# [181. Employees Earning More Than Their Managers](https://leetcode.com/problems/employees-earning-more-than-their-managers) [中文文档](/solution/0100-0199/0181.Employees%20Earning%20More%20Than%20Their%20Managers/README.md) ## 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 ### Solution 1 ```python 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}) ``` ```sql SELECT Name AS Employee FROM Employee AS Curr WHERE Salary > ( SELECT Salary FROM Employee WHERE Id = Curr.ManagerId ); ``` ### Solution 2 ```sql # Write your MySQL query statement below SELECT e1.name AS Employee FROM Employee AS e1 JOIN Employee AS e2 ON e1.managerId = e2.id WHERE e1.salary > e2.salary; ```