Solving the most complex SQL Problem on LeetCode | SQL Interview Question and Answer

[@kyleebrahim8061]

This guy is a SQL genius

[@vanshagrawal6896]

instead of l1.user_id < l2.user_id we can write l1.user_id !=l2.user_id so that we do not have to union the table at the end

[@derek713]

I don't think this works, this method means you will have to union friendship table within the "where clause" ......... where (u1.user_id,u2.user_id) not in (select user1_id,user2_id from friendship union select user2_id,user1_id from friendship) group by u1.user_id, u2.user_id, u1.day,u2.day having count(*)>=3 )

[@hassamulhaq7762]

because of previous sql videos I easily understand thelogic he used to write queries in this video. really thanks for such educational video.

[@maradanaajaykumar3657]

Please do more leet code problems bro it helps us to understand complex problems

[@anonymousgreen5080]

Sir leetcode sql premium questions are difficult. Would appreciate few more videos on this topic.

[@shafaligarg1796]

He's crazy talented man 🤯 kudos

[@jirikalina1890]

You should definitely do another. Very good content Sir.

[@vikaskumar-oj2fr]

Your database skills are great 👍

[@GarvitaSaraswat-t9r]

looking forward to more such videos! thanks in advance!

[@deepakdas4513]

crystal clear explanation 👍

[@sujaa1000]

Thank you very much! very very helpful. Look forward to more such problems.

[@iamdushyanta]

Could you please do more leet code problems?

[@RRR5670]

sir please come up with a series❤

[@mainsocial1105]

Really enjoyed this video!! Thanks :)

[@roz_world]

@techtfq sir please leetcode SQL problem solution series please

[@FabioAlmeida-k6t]

well illustrated and explained

[@saadmultani7251]

Please do more leetcode complex problem

[@lio-s4n7]

Nice. Be posting leetcode solutions for sql.

[@mohansai9109]

Please do more leet code problems

[@rohit_vora]

hello everyone , i am having a tiny problem while writing query , if anyone could help me that would be so great. so, when i write query and press enter to go to next line, it automaticaly add some space at the begining which i don't want for ex: (select * from abc where aaa group by bbb) somthin like this. if anyone knows what settings should i chaneg to write just like sir writing in the video, i would be so greatfull to you.

[@adesholaolaitan2257]

Please which tool are you using to record your screen?

[@bestquotes2765]

It's a good approach, but have one doubt do we need the valid_user cte?. Because we are anyway filtering it using having in a freinds cte.

[@sunetala2957]

How to FCFS for insert or update in code sql😢

[@JoelJeremiah-v5t]

For what roles are such questions asked in the interview, sir?

[@Naveenvuppala]

When you understand the logic and know the cte logic it's like playing game

[@MohdSaeed-g6l]

Sir, I am clicking this question then require subscribe to unlock. Can you suggest me how it will open.

[@marrapumani]

sir try to make videos on transactions and cursors

[@shadeditz181]

sir do you have any paid course of sql

[@Malaika_ki_rahein]

sir please teach transaction management with real world exampless

[@arunsam6830]

Can any one suggest a playlists for Terraform for IAC?

[@rajeshmajumdar4999]

Champion !!!

[@Mr.nikk62]

Hii sir thanku for this quality video but sir i have a big question to solve in SQL sir there any way to connect with you and tell you the problem and other student also learn form the question Kindly provide me the way to connect with you

[@Malaika_ki_rahein]

please teach us by reference of KORTH BOOK

[@haleylearn]

Here my solution WITH cte AS ( SELECT l1.user_id AS user_id1, l2.user_id AS user_id2, COUNT(l2.user_id) AS cnt FROM listens l1 JOIN listens l2 ON l1.user_id l2.user_id AND l1.song_id = l2.song_id AND l1.day = l2.day GROUP BY l1.user_id, l2.user_id HAVING COUNT(l2.user_id) >= 3 ) , cte_concat_listen AS ( SELECT *, CASE WHEN user_id1 < user_id2 THEN CONCAT(user_id1, user_id2) ELSE CONCAT(user_id2, user_id1) END AS concat_userid FROM cte ) -- Result SELECT user_id1 AS user_id, user_id2 AS recommended_id FROM cte_concat_listen WHERE concat_userid NOT IN (SELECT CONCAT(user1_id, user2_id) AS concat_friend FROM friendship) ORDER BY user_id1;