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185. Department Top Three Salaries

Description

Table: Employee

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| id           | int     |
| name         | varchar |
| salary       | int     |
| departmentId | int     |
+--------------+---------+
id is the primary key (column with unique values) for this table.
departmentId is a foreign key (reference column) of the ID from the Department table.
Each row of this table indicates the ID, name, and salary of an employee. It also contains the ID of their department.

 

Table: Department

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| name        | varchar |
+-------------+---------+
id is the primary key (column with unique values) for this table.
Each row of this table indicates the ID of a department and its name.

 

A company's executives are interested in seeing who earns the most money in each of the company's departments. A high earner in a department is an employee who has a salary in the top three unique salaries for that department.

Write a solution to find the employees who are high earners in each of the departments.

Return the result table in any order.

The result format is in the following example.

 

Example 1:

Input: 
Employee table:
+----+-------+--------+--------------+
| id | name  | salary | departmentId |
+----+-------+--------+--------------+
| 1  | Joe   | 85000  | 1            |
| 2  | Henry | 80000  | 2            |
| 3  | Sam   | 60000  | 2            |
| 4  | Max   | 90000  | 1            |
| 5  | Janet | 69000  | 1            |
| 6  | Randy | 85000  | 1            |
| 7  | Will  | 70000  | 1            |
+----+-------+--------+--------------+
Department table:
+----+-------+
| id | name  |
+----+-------+
| 1  | IT    |
| 2  | Sales |
+----+-------+
Output: 
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| IT         | Joe      | 85000  |
| IT         | Randy    | 85000  |
| IT         | Will     | 70000  |
| Sales      | Henry    | 80000  |
| Sales      | Sam      | 60000  |
+------------+----------+--------+
Explanation: 
In the IT department:
- Max earns the highest unique salary
- Both Randy and Joe earn the second-highest unique salary
- Will earns the third-highest unique salary

In the Sales department:
- Henry earns the highest salary
- Sam earns the second-highest salary
- There is no third-highest salary as there are only two employees

Solutions

  • import pandas as pd
    
    
    def top_three_salaries(
        employee: pd.DataFrame, department: pd.DataFrame
    ) -> pd.DataFrame:
        salary_cutoff = (
            employee.drop_duplicates(["salary", "departmentId"])
            .groupby("departmentId")["salary"]
            .nlargest(3)
            .groupby("departmentId")
            .min()
        )
        employee["Department"] = department.set_index("id")["name"][
            employee["departmentId"]
        ].values
        employee["cutoff"] = salary_cutoff[employee["departmentId"]].values
        return employee[employee["salary"] >= employee["cutoff"]].rename(
            columns={"name": "Employee", "salary": "Salary"}
        )[["Department", "Employee", "Salary"]]
    
    
  • # Write your MySQL query statement below
    SELECT
        d.Name AS 'Department', e1.Name AS 'Employee', e1.Salary
    FROM
        Employee e1
    JOIN
        Department d
    ON e1.DepartmentId = d.Id
    WHERE
        3 > (
            SELECT
                COUNT(DISTINCT e2.Salary)
            FROM
                Employee e2
            WHERE
                e2.Salary > e1.Salary
            AND e1.DepartmentId = e2.DepartmentId
            )
    ;
    
    
    -- use group by
    
    SELECT
    D.Name as Department, E.Name as Employee, E.Salary
    FROM
    Department D, Employee E, Employee E2
    WHERE
        D.ID = E.DepartmentId
        AND E.DepartmentId = E2.DepartmentId
        AND E.Salary <= E2.Salary
    GROUP BY D.ID,E.Name
    HAVING COUNT(DISTINCT E2.Salary) <= 3
    ORDER BY D.Name, E.Salary DESC
    
    -- 
    
    WITH
        T AS (
            SELECT
                *,
                DENSE_RANK() OVER (
                    PARTITION BY departmentId
                    ORDER BY salary DESC
                ) AS rk
            FROM Employee
        )
    SELECT d.name AS Department, t.name AS Employee, salary AS Salary
    FROM
        T AS t
        JOIN Department AS d ON t.departmentId = d.id
    WHERE rk <= 3;
    
    

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