15 days of Sql From hackerrank - mysql

Question

I am unable to understand the use of this line in a code can someone please explain me about this or give some different way to approach to this question
Link to the question:https://www.hackerrank.com/challenges/15-days-of-learning-sql
Code:
select
submission_date ,
( SELECT COUNT(distinct hacker_id)
FROM Submissions s2
WHERE s2.submission_date = s1.submission_date
AND ( SELECT COUNT(distinct s3.submission_date)
FROM Submissions s3
WHERE
s3.hacker_id = s2.hacker_id
AND s3.submission_date < s1.submission_date
) = dateDIFF(s1.submission_date , '2016-03-01'))
, ( select hacker_id
from submissions s2
where s2.submission_date = s1.submission_date
group by hacker_id
order by count(submission_id) desc , hacker_id limit 1
) as shit
, ( select name
from hackers where hacker_id = shit
)
FROM
( select distinct submission_date
from submissions) s1
group by submission_date
Unable to understand why they have used this line from this part of the code:
(s3.submission_date < s1.submission_date) = dateDIFF(s1.submission_date , '2016-03-01'))

Answer 1

CREATE TABLE #max_submissions (
submission_date date,
hacker_id integer,
submission_count integer,
ordering_row integer
)
insert into #max_submissions
select
submission_date,
hacker_id,
submission_count,
row_number() over(partition by submission_date order by submission_count desc, hacker_id) as ordering_row
from (
select submission_date,
hacker_id,
count(hacker_id) as submission_count
from submissions
group by submission_date, hacker_id
) tbl_submission_count
CREATE TABLE #hacker_counts (
submission_date date,
hacker_count integer
)
insert into #hacker_counts
select tbl.submission_date,
COUNT(distinct tbl.hacker_id) as cc
from (
select *,
(case when (
(select count(*)
from (select distinct *
from (select s1.hacker_id,
s1.submission_date
from Submissions s1
where s1.hacker_id = s.hacker_id and
(s1.submission_date >= '2016-03-01' and
s1.submission_date <= s.submission_date)) t1
) t2
) >= (DATEDIFF(day, '2016-03-01', s.submission_date) + 1) )
then 1
else 0
end) as logic
from Submissions s
) tbl
where tbl.logic = 1
group by tbl.submission_date
select max_submissions.submission_date,
hacker_counts.hacker_count,
max_submissions.hacker_id,
h.name
from #max_submissions max_submissions
inner join hackers h on max_submissions.hacker_id = h.hacker_id
left join #hacker_counts hacker_counts on max_submissions.submission_date = hacker_counts.submission_date
where max_submissions.ordering_row = 1
order by max_submissions.submission_date
drop table #max_submissions
drop table #hacker_counts

Answer 2

To understand this line
( SELECT COUNT(distinct s3.submission_date)
FROM Submissions s3
WHERE
s3.hacker_id = s2.hacker_id
AND s3.submission_date < s1.submission_date)
= dateDIFF(s1.submission_date , '2016-03-01')
First understand left hand side:
(SELECT COUNT(distinct s3.submission_date) FROM Submissions s3 WHERE s3.hacker_id = s2.hacker_id AND s3.submission_date < s1.submission_date)
This line counts the unique submission dates for each hacker_id uptil the current date,
So if the date for one row is 2016-03-05, it will count unique submissions for a hacker_id uptil this date (note it will count multiple submissions by a single hacker on a day as 1 count only)
In other words, this takes a hacker_id and start checking if there is a submission by this hacker_id for each day from 1st day uptil this day,it will do this for each submission date
Then Understand Right Hand Side:
dateDIFF(s1.submission_date , '2016-03-01')
this will take the difference of this current date 2016-03-05 to first day 2016-03-01,
Understanding the whole statement now:
So if a hacker made at least one submission each day from 2016-03-05 to 2016-03-01, then both sides of the above code will be equal,
that is date difference from 5th to 1st will be 5 (Right Hand Side) and distinct submission date for a hacker who made at least one submission each day from 1st to 5th will also be 5 (left hand side)