데이터 분석/[SQL] 기초
[hackerrank] MySQL - Challenges
INCHELIN
2025. 3. 20. 14:19
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https://www.hackerrank.com/challenges/challenges/problem
Challenges | HackerRank
Print the total number of challenges created by hackers.
www.hackerrank.com
with t1 as(
select hacker_id, count(distinct challenge_id) as c_cnt
from challenges
group by hacker_id
)
select t1.hacker_id, hackers.name, c_cnt
from t1
join hackers on t1.hacker_id = hackers.hacker_id
where 1=1
and (t1.c_cnt = (select max(c_cnt) from t1)
or
t1.c_cnt not in (select c_cnt from t1
group by c_cnt having count(c_cnt)>=2)
)
order by c_cnt desc, hacker_id asc
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