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sir last 2 months i followed ur videos . i am also a datascience seeker. ur sql playlists are too good in the same way postother topics also . plz post other datascience related topics also like pandas , power bi etc....❤‍🔥❤‍🔥❤‍🔥❤‍🔥❤‍🔥❤‍🔥❤‍🔥❤‍🔥❤‍🔥❤‍🔥❤‍🔥❤‍🔥❤‍🔥❤‍🔥

So useful. Thankyouu

Solved

cool one thanks

This is the perfect explaination on youtube

Thanks for making such a wonderful video

y cant we use groupby email here and can take min of id

For those getting an error on this one , since rank is a keyword, use it in single quotes SELECT score , dense_rank() OVER (ORDER BY score DESC) AS 'rank' FROM Scores

is this leetcode playlist useful for data engineering?

bro are you tamil

bhai sari premium problem kyo kra rahe ho, sab ke pas leetcode ka premium nhi hai

Using dense rank instead of rank my solution was accepted

Excellent Logic! keep up the good work .. waiting for more videos!

Nice explanation sir🎉

Thanks a lot for this video! Great explanation.

Thank you sir!

Code failed with following error "BIGINT UNSIGNED value is out of range in '(`test`.`logs`.`id` - `row_number() OVER (PARTITION BY ``logs``.num ORDER BY ``logs``.id ) `)' " The code I used was ` with cte as ( select id, num , id - row_number() over(partition by num order by id) row_diff from logs ) select distinct num ConsecutiveNums from cte group by num, row_diff having count(*) >= 3 `

Thanks for this great explanation!

Great Explanation. I was searching for the good explanation video for this sql problem. Gladd!! i found it.

Thank You for SQL SERVER Answer

Nice! i have further optimised this and found out that this can be done without using subquery SELECT a.machine_id, ROUND(AVG(b.timestamp-a.timestamp),3) as processing_time from Activity a join Activity b on a.machine_id= b.machine_id and a.process_id = b.process_id and a.timestamp < b.timestamp Group by a.machine_id;

Hi Prasad, Im getting this error as "Incorrect parameter count in the call to native function 'DATEDIFF'" I gave the exact code like you gave select b.id as id from Weather a inner join Weather b on DATEDIFF(Day, a.recordDate, b.recordDate) = 1 and a.temperature < b.temperature

very helpful video .. i skip this question

select player_id, device_id from Activity where event_date in (select min(event_date) as event_date group by player_id from Activity) is the above query achive the same results i cannot test it bcz the problem is for premium subscribers. pls tell if possible

hi MLV Prasad kaisa hai bhai ?

this ques is should be marked very easy !!

it was helpful

NIce work❤

Please explain more of why you are doing particular thing i started from basics and its really hard to understood

When I run the solution it shows "accepted" with 0.33 as output but when I submit I get wrong answer displayed. WHy like that?Sir please help

Thank you for your video

Good work and clear explanation! Thank you Prasad

Would be better for freshers if more explanation is shown before doing each step

Best Explanation. Thanks

I can’t see these questions now on leet code . Please help

nice explanation

with cte as ( select seat_id, free, lead(free, 1) over( order by seat_id ) as next lag(free, 1) over( order by seat_id ) as prev from cinema ) select seat_id from cte where ( free = next or free = prev) and free = 1

hi MLV Prasad can you please explain this below hierarchy query (or) make a video on it with your own style......... I was unable to crack this need your support thanks in advance......❤❤✌✌ CREATE TABLE company ( employee varchar(10) primary key, manager varchar(10) ); INSERT INTO company values ('Elon', null); INSERT INTO company values ('Ira', 'Elon'); INSERT INTO company values ('Bret', 'Elon'); INSERT INTO company values ('Earl', 'Elon'); INSERT INTO company values ('James', 'Ira'); INSERT INTO company values ('Drew', 'Ira'); INSERT INTO company values ('Mark', 'Bret'); INSERT INTO company values ('Phil', 'Mark'); INSERT INTO company values ('Jon', 'Mark'); INSERT INTO company values ('Omid', 'Earl'); SELECT * FROM company; /* Given graph shows the hierarchy of employees in a company. Write an SQL query to split the hierarchy and show the employees corresponding to their team.*/ WITH RECURSIVE cte_teams AS ( SELECT mng.employee, CONCAT('Team', ROW_NUMBER() OVER (ORDER BY mng.employee)) AS teams FROM company root JOIN company mng ON root.employee = mng.manager WHERE root.manager IS NULL ), cte AS ( SELECT c.employee, c.manager, t.teams FROM company c CROSS JOIN cte_teams t WHERE c.manager IS NULL UNION SELECT c.employee, c.manager, COALESCE(t.teams, cte.teams) AS teams FROM company c JOIN cte ON cte.employee = c.manager LEFT JOIN cte_teams t ON t.employee = c.employee ) SELECT teams, GROUP_CONCAT(employee, ', ') AS members FROM cte GROUP BY teams ORDER BY teams;

how about filtering the salary using desc use distinct select the position 2

[42000] [Microsoft][ODBC Driver 17 for SQL Server][SQL Server]An expression of non-boolean type specified in a context where a condition is expected, near ','. (4145) (SQLExecDirectW) getting this following error near where clause can you help.. select * from activity a inner join activity b on dateadd(day,1,a.event_date) = b.event_date and a.player_id = b.player_id where (a.event_date,a.player_id) in (select min(event_date),player_id from activity group by player_id) this is my partial query

You may have to mentioned from where you are taking ex: (p.firstName, p.lastName, a.city, a.state) because sql doesn’t know what to take from where. Its always a good practice to put table name first. In this scenario it was acceptable if not mentioned

in the case stmt why did you use sum and not count, what is affected by using count/sum in that stmt?

I really like your content and the way of explanation by showing schematic way using notepad plz keep it up bro.....my support to u always

i am not able to understand isnull functionality over here

nice explanation prasad

DELETE p1.* FROM Person AS p1 INNER JOIN Person AS p2 WHERE p1.email = p2.email AND p1.id > p2.id

very helpful playlists, thanks!

Hi Prasad , from GitHub not able to download or open the Query , can you pls check once . Thanks for your Explanations and help

How r u thinking..which should we use and which logic should use? Does all comes under practice?

select r.contest_id , Round(((count( r.user_id))/ (SELECT count(*) FROM users)) *100, 2) as percentage from Users u Left Join Register r on u.user_id = r.user_id group by r.contest_id order by percentage desc, contest_id asc; sir i have written like this. getting error in case 2, getting null values with answer...