Last updated on October 10th, 2024 at 12:03 am
This Leetcode problem Project Employees I LeetCode Solution is done in SQL.
List of all LeetCode Solution
Level of Question
Easy
Project Employees I LeetCode Solution
Table of Contents
Problem Statement
| Column Name | Type |
| project_id | int |
| employee_id | int |
Project(project_id, employee_id) is the primary key of this table. employee_id is a foreign key to Employee table. Each row of this table indicates that the employee with employee_id is working on the project with project_id.
| Column Name | Type |
| employee_id | int |
| name | varchar |
| experience_years | int |
Employeeemployee_id is the primary key of this table. It’s guaranteed that experience_years is not NULL. Each row of this table contains information about one employee.
Write an SQL query that reports the average experience years of all the employees for each project, rounded to 2 digits.
Return the result table in any order.
The query result format is in the following example.
Example 1:
Input:
| project_id | employee_id |
| 1 | 1 |
| 1 | 2 |
| 1 | 3 |
| 2 | 1 |
| 2 | 4 |
| employee_id | name | experience_years |
| 1 | Khaled | 3 |
| 2 | Ali | 2 |
| 3 | John | 1 |
| 4 | Doe | 4 |
Output:
| project_id | average_years |
| 1 | 2.00 |
| 2 | 2.50 |
Explanation: The average experience years for the first project is (3 + 2 + 1) / 3 = 2.00 and for the second project is (3 + 2) / 2 = 2.50
1. Project Employees I LeetCode Solution MySQL
select
project_id,
round(
avg(experience_years),
2
) as average_years
from
Project as p
left join Employee as e on p.employee_id = e.employee_id
group by
project_id;