Problem Name: Sql -
In this HackerRank Functions in SQL problem solution,
Samantha interviews many candidates from different colleges using coding challenges and contests. Write a query to print the contest_id, hacker_id, name, and the sums of total_submissions, total_accepted_submissions, total_views, and total_unique_views for each contest sorted by contest_id. Exclude the contest from the result if all four sums are 0.
Note: A specific contest can be used to screen candidates at more than one college, but each college only holds 1 screening contest.
The following tables hold interview data:
Contests: The contest_id is the id of the contest, hacker_id is the id of the hacker who created the contest, and name is the name of the hacker.
Colleges: The college_id is the id of the college, and contest_id is the id of the contest that Samantha used to screen the candidates.
Challenges: The challenge_id is the id of the challenge that belongs to one of the contests whose contest_id Samantha forgot, and college_id is the id of the college where the challenge was given to candidates.
View_Stats: The challenge_id is the id of the challenge, total_views is the number of times the challenge was viewed by candidates, and total_unique_views is the number of times the challenge was viewed by unique candidates.
Submission_Stats: The challenge_id is the id of the challenge, total_submissions is the number of submissions for the challenge, and total_accepted_submission is the number of submissions that achieved full scores.
Contests Table: Colleges Table: Challenges Table: View_Stats Table: Submission_Stats Table:
66406 17973 Rose 111 39 156 56 66556 79153 Angela 0 0 11 10 94828 80275 Frank 150 38 41 15
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SELECT A.CONTEST_ID, A.HACKER_ID, A.NAME, SUM(TOTAL_SUBMISSIONS) AS TOTAL_SUBMISSIONS, SUM(TOTAL_ACCEPTED_SUBMISSIONS) AS TOTAL_ACCEPTED_SUBMISSIONS, SUM(TOTAL_VIEWS) AS TOTAL_VIEWS, SUM(TOTAL_UNIQUE_VIEWS) AS TOTAL_UNIQUE_VIEWS FROM CONTESTS AS A LEFT JOIN COLLEGES AS B ON A.CONTEST_ID = B.CONTEST_ID LEFT JOIN CHALLENGES AS C ON B.COLLEGE_ID = C.COLLEGE_ID LEFT JOIN (SELECT CHALLENGE_ID, SUM(TOTAL_VIEWS) AS TOTAL_VIEWS, SUM(TOTAL_UNIQUE_VIEWS) AS TOTAL_UNIQUE_VIEWS FROM VIEW_STATS GROUP BY CHALLENGE_ID) AS D ON C.CHALLENGE_ID = D.CHALLENGE_ID LEFT JOIN (SELECT CHALLENGE_ID, SUM(TOTAL_SUBMISSIONS) AS TOTAL_SUBMISSIONS, SUM(TOTAL_ACCEPTED_SUBMISSIONS) AS TOTAL_ACCEPTED_SUBMISSIONS FROM SUBMISSION_STATS GROUP BY CHALLENGE_ID) AS E ON C.CHALLENGE_ID = E.CHALLENGE_ID GROUP BY A.CONTEST_ID, A.HACKER_ID, A.NAME HAVING (TOTAL_SUBMISSIONS + TOTAL_ACCEPTED_SUBMISSIONS + TOTAL_VIEWS + TOTAL_UNIQUE_VIEWS) > 0 ORDER BY A.CONTEST_ID;