Publish Date - January 26th, 2023|
Last Modified - March 7th, 2023
Joins, joins, joins – nothing but joins! Four tables that need to be joined together and uniquely filtered for specific things. Really think about how you want to structure your joins as if you’re joining. While this was not as hard as occupations, it’s easier than placement. However, overall of these queries are great ways to learn more complex querying in MySQL, SQL server or Oracle.
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.
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.
Hackers Table:Difficulty Table:Challenges Table:Submissions Table:
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.
The solution is an interesting one because you need to make sure you’re joining appropriately.
/* hacker_id, name of hackers WHERE score = max(score) AND score > 2 ORDER BY DESC COUNT of challenges of hacker_id, score = max(score) COUNT(score) in same challenges, sort by ASC hacker_id t1 = Hackers t2 = Submissions t3 = Challenges t4 = Difficulty */ SELECT t1.hacker_id, t1.name FROM hackers AS t1 JOIN submissions AS t2 ON t1.hacker_id = t2.hacker_id JOIN challenges AS t3 ON t3.challenge_id = t2.challenge_id JOIN difficulty AS t4 ON t4.difficulty_level = t3.difficulty_level WHERE t4.score = t2.score GROUP BY t1.hacker_id,t1.name HAVING COUNT(*) > 1 ORDER BY COUNT(*) DESC,t1.hacker_id ASC
As always, I start with some pseudo code to outline everything.
/* hacker_id, name of hackers WHERE score = max(score) AND score > 2 ORDER BY DESC // 1 COUNT of challenges of hacker_id, score = max(score) // 2 COUNT(score) in same challenges, sort by ASC hacker_id // 3 t1 = Hackers t2 = Submissions t3 = Challenges t4 = Difficulty */
Note, how I label my table prefixes (almost like variables) and provide some logic on the WHERE, GROUP BY and ORDER BY clause at the end (marked 1, 2, 3 respectively).
Next, here are the JOINs. You need to figure out a logical way to JOIN these tables together since there’s not one primary key.
SELECT t1.hacker_id, t1.name FROM hackers AS t1 JOIN submissions AS t2 ON t1.hacker_id = t2.hacker_id JOIN challenges AS t3 ON t3.challenge_id = t2.challenge_id JOIN difficulty AS t4 ON t4.difficulty_level = t3.difficulty_level
With hackers as your main table:
- Submissions joins on hackers by hacker_id
- Challenges joins on submissions by challenge_id
- difficulty joins on challenges by difficulty_level
By doing this, you ensure that all of values are reported.
After that the crucial point is this:
WHERE t4.score = t2.score GROUP BY t1.hacker_id,t1.name HAVING COUNT(*) > 1 ORDER BY COUNT(*) DESC,t1.hacker_id ASC
With those joins, now you can ask MySQL to match scores from the submissions table and the difficulty table. When you do that, you’re ensuring rows in the query are basically people who got 100% in the competition, since t2.score = max(score) possible for that specific challenge.
The rest of the query is just clean-up and organization of that data with a GROUP BY for the HAVING CLAUSE, which provides you the ability to showcase Hacker_id’s and names that appear more than once.
Here’s the final query result:
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
If you’re working on a potato like me (computer), your query may take a long time to run (those JOINs and order clauses are RAM killers). However, this was a nice change of pace from other MySQL courses on Hackerrank as it’s very similar to queries you may encounter in data science or on the job.
Check out some of my articles on where I’m learning how to code: