[Hacker Rank] Top Competitors

📢 본 포스팅에 활용되는 기본 문제 및 자료 출처는 HackerRank 임을 밝힙니다.

https://www.hackerrank.com

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■ 문제

https://www.hackerrank.com/challenges/full-score/problem

Top Competitors | HackerRank

The following tables contain contest data:

• Hackers: The hacker_id is the id of the hacker, and name is the name of the hacker. 
  
  
• Difficulty: The difficult_level is the level of difficulty of the challenge, and score is the score of the challenge for the difficulty level. 
  
  
• Challenges: The challenge_id is the id of the challenge, the hacker_id is the id of the hacker who created the challenge, and difficulty_level is the level of difficulty of the challenge. 
• Submissions: The submission_id is the id of the submission, hacker_id is the id of the hacker who made the submission, challenge_id is the id of the challenge that the submission belongs to, and score is the score of the submission. 

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Julia just finished conducting a coding contest, and she needs your help assembling the leaderboard!

Write a query to print the respective hacker_id and name of hackers who achieved full scores for more than one challenge.

Order your output in descending order by the total number of challenges in which the hacker earned a full score.

If more than one hacker received full scores in same number of challenges, then sort them by ascending hacker_id.

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■ 풀이

문제 요구사항

두 개 이상의 챌린지에서 전체 점수를 얻은 해커의 이름과 각각의 해커 ID를 출력하는 쿼리를 작성

> 해커의 id와 이름으로 그룹핑이 필요함

> Submission의 score와 Difficulty의 score가 동일해야함.

> HAVING COUNT(Sub.submission_id) >= 2

 

해커가 전체 점수를 획득한 총 챌린지 수를 기준으로 출력을 내림차순으로 정렬

두 명 이상의 해커가 동일한 수의 챌린지에서 전체 점수를 받은 경우 해커 ID 오름차순으로 정렬

> ORDER BY COUNT(Sub.submission_id) DESC, H.hacker_id;

SELECT H.hacker_id  
      ,H.name
 FROM Submissions Sub

 INNER JOIN Hackers H
 ON Sub.hacker_id = H.hacker_id

 INNER JOIN Challenges C
 ON Sub.challenge_id = C.challenge_id

 INNER JOIN Difficulty D
 ON C.difficulty_level = D.difficulty_level

 WHERE Sub.Score = D.score
 GROUP BY 1, 2
 HAVING COUNT(Sub.submission_id) >= 2
 ORDER BY COUNT(Sub.submission_id) DESC, H.hacker_id;