Counter Example to the Following Solution? - sql

Question

I recently completed a coding challenge with the following prompt:
The total score of a hacker is the sum of their maximum scores for all
of the challenges. Write a query to print the hacker_id, name, and
total score of the hackers ordered by the descending score. If more
than one hacker achieved the same total score, then sort the result by
ascending hacker_id. Exclude all hackers with a total score of 0 from
your result.
The following tables contain contest data:
Hackers: The hacker_id is the id of the hacker, and name is the name
of the hacker.
Submissions: The submission_id is the id of the submission, hacker_id
is the id of the hacker who made the submission, challenge_id is the
id of the challenge for which the submission belongs to, and score is
the score of the submission.
My solution passed the test cases, but it took me many iterations to get there.
I get the feeling that there may be an edge case / specific input which would not pass using my solution, but I'm having trouble figuring it out.
Any guesses or counter examples?
My solution:
Select ID, Name, sum(maxscore) as tot From
(Select ID, Name, chal, Max(score) as maxscore From
(Select Submissions.hacker_id as ID, Hackers.name as Name, Submissions.score as score, Submissions.challenge_id as chal
From Submissions
Inner Join Hackers on Submissions.hacker_id = Hackers.hacker_id
Where Submissions.score <> 0)
Group by chal, ID, Name)
Group by ID, Name Order by tot desc, ID asc;

Answer 1

Select a.hacker_id,a.name,x.sum_max
from Hackers a
inner join
(SELECT hacker_id,sum(max_score) as sum_max from (
Select hacker_id,challenge_id,max(Score) as max_score
from submissions
group by hacker_id,challenge_id)b
group by hacker_id
having sum_max>0)x on a.hacker_id=x.hacker_id
order by sum_max desc,hacker_id

Answer 2

Here is mine
select T.hacker_id, T.name , sum(T.max_score) sum_score from (
Select h.hacker_id, name , challenge_id, max(score) max_score from Hackers h
inner join submissions s on h.hacker_id = s.hacker_id
group by h.hacker_id, name , challenge_id) as T
group by T.hacker_id, T.name
having sum(T.max_score) > 0
order by sum_score desc , T.hacker_id

Answer 3

This code for MySQL successfully ran the test cases.
select m.id, h.name, m.total from
(select x.hacker_id id, sum(x.max_score) total from
(select h.hacker_id, s.challenge_id, max(s.score) as max_score from submissions as s
join hackers as h on h.hacker_id = s.hacker_id
group by h.hacker_id, s.challenge_id) as x
group by x.hacker_id
having total <> 0)as m,
hackers as h
where m.id = h.hacker_id
order by m.total desc, m.id;
enter image description here

Answer 4

First try this query than vote:
select h.hacker_id,h.name,sum(s.score) as scores from Hackers h
inner join
(select hacker_id,max(score) as score from Submissions group by hacker_id,challenge_id) s
on h.hacker_id=s.hacker_id
group by h.hacker_id,h.name having scores>0
order by scores desc,h.hacker_id;

Answer 5

SELECT A, B, sum(C) FROM
(SELECT h.hacker_id AS A, h.name AS B, max(s.score)AS C FROM
Hackers h
JOIN Submissions s ON s.hacker_id=h.hacker_id
GROUP BY h.hacker_id,h.name, s.challenge_id) Sub
GROUP BY A, B
HAVING sum(C)>0
ORDER BY sum(C) DESC, A ASC;

Answer 6

That's how I did it:
WITH NOW AS (
SELECT CHALLENGE_ID AS CI, HACKER_ID AS HI, MAX(SCORE) AS TSCORE FROM SUBMISSIONS
WHERE HACKER_ID = HACKER_ID GROUP BY CHALLENGE_ID, HACKER_ID
)
SELECT C.HI, H.NAME, SUM(C.TSCORE) AS [TOTAL] FROM NOW AS C, HACKERS AS H
WHERE C.HI = H.HACKER_ID AND C.TSCORE > 0
GROUP BY C.HI, H.NAME ORDER BY TOTAL DESC, C.HI

Answer 7

SELECT
res.hacker_id as id,
res.name as name,
SUM(res.max_score) as score
from (
SELECT s.challenge_id as challenge_id, h.hacker_id as hacker_id ,h.name as name,Max(score) as max_score
FROM Hackers as h JOIN Submissions as s
ON h.hacker_id = s.hacker_id
GROUP BY challenge_id ,hacker_id,name
having max_score <> 0
) as res
GROUP BY hacker_id,name order by score desc,id asc;

Answer 8

Try this for mysql
SELECT t.hacker_id,t.name,sum(t.score) as score FROM
(select s.hacker_id, h.name, s.challenge_id, max(s.score) as score
FROM submissions s join Hackers h ON s.hacker_id = h.hacker_id
GROUP BY hacker_id, h.name, challenge_id ) as t
where score != 0 group by t.hacker_id,t.name order by sum(t.score) desc,t.hacker_id

Answer 9

oracle solution:
with temp as (
select tp.hacker_id,h.name, sum(tp.score) s from (
select hacker_id ,submission_id, challenge_id ,score, rank() over (partition by
hacker_id, challenge_id order by score desc,submission_id )as rank from
submissions)tp
join hackers h on h.hacker_id=tp.hacker_id
where tp.rank=1
group by tp.hacker_id,h.name
order by s desc,tp.hacker_id)
select hacker_id,name,s from temp where s <> 0 order by s desc,hacker_id ;

Answer 10

i tried this and worked with me
select H.Hacker_id, h.Name, sum(x.Score)
from hackers h
inner join (
select hacker_id, challenge_id, max(Score) as Score
from Submissions
group by hacker_id, challenge_id
) x
on H.Hacker_id = x.hacker_Id
group by H.Hacker_Id, h.Name
having sum(x.score) > 0
order by 3 desc, h.hacker_id