# $1789. 员工的直属部门
# https://leetcode.cn/problems/primary-department-for-each-employee/

# SQL架构
Create table If Not Exists Employee (employee_id int, department_id int, primary_flag ENUM('Y','N'));
Truncate table Employee;
insert into Employee (employee_id, department_id, primary_flag) values ('1', '1', 'N');
insert into Employee (employee_id, department_id, primary_flag) values ('2', '1', 'Y');
insert into Employee (employee_id, department_id, primary_flag) values ('2', '2', 'N');
insert into Employee (employee_id, department_id, primary_flag) values ('3', '3', 'N');
insert into Employee (employee_id, department_id, primary_flag) values ('4', '2', 'N');
insert into Employee (employee_id, department_id, primary_flag) values ('4', '3', 'Y');
insert into Employee (employee_id, department_id, primary_flag) values ('4', '4', 'N');

# Write your MySQL query statement below
select
    employee_id,
    department_id
from
    Employee
where
    primary_flag = 'Y'
union
select
    employee_id,
    # sql_mode=only_full_group_by
    any_value(department_id)
from
    Employee
group by
    employee_id
having
    count(department_id) = 1;

# clean-up
drop table Employee;