15 Days to learn SQL Hard SQL(Advanced)- Solution SQL, Best and Optimal Solutions, All you need.

Solutions of SQL all HackerRank:

Solution – 15 Days of Learning SQL

MySQL

SELECT t1.submission_date, hkr_cnt, t2.hacker_id, name
FROM (SELECT p1.submission_date, 
             COUNT(DISTINCT p1.hacker_id) AS hkr_cnt
      FROM (SELECT submission_date, hacker_id, 
                   @h_rnk := CASE WHEN @h_grp != hacker_id THEN 1 ELSE @h_rnk+1 END AS hacker_rank,
                   @h_grp := hacker_id AS hacker_group
            FROM (SELECT DISTINCT submission_date, hacker_id 
                  FROM submissions
                  ORDER BY hacker_id, submission_date) AS a, 
                 (SELECT @h_rnk := 1, @h_grp := 0) AS r) AS p1
      JOIN (SELECT submission_date, 
                   @d_rnk := @d_rnk + 1 AS date_rank
            FROM (SELECT DISTINCT submission_date
                  FROM submissions 
                  ORDER BY submission_date) AS b, 
                 (SELECT @d_rnk := 0) r) AS p2
      ON p1.submission_date = p2.submission_date 
         AND hacker_rank = date_rank
      GROUP BY p1.submission_Date) AS t1
JOIN (SELECT submission_date, hacker_id, sub_cnt,
             @s_rnk := CASE WHEN @d_grp != submission_date THEN 1 ELSE @s_rnk+1 END AS max_rnk,
             @d_grp := submission_date AS date_group
      FROM (SELECT submission_date, hacker_id, COUNT(*) AS sub_cnt
            FROM submissions AS s
            GROUP BY submission_date, hacker_id
            ORDER BY submission_date, sub_cnt DESC, hacker_id) AS c,
           (SELECT @s_rnk := 1, @d_grp := 0) AS r) AS t2                            
ON t1.submission_date = t2.submission_date AND max_rnk = 1
JOIN hackers AS h ON h.hacker_id = t2.hacker_id            
ORDER BY t1.submission_date
;

Here are all the Solutions of All SQL of Hacker Rank, Leave a comment for similar posts

SQL 15 Days to learn SQL HardSQL (Advanced) HackerRank Solution

with Sub1 as (
    select s1.submission_date, s1.hacker_id,
    count (distinct s1.submission_id) as date_submissions,
    1 + datediff(day, 'March 1, 2016', s1.submission_date) as contest_day,
    count (distinct s2.submission_date) as submission_days
    from Submissions s1 join Submissions s2
        on s1.hacker_id = s2.hacker_id and s1.submission_date >= s2.submission_date
    group by s1.submission_date, s1.hacker_id
),
Sub2 as (
    select submission_date, hacker_id, date_submissions
    from Sub1 where submission_days = contest_day
),
Sub3a as (
    select submission_date, count(hacker_id) as hackers
    from Sub2 group by submission_date
),
Sub3b as (
    select submission_date, max(date_submissions) as max_submissions
    from Sub1 group by submission_date
),
Sub4 as (
    select s1.submission_date, s3.hackers,
        (select top 1 s2.hacker_id from Sub1 s2
         where s1.submission_date = s2.submission_date and s1.max_submissions = s2.date_submissions
         order by s2.hacker_id) as hacker_id
    from Sub3b s1
    join Sub3a s3 on s1.submission_date = s3.submission_date
)
select s.*, h.name from Sub4 s join Hackers h on s.hacker_id = h.hacker_id order by submission_date;

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