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

Solution 1: Self-Join + Conditional Filtering

We can find employees' salaries and their managers' salaries by self-joining the Employee table, then filter out employees whose salaries are higher than their managers' salaries.

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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

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# Write your MySQL query statement below
SELECT e1.name Employee
FROM
    Employee e1
    JOIN Employee e2 ON e1.managerId = e2.id
WHERE e1.salary > e2.salary;

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