SQL Advanced Join Interviews question hacker rank

 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 .

Note: A specific contest can be used to screen candidates at more than one college, but each college only holds screening contest.

Input Format

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.

Sample Input

Contests Table:

Colleges Table:

Challenges Table:

View_Stats Table:

Submission_Stats Table:

Sample Output

66406 17973 Rose 111 39 156 56
66556 79153 Angela 0 0 11 10
94828 80275 Frank 150 38 41 15

Explanation

The contest is used in the college . In this college , challenges and are asked, so from the view and submission stats:

• Sum of total submissions
• Sum of total accepted submissions
• Sum of total views
• Sum of total unique views

Similarly, we can find the sums for contests and .

Solution in MYSQL

select con.contest_id,
con.hacker_id,
con.name,
sum(total_submissions),
sum(total_accepted_submissions),
sum(total_views), sum(total_unique_views)
from contests con
join colleges col on con.contest_id = col.contest_id
join challenges cha on col.college_id = cha.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) vs on cha.challenge_id = vs.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) ss on cha.challenge_id = ss.challenge_id
group by con.contest_id, con.hacker_id, con.name
having sum(total_submissions)!=0 or
sum(total_accepted_submissions)!=0 or
sum(total_views)!=0 or
sum(total_unique_views)!=0

order by contest_id;