Interviews in SQL | HackerRank Programming Solutions | HackerRank SQL Solutions

Hello Programmers/Coders, Today we are going to share solutions of Programming problems of HackerRank of Programming Language SQL. At Each Problem with Successful submission with all Test Cases Passed, you will get an score or marks. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for Interviews in SQL-HackerRank Problem. We are providing the correct and tested solutions of coding problems present on HackerRank. If you are not able to solve any problem, then you can take help from our Blog/website.

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Introduction To SQL

SQL stands for Structured Query Language. SQL is used to create, remove, alter the database and database objects in a database management system and to store, retrieve, update the data in a database. SQL is a standard language for creating, accessing, manipulating database management system. SQL works for all modern relational database management systems, like SQL Server, Oracle, MySQL, etc.

  • It is a standard language for Relational Database System. It enables a user to create, read, update and delete relational databases and tables.
  • All the RDBMS like MySQL, Informix, Oracle, MS Access and SQL Server use SQL as their standard database language.
  • SQL allows users to query the database in a number of ways, using English-like statements.

Link for the ProblemInterviews 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. 1458517426 e017c3460e ScreenShot2016 03 21at4.57.47AM
  • 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. 1458517503 fd4aa63111 ScreenShot2016 03 21at4.57.56AM
  • 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. 1458517661 a642f750ce ScreenShot2016 03 21at4.58.04AM
  • 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. 1458517983 b4302286a8 ScreenShot2016 03 21at4.58.15AM
  • 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. 1458518090 80983c916a ScreenShot2016 03 21at4.58.27AM

Sample Input

Contests Table: 1458519044 d788f8a6ee ScreenShot2016 03 21at4.58.39AM Colleges Table: 1458519098 912836d6ac ScreenShot2016 03 21at4.59.22AM Challenges Table: 1458519120 c531743caf ScreenShot2016 03 21at4.59.32AM View_Stats Table: 1458519152 107a67866b ScreenShot2016 03 21at4.59.43AM Submission_Stats Table: 1458519173 091aba871a ScreenShot2016 03 21at4.59.55AM0

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;

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