Last updated on October 9th, 2024 at 09:58 pm

This Leetcode problem Department Top Three Salaries LeetCode Solution is done in SQL.

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Department Top Three Salaries LeetCode Solution

Problem Statement

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.

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:

Output:

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

1. Department Top Three Salaries LeetCode Solution MySQL

select 
  d.name as department, 
  e1.name as employee, 
  e1.salary as 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
  )

2. Department Top Three Salaries LeetCode Solution Pandas

import pandas as pd

def top_three_salaries(employee: pd.DataFrame, department: pd.DataFrame) -> pd.DataFrame:
    return employee[employee.groupby('departmentId').salary.rank(method='dense',ascending=False)<=3].merge(department,
    left_on='departmentId',right_on='id').rename(columns={'name_y':'Department','name_x':'Employee'}).iloc[:,[5,1,2]]