README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/1000-1099/1076.Project%20Employees%20II/README_EN.md	Database

1076. Project Employees II 🔒

中文文档

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

Solution 1

MySQL

# Write your MySQL query statement below
SELECT project_id
FROM Project
GROUP BY 1
HAVING
    COUNT(1) >= all(
        SELECT COUNT(1)
        FROM Project
        GROUP BY project_id
    );

Solution 2

MySQL

# Write your MySQL query statement below
WITH
    T AS (
        SELECT
            project_id,
            RANK() OVER (ORDER BY COUNT(employee_id) DESC) AS rk
        FROM Project
        GROUP BY 1
    )
SELECT project_id
FROM T
WHERE rk = 1;