Employees Earning More Than Their Managers
<p>Table: <code>Employee</code></p>

<pre>
+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| name        | varchar |
| salary      | int     |
| managerId   | int     |
+-------------+---------+
id is the primary key (column with unique values) for this table.
Each row of this table indicates the ID of an employee, their name, salary, and the ID of their manager.
</pre>

<p>&nbsp;</p>

<p>Write a solution&nbsp;to find the employees who earn more than their managers.</p>

<p>Return the result table in <strong>any order</strong>.</p>

<p>The result format is in the following example.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> 
Employee table:
+----+-------+--------+-----------+
| id | name  | salary | managerId |
+----+-------+--------+-----------+
| 1  | Joe   | 70000  | 3         |
| 2  | Henry | 80000  | 4         |
| 3  | Sam   | 60000  | Null      |
| 4  | Max   | 90000  | Null      |
+----+-------+--------+-----------+
<strong>Output:</strong> 
+----------+
| Employee |
+----------+
| Joe      |
+----------+
<strong>Explanation:</strong> Joe is the only employee who earns more than his manager.
</pre>
Database
import pandas as pd


def find_employees(employee: pd.DataFrame) -> pd.DataFrame:
    merged = employee.merge(
        employee, left_on="managerId", right_on="id", suffixes=("", "_manager")
    )
    result = merged[merged["salary"] > merged["salary_manager"]][["name"]]
    result.columns = ["Employee"]
    return result
