Hackerrank / Basic Join / Top Competitors / MySQL

📚문제

https://www.hackerrank.com/challenges/full-score/problem?isFullScreen=true

Top Competitors | HackerRank

이번 문제는 너무 길어서 캡쳐하기 번거로워서 링크 첨부

📝풀이

SELECT id   
     , name
FROM (SELECT h.hacker_id as id
     , h.name as name
     , COUNT(DISTINCT s.challenge_id) AS chal
FROM Submissions s
INNER JOIN Challenges c ON s.challenge_id = c.challenge_id
INNER JOIN Difficulty d ON c.difficulty_level = d.difficulty_level
INNER JOIN Hackers h ON s.hacker_id = h.hacker_id
WHERE s.score = d.score
GROUP BY h.hacker_id,h.name
HAVING chal >= 2) etc
ORDER BY chal DESC,id