Top Competitors

Sort by

recency

|

2500 Discussions

|

  • + 0 comments

    select h.hacker_id, h.name from hackers h

    Inner Join Submissions s on h.hacker_id = s.hacker_id Inner Join Challenges c on s.challenge_id = c.challenge_id Inner join Difficulty d on d.difficulty_level = c.difficulty_level

    where s.score = d.score

    group by h.hacker_id, h.name

    having count(s.challenge_id ) > 1

    Order By count(s.challenge_id) desc, hacker_id asc

  • + 0 comments
    SELECT S.hacker_id, H.name FROM Submissions S
    JOIN Challenges C on C.challenge_id = S.challenge_id 
    JOIN Difficulty D on D.difficulty_level = C.difficulty_level
    JOIN Hackers H on H.hacker_id = S.hacker_id
    WHERE D.difficulty_level = C.difficulty_level AND D.score = S.score
    GROUP BY S.hacker_id, H.name
    HAVING COUNT(S.hacker_id) > 1
    ORDER BY COUNT(S.hacker_id) DESC, S.hacker_id ASC
    
  • + 0 comments

    select h.hacker_id, h.name from hackers h join (select sub.hacker_id hackerId, count(sub.hacker_id) hacker_count from submissions sub join challenges ch on sub.challenge_id=ch.challenge_id join difficulty diff on ch.difficulty_level=diff.difficulty_level where diff.score=sub.score group by(sub.hacker_id)) sl on sl.hackerId=h.hacker_id where hacker_count > 1 order by hacker_count desc, h.hacker_id;

  • + 0 comments

    SELECT h.hacker_id, h.name FROM Hackers h JOIN Submissions s ON h.hacker_id = s.hacker_id JOIN Challenges c ON s.challenge_id = c.challenge_id JOIN Difficulty d ON c.difficulty_level = d.difficulty_level WHERE s.score = d.score -- full 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;

  • + 0 comments

    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