Interviews — HackerRank Advanced SQL Question ( cool )

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

Solution

1. Very slow join solution

SELECT   contests.contest_id, 
         hacker_id, 
         NAME, 
         Sum(total_submissions)          AS tot_subs, 
         Sum(total_accepted_submissions) AS tot_act_subs, 
         Sum(total_views)                AS tot_vw, 
         Sum(total_unique_views)         AS tot_uni_vw 
FROM     view_stats vs 
JOIN     (submission_stats, challenges, colleges, contests) 
ON       ( 
                  vs.challenge_id = submission_stats.challenge_id 
         AND      vs.challenge_id = challenges.challenge_id 
         AND      challenges.college_id = colleges.college_id 
         AND      colleges.contest_id = contests.contest_id) 
GROUP BY contests.contest_id, 
         hacker_id, 
         NAME 
HAVING   (tot_subs + tot_act_subs +tot_vw + tot_uni_vw) != 0 
ORDER BY contests.contest_id

3.

/*
Enter your query here.
Please append a semicolon ";" at the end of the query and enter your query in a single line to avoid error.
*/
WITH SUM_View_Stats AS (
SELECT challenge_id
    , sum(total_views) as total_views 
    , sum(total_unique_views) as total_unique_views 
FROM View_Stats 
GROUP BY challenge_id
)
,SUM_Submission_Stats AS (
SELECT challenge_id
    , sum(total_submissions) as total_submissions 
    , sum(total_accepted_submissions) as total_accepted_submissions  
FROM Submission_Stats 
GROUP BY challenge_id
)
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
INNER JOIN Colleges col
    ON con.contest_id = col.contest_id
INNER JOIN Challenges cha
    ON cha.college_id = col.college_id
LEFT JOIN SUM_View_Stats vs
    ON vs.challenge_id = cha.challenge_id
LEFT JOIN SUM_Submission_Stats ss
    ON ss.challenge_id = cha.challenge_id
GROUP BY con.contest_id,con.hacker_id,con.name
HAVING (SUM(total_submissions)
        +sum(total_accepted_submissions)
        +sum(total_views)
        +sum(total_unique_views)) <> 0
ORDER BY con.contest_ID;