Top Competitors

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    select hacker_id, name from (select h.name, h.hacker_id from hackers h join submissions s on h.hacker_id=s.hacker_id join challenges ch on s.challenge_id=ch.challenge_id join difficulty df on ch.difficulty_level=df.difficulty_level where s.score=df.score order by h.hacker_id) as maxxx group by hacker_id, name having count(name) > 1 order by count(name)desc, hacker_id asc