# 181. Employees Earning More Than Their Managers

## Description

Table: Employee

+-------------+---------+
| 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.


Write a solution to find the employees who earn more than their managers.

Return the result table in any order.

The result format is in the following example.

Example 1:

Input:
Employee table:
+----+-------+--------+-----------+
| id | name  | salary | managerId |
+----+-------+--------+-----------+
| 1  | Joe   | 70000  | 3         |
| 2  | Henry | 80000  | 4         |
| 3  | Sam   | 60000  | Null      |
| 4  | Max   | 90000  | Null      |
+----+-------+--------+-----------+
Output:
+----------+
| Employee |
+----------+
| Joe      |
+----------+
Explanation: Joe is the only employee who earns more than his manager.


## Solutions

• import pandas as pd

def find_employees(employee: pd.DataFrame) -> pd.DataFrame:
df = employee.merge(right=employee, how="left", left_on="managerId", right_on="id")
emp = df[df["salary_x"] > df["salary_y"]]["name_x"]

return pd.DataFrame({"Employee": emp})


• SELECT e1.Name FROM Employee e1
JOIN Employee e2 ON e1.ManagerId = e2.Id
WHERE e1.Salary > e2.Salary;

--

SELECT e1.Name FROM Employee e1, Employee e2
WHERE e1.ManagerId = e2.Id AND e1.Salary > e2.Salary;

--

SELECT Name AS Employee
FROM Employee AS Curr
WHERE
Salary > (
SELECT Salary
FROM Employee
WHERE Id = Curr.ManagerId
);