SQL Query optimization written for an Hacker Rank Challenge [closed] - sql

Question

Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 8 months ago.
Improve this question
For a hacker rank SQL challenge, I have written SQL script that is producing the required result. But I overused too many subqueries, I would like to know if the code can be optimized. Below is the link for the SQL challenge
https://www.hackerrank.com/challenges/challenges/problem .
Challenge details in brief :
"Write a query to print the hacker_id, name, and the total number of challenges created by each student. Sort your results by the total number of challenges in descending order. If more than one student created the same number of challenges, then sort the result by hacker_id. If more than one student created the same number of challenges and the count is less than the maximum number of challenges created, then exclude those students from the result."
Here the code I have written for above challenge:
SELECt
h.hacker_id,
h.name,
t.tot_ch
from
hackers h,
(
Select
c.hacker_id,
count(c.challenge_id) Tot_ch
from
challenges c
Group by
hacker_id
)
T,
(
SELect
tot_ch,
count(tot_ch) DUPS
from
(
Select
c.hacker_id,
count(c.challenge_id) Tot_ch
from
challenges c
Group by
hacker_id
)
group by tot_ch
)
D
Where
h.hacker_id = t.hacker_id
And d.tot_ch = t.tot_ch
AND
(
CASe
when
d.dups < 2
then
1
ELSE
(
case when
t.tot_ch =
(
select
MAX(T1.tot_ch)
from
(
Select
c.hacker_id,
count(c.challenge_id) Tot_ch
from
challenges c
Group by
hacker_id
)
T1
)
then
1
End
)
end
)
= 1
ORDER BY
t.tot_ch desc, h.hacker_id;

Answer 1

You can use common table expressions to turn your repeated subqueries into inline views which you can query just as you would a normal view:
WITH cteCount AS (SELECT c.HACKER_ID,
COUNT(c.CHALLENGE_ID) TOT_CH
FROM CHALLENGES c
GROUP BY HACKER_ID),
cteDups AS (SELECT TOT_CH,
COUNT(TOT_CH) AS DUPS
FROM cteCount
GROUP BY TOT_CH)
SELECT h.HACKER_ID,
h.NAME,
t.TOT_CH
FROM HACKERS h
INNER JOIN cteCount t
ON t.HACKER_ID = h.HACKER_ID
INNER JOIN cteDups d
ON d.TOT_CH = t.TOT_CH
WHERE CASE
WHEN d.DUPS < 2
THEN 1
WHEN t.TOT_CH = (SELECT MAX(TOT_CH)
FROM cteCount)
THEN 1
END = 1
ORDER BY t.TOT_CH DESC,
h.HACKER_ID;

Answer 2

You can use the combination of the analytical function as following:
Select hacker_id, name, cnt
from
(Select t.*,
count(1) over (partition by cnt) as same_cnt,
Max(cnt) over () as max_cnt
from
(Select h.hacker_id,
h.name,
Count(1) as cnt
From hackers h
Join challenges c
On h.hacker_id = c.hacker_id
Group by h.hacker_id, h.name) t)
Where same_cnt = 1
or cnt = max_cnt
Cheers!!

Answer 3

with
cte1 as (select hacker_id, count(challenge_id) as challenges_created
from Challenges
group by hacker_id
order by hacker_id
),
cte2 as (select c.hacker_id, h.name, c.challenges_created
from cte1 c
left join Hackers h
on c.hacker_id = h.hacker_id
),
cte3 as (select max(challenges_created) as max_challenges
from cte2
),
cte4 as (select challenges_created, count(hacker_id) as hack_count_same
from cte2
where challenges_created not in(select max_challenges from cte3)
group by challenges_created
order by challenges_created
),
cte5 as (select hacker_id, name, challenges_created
from cte2
where challenges_created not in(select challenges_created from cte4 where hack_count_same > 1)
),
cte6 as (select hacker_id, name, challenges_created
from cte5
union all
select hacker_id, name, challenges_created
from cte2
where challenges_created in(select max_challenges from cte3)
and hacker_id not in(select hacker_id from cte5)
)
select *
from cte6
order by challenges_created desc, hacker_id;

Answer 4

select a.hacker_id,a.name,count(b.hacker_id) As challenges_created
from Hackers a, Challenges b
WHERE a.hacker_id = b.hacker_id
GROUP BY a.hacker_id,a.name
HAVING count(b.hacker_id) not in (select distinct count(hacker_id) from Challenges
WHERE hacker_id <> a.hacker_id
group by hacker_id
having count(hacker_id) < (select max(x.challenge_count)
from (select count(b.challenge_id) as challenge_count from Challenges b GROUP BY b.hacker_id) as x ))
ORDER BY count(b.hacker_id) desc, a.hacker_id