## Algorithm

Problem Name: Sql - Interviews

In this HackerRank Functions in SQL problem solution,

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 0.

Note: A specific contest can be used to screen candidates at more than one college, but each college only holds 1 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
``````

## Code Examples

### #1 Code Example with MySQL

```Code - MySQL```

``````
SELECT A.CONTEST_ID, A.HACKER_ID, A.NAME,
SUM(TOTAL_SUBMISSIONS) AS TOTAL_SUBMISSIONS,
SUM(TOTAL_ACCEPTED_SUBMISSIONS) AS TOTAL_ACCEPTED_SUBMISSIONS,
SUM(TOTAL_VIEWS) AS TOTAL_VIEWS,
SUM(TOTAL_UNIQUE_VIEWS) AS TOTAL_UNIQUE_VIEWS
FROM CONTESTS AS A
LEFT JOIN COLLEGES AS B
ON A.CONTEST_ID = B.CONTEST_ID
LEFT JOIN CHALLENGES AS C
ON B.COLLEGE_ID = C.COLLEGE_ID
LEFT JOIN (SELECT CHALLENGE_ID, SUM(TOTAL_VIEWS) AS TOTAL_VIEWS,
SUM(TOTAL_UNIQUE_VIEWS) AS TOTAL_UNIQUE_VIEWS
FROM VIEW_STATS
GROUP BY CHALLENGE_ID) AS D
ON C.CHALLENGE_ID = D.CHALLENGE_ID
LEFT JOIN (SELECT CHALLENGE_ID, SUM(TOTAL_SUBMISSIONS) AS TOTAL_SUBMISSIONS,
SUM(TOTAL_ACCEPTED_SUBMISSIONS) AS TOTAL_ACCEPTED_SUBMISSIONS
FROM SUBMISSION_STATS
GROUP BY CHALLENGE_ID) AS E
ON C.CHALLENGE_ID = E.CHALLENGE_ID
GROUP BY A.CONTEST_ID, A.HACKER_ID, A.NAME
HAVING (TOTAL_SUBMISSIONS + TOTAL_ACCEPTED_SUBMISSIONS + TOTAL_VIEWS + TOTAL_UNIQUE_VIEWS) > 0
ORDER BY A.CONTEST_ID;
``````
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