Interviews SQL Hacker Rank Solution

Hello coders, In this post, you will learn how to solve the Interviews SQL Hacker Rank Solution. This problem is a part of the SQL Hacker Rank series.

Interviews SQL Hacker Rank Solution

Problem

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: Interviews SQL Hacker Rank Solution 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

Interviews SQL Hacker Rank Solution

SELECT con.contest_id, con.hacker_id, con.name,
SUM(sg.total_submissions), SUM(sg.total_accepted_submissions),
SUM(vg.total_views), SUM(vg.total_unique_views)
FROM Contests AS con
JOIN Colleges AS col ON con.contest_id = col.contest_id
JOIN Challenges AS cha ON cha.college_id = col.college_id
LEFT JOIN
(SELECT ss.challenge_id, SUM(ss.total_submissions) AS total_submissions, SUM(ss.total_accepted_submissions) AS total_accepted_submissions FROM Submission_Stats AS ss GROUP BY ss.challenge_id) AS sg
ON cha.challenge_id = sg.challenge_id
LEFT JOIN
(SELECT vs.challenge_id, SUM(vs.total_views) AS total_views, SUM(vs.total_unique_views) AS total_unique_views
FROM View_Stats AS vs GROUP BY vs.challenge_id) AS vg
ON cha.challenge_id = vg.challenge_id
GROUP BY con.contest_id, con.hacker_id, con.name
HAVING SUM(sg.total_submissions) +
       SUM(sg.total_accepted_submissions) +
       SUM(vg.total_views) +
       SUM(vg.total_unique_views) > 0
ORDER BY con.contest_id;

Disclaimer: The above Problem (Interviews) generated by Hackerrank but the Solution is Provided by Chase2Learn. This tutorial is only for Educational and Learning purposes.