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๐Ÿ”ฐAuthentic Coding Solutions(with Outputs)
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โœ…โœ…โœ…#HackerRank Department Summary - SQL HackerRank Solution

select d.name, count(unique(e.id)) as c
from department d, employee e
where d.id = e.dept_id
group by d.name
order by c desc, d.name;

select d.name, 0
from department d
left outer join employee e on d.id = e.dept_id
where e.id is null
group by d.name
order by d.name;

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