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.