Leetcode SQL Interview Questions | Practise SQL questions

[@sumitmittal07]

Checkout the Big Data course details here: trendytech.in/?referrer=youtube_lc4

[@sakshijain3534]

with cte as(select email ,row_number() over(partition by email order by email) as rn from person) select distinct email from cte where rn > 1;

[@mohammadshahfaishal2747]

#duplicate email.You don't need to write a derived table. You can do it one email. "Select email from table_name group by 1 Having count(*)> 1"

[@souvik5560]

Loved it 👍🏼 Thank you for the session. It would be really great if you could explain why one query is better than the other query. The interviews do ask for optimisation and the basis for it.

[@sumitmittal07]

Glad that you are liking the sessions. Yes, will cover it in the upcoming videos.

[@abhisheksuman6592]

Very Informative 👍 Continue this series as it will help a lot to people looking to crack data analyst interview.

[@manoharmanu399]

select name from customers where id not in(select customerID from orders)

[@Anonymous_it_is]

Thanks for this video sir. Alternate solution using Partition : SELECT DISTINCT email FROM (SELECT email, COUNT(email) OVER (PARTITION BY email) AS part FROM Person) AS temp WHERE temp.part > 1; Sir, could you please explain - is there any performance differentiation between the above query and the query shown in video. Thank You!

[@arunsundar3739]

able to solve sql problems on my own now :)

[@mallikarjunpatil7553]

Thanks for very nice explanation.. Waiting for the premium questions

[@shananwar3199]

Thanks a lot, waiting for next session.

[@sumitmittal07]

Very soon :)

[@shivangikhanna6163]

I got this question in one of the interviews last week. You are just late by few days 😅. Just kidding! Incredible content Sumit sir.

[@sayantansarbadhikary6917]

🎉

[@vijaybabukommuri5125]

select name as Customers from Customers where id not in (select customerId from Orders group by customerId)

[@vijaybabukommuri5125]

I think this is more simpler way SELECT email FROM person GROUP BY email HAVING COUNT(email) > 1;

[@akkiwadhwa6618]

Jb nhi pta kyu btara hai..Chu

[@pragyajaiswal5754]

select Customers.name as "Customers" from Customers join Orders ON Customers.id != Orders.customerid; Can someone point out where i am doing wrong?

[@HARSHRAJ-gp6ve]

select email FROM Person GROUP BY email HAVING COUNT(id)>1; select name as Customers FROM Customers LEFT JOIN Orders ON Customers. id=Orders.CustomerID where CustomerId is null;