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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 Problem – Interviews SQL – Hacker Rank Solution
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: 
Colleges Table: 
Challenges Table: 
View_Stats Table: 
Submission_Stats Table: 
0
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;
