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1076. Project Employees II

Description

Table: Project

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| project_id  | int     |
| employee_id | int     |
+-------------+---------+
(project_id, employee_id) is the primary key (combination of columns with unique values) of this table.
employee_id is a foreign key (reference column) to Employee table.
Each row of this table indicates that the employee with employee_id is working on the project with project_id.

 

Table: Employee

+------------------+---------+
| Column Name      | Type    |
+------------------+---------+
| employee_id      | int     |
| name             | varchar |
| experience_years | int     |
+------------------+---------+
employee_id is the primary key (column with unique values) of this table.
Each row of this table contains information about one employee.

 

Write a solution to report all the projects that have the most employees.

Return the result table in any order.

The result format is in the following example.

 

Example 1:

Input: 
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                |
+-------------+--------+------------------+
Output: 
+-------------+
| project_id  |
+-------------+
| 1           |
+-------------+
Explanation: The first project has 3 employees while the second one has 2.

Solutions

• SQL

• # Write your MySQL query statement below
  
  select project_id 
  from Project 
  group by project_id
  having count(distinct employee_id) = (
      select max(count_employee_id) from (
          select project_id, count(employee_id) as count_employee_id
          from Project
          group by project_id
      ) as max_employee
  );
  
  
  -- same as above, but separated into 2 parts
  WITH max_employee AS (
      SELECT project_id, COUNT(employee_id) AS count_employee_id
      FROM Project
      GROUP BY project_id
  )
  
  SELECT project_id 
  FROM Project 
  GROUP BY project_id
  HAVING COUNT(DISTINCT employee_id) = (
      SELECT MAX(count_employee_id) 
      FROM max_employee
  );
  
  --
  
  SELECT project_id
  FROM Project
  GROUP BY 1
  HAVING
      COUNT(1) >= ALL (
          SELECT COUNT(1)
          FROM Project
          GROUP BY project_id
      );
  
  

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