# Project Employees II Problem

## Description

LeetCode Problem 1076.

Table: Project

``````1
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+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| project_id  | int     |
| employee_id | int     |
+-------------+---------+
(project_id, employee_id) is the primary key of this table.
employee_id is a foreign key to Employee table.
``````

Table: Employee

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+------------------+---------+
| Column Name      | Type    |
+------------------+---------+
| employee_id      | int     |
| name             | varchar |
| experience_years | int     |
+------------------+---------+
employee_id is the primary key of this table.
``````

Write an SQL query that reports all the projects that have the most employees.

The query result format is in the following example:

``````1
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Project table:
+-------------+-------------+
| project_id  | employee_id |
+-------------+-------------+
| 1           | 1           |
| 1           | 2           |
| 1           | 3           |
| 2           | 1           |
| 2           | 4           |
+-------------+-------------+

Employee table:
+-------------+--------+------------------+
| employee_id | name   | experience_years |
+-------------+--------+------------------+
| 1           | Khaled | 3                |
| 2           | Ali    | 2                |
| 3           | John   | 1                |
| 4           | Doe    | 2                |
+-------------+--------+------------------+

Result table:
+-------------+
| project_id  |
+-------------+
| 1           |
+-------------+
The first project has 3 employees while the second one has 2.
``````

## MySQL Solution

``````1
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select p.project_id
from Project p
group by p.project_id
having count(p.employee_id) =
(select max(p1.employee_cnt)
from (select project_id, count(employee_id) as employee_cnt
from Project
group by project_id) p1
)
``````