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
Interviews SQL Hacker Rank Solution

Interviews SQL Hacker Rank Solution


Samantha interviews many candidates from different colleges using coding challenges and contests. Write a query to print the contest_idhacker_idname, and the sums of total_submissionstotal_accepted_submissionstotal_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. Interviews SQL Hacker Rank Solution
  • 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. Interviews SQL Hacker Rank Solution
  • 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. Interviews SQL Hacker Rank Solution
  • 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. Interviews SQL Hacker Rank Solution
  • 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. Interviews SQL Hacker Rank Solution

Sample Input

Contests Table: Interviews SQL Hacker Rank Solution Colleges Table: Interviews SQL Hacker Rank Solution Challenges Table: Interviews SQL Hacker Rank Solution View_Stats Table: Interviews SQL Hacker Rank Solution Submission_Stats Table: Interviews SQL Hacker Rank Solution

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,,
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
(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
(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,
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.

Sharing Is Caring

Leave a Comment