By far, this is the most attention-seeking query I faced in this platform!

SELECT h.hacker_id, h.name FROM hackers AS h JOIN submissions AS s ON h.hacker_id = s.hacker_id JOIN challenges AS c ON c.challenge_id = s.challenge_id JOIN difficulty AS d ON d.difficulty_level = c.difficulty_level WHERE d.score = s.score GROUP BY h.hacker_id, h.name HAVING COUNT(DISTINCT s.challenge_id) > 1 ORDER BY COUNT(DISTINCT s.challenge_id) DESC, h.hacker_id ASC

SELECT tt.hacker_id, tt.name FROM ( SELECT t.hacker_id, t.name, COUNT(hacker_id) AS times_max FROM ( SELECT h.name AS name, c.challenge_id AS challenge_id, c.difficulty_level AS difficulty_level, d.score AS max_score, s.hacker_id AS hacker_id, s.score AS user_score FROM challenges AS c JOIN difficulty AS d ON d.difficulty_level = c.difficulty_level JOIN submissions AS s ON c.challenge_id = s.challenge_id JOIN hackers AS h ON h.hacker_id = s.hacker_id WHERE s.score = d.score ) AS t GROUP BY hacker_id, name HAVING times_max > 1 ORDER BY times_max DESC, hacker_id ASC )AS tt;

this problem is tricky and waste of time, make sure, you know that you need hacker_id to find submission_id first to get score, then use challenge id to get the full score with difficulty level; not the other way around..

SELECT h.hacker_id, name FROM hackers AS h join submissions AS s ON h.hacker_id = s.hacker_id join challenges AS c ON c.challenge_id = s.challenge_id join difficulty AS d ON d.difficulty_level = c.difficulty_level WHERE d.score = s.score GROUP BY h.hacker_id, h.name HAVING COUNT(DISTINCT s.challenge_id) > 1 ORDER BY COUNT(DISTINCT s.challenge_id) DESC, h.hacker_id asc

select s.hacker_id,--d.score, c.difficulty_level, h.name from Submissions s inner join Challenges c on s.challenge_id = c.challenge_id inner join difficulty d on d.difficulty_level = c.difficulty_level inner join hackers h on h.hacker_id = s.hacker_id where d.score = s.score --and s.hacker_id = 10857 group by s.hacker_id, h.name having count()>1 order by count() desc, s.hacker_id asc;