HackerRank SQL: Top Competitors

Julia just finished conducting a coding contest, and she needs your help assembling the leaderboard! Write a query to print the respective hacker_id and name of hackers who achieved full scores for more than one challenge. Order your output in descending order by the total number of challenges in which the hacker earned a full score. If more than one hacker received full scores in same number of challenges, then sort them by ascending hacker_id.

Input Format

The following tables contain contest data:

• Hackers: The hacker_id is the id of the hacker, and name is the name of the hacker.

• Difficulty: The difficult_level is the level of difficulty of the challenge, and score is the score of the challenge for the difficulty level.

• Challenges: The challenge_id is the id of the challenge, the hacker_id is the id of the hacker who created the challenge, and difficulty_level is the level of difficulty of the challenge.

• 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 that the submission belongs to, and score is the score of the submission.

Sample Input

Hackers Table:

Difficulty Table:

Challenges Table:

Submissions Table:

Sample Output

90411 Joe

Explanation

Hacker 86870 got a score of 30 for challenge 71055 with a difficulty level of 2, so 86870 earned a full score for this challenge.

Hacker 90411 got a score of 30 for challenge 71055 with a difficulty level of 2, so 90411 earned a full score for this challenge.

Hacker 90411 got a score of 100 for challenge 66730 with a difficulty level of 6, so 90411 earned a full score for this challenge.

Only hacker 90411 managed to earn a full score for more than one challenge, so we print the their hacker_id and name as space-separated values.

Solution:

Selecting ids and names nad order it is simple enough.

 select   h.hacker_id , h.name
from hackers as h
order h.hacker_id ASC

The twist lies in getting Full Scores at the same time as Number of Challenges. Those are in different tables. Therefore we need use eg. a subquerry, to aggregate data and JOINSto connected them. The first JOIN is simple enough. We can already set aggregation as well.

select s.hacker_id, count(s.challenge_id) c_id  from submissions s
left join challenges c on s.challenge_id = c.challenge_id
group by  s.hacker_id 
having count(s.challenge_id) > 1

Second is a bit complicated, as we want to “imprison/lock” two sectors at a same time to force Full Scores with WHERE.

left join difficulty d on s.score = d.score  and  d.difficulty_level = c.difficulty_level  
where s.score = d.score 

Finally the subquerry should look like this, and deliver hackers ids + number of Full challenges.

 ( select s.hacker_id, count(s.challenge_id) c_id  from submissions s
left join challenges c on s.challenge_id = c.challenge_id
left join difficulty d on s.score = d.score  and  d.difficulty_level = c.difficulty_level  

               where s.score = d.score 
                group by  s.hacker_id 
                 having count(s.challenge_id) > 1) as sub

The ids are a perfect link for the main querry.

Final looks like this:

select   h.hacker_id , h.name
from hackers as h
inner join 
                ( select s.hacker_id, count(s.challenge_id) c_id  from submissions s
left join challenges c on s.challenge_id = c.challenge_id
left join difficulty d on s.score = d.score  and  d.difficulty_level = c.difficulty_level  

               where s.score = d.score 
                group by  s.hacker_id 
                 having count(s.challenge_id) > 1) as sub
                 on h.hacker_id = sub.hacker_id
order by c_id DESC, h.hacker_id ASC

The output should be:

27232 Phillip 
28614 Willie 
15719 Christina 
43892 Roy 
14246 David 
14372 Michelle 
18330 Lawrence 
26133 Jacqueline 
26253 John 
30128 Brandon 
35583 Norma 
13944 Victor 
17295 Elizabeth 
19076 Matthew 
26895 Evelyn 
32172 Jonathan 
41293 Robin 
45386 Christina 
45785 Jesse 
49652 Christine 
13391 Robin 
14366 Donna 
14777 Gerald 
16259 Brandon 
17762 Joseph 
28275 Debra 
36228 Nancy 
37704 Keith 
40226 Anna 
49307 Brian 
12539 Paul 
14363 Joyce 
14658 Stephanie 
19448 Jesse 
20504 John 
20534 Martha 
22196 Anthony 
23678 Kimberly 
28299 David 
30721 Ann 
32254 Dorothy 
46205 Joyce 
47641 Patricia 
13122 James 
13762 Gloria 
14863 Walter 
18690 Marilyn 
18983 Lori 
21212 Timothy 
25732 Antonio 
28250 Evelyn 
30755 Emily 
38852 Benjamin 
42052 Andrew 
44188 Diana 
48984 Gregory 
13380 Kelly 
13523 Ralph 
21463 Christine 
24663 Louise 
26243 Diana 
26289 Dorothy 
39277 Charles 
23278 Paula 
25184 Martin 
32121 Dorothy 
36322 Andrew 
39782 Tammy 
40257 James 
41319 Jean 
10857 Kevin 
25238 Paul 
34242 Marilyn 
39771 Alan 
49789 Lillian 
57947 Justin 
74413 Harry

The compiler will truncate it on line 20.