Leetcode SQL Interview Questions | Practise SQL questions

𝐖𝐚𝐧𝐭 𝐭𝐨 𝐌𝐚𝐬𝐭𝐞𝐫 𝐒𝐐𝐋? 𝐋𝐞𝐚𝐫𝐧 𝐒𝐐𝐋 𝐭𝐡𝐞 𝐫𝐢𝐠𝐡𝐭 𝐰𝐚𝐲 𝐭𝐡𝐫𝐨𝐮𝐠𝐡 𝐭𝐡𝐞 𝐦𝐨𝐬𝐭 𝐬𝐨𝐮𝐠𝐡𝐭 𝐚𝐟𝐭𝐞𝐫 𝐜𝐨𝐮𝐫𝐬𝐞 - 𝐒𝐐𝐋 𝐂𝐡𝐚𝐦𝐩𝐢𝐨𝐧𝐬 𝐏𝐫𝐨𝐠𝐫𝐚𝐦 𝐛𝐲 𝐒𝐮𝐦𝐢𝐭 𝐒𝐢𝐫!
"𝐀 8 𝐰𝐞𝐞𝐤 𝐏𝐫𝐨𝐠𝐫𝐚𝐦 𝐝𝐞𝐬𝐢𝐠𝐧𝐞𝐝 𝐭𝐨 𝐡𝐞𝐥𝐩 𝐲𝐨𝐮 𝐜𝐫𝐚𝐜𝐤 𝐭𝐡𝐞 𝐢𝐧𝐭𝐞𝐫𝐯𝐢𝐞𝐰𝐬 𝐨𝐟 𝐭𝐨𝐩 𝐩𝐫𝐨𝐝𝐮𝐜𝐭 𝐛𝐚𝐬𝐞𝐝 𝐜𝐨𝐦𝐩𝐚𝐧𝐢𝐞𝐬 𝐛𝐲 𝐝𝐞𝐯𝐞𝐥𝐨𝐩𝐢𝐧𝐠 𝐚 𝐭𝐡𝐨𝐮𝐠𝐡𝐭 𝐩𝐫𝐨𝐜𝐞𝐬𝐬 𝐚𝐧𝐝 𝐚𝐧 𝐚𝐩𝐩𝐫𝐨𝐚𝐜𝐡 𝐭𝐨 𝐬𝐨𝐥𝐯𝐞 𝐚𝐧 𝐮𝐧𝐬𝐞𝐞𝐧 𝐏𝐫𝐨𝐛𝐥𝐞𝐦."
𝐇𝐞𝐫𝐞 𝐢𝐬 𝐡𝐨𝐰 𝐲𝐨𝐮 𝐜𝐚𝐧 𝐫𝐞𝐠𝐢𝐬𝐭𝐞𝐫 𝐟𝐨𝐫 𝐭𝐡𝐞 𝐏𝐫𝐨𝐠𝐫𝐚𝐦 -
𝐑𝐞𝐠𝐢𝐬𝐭𝐫𝐚𝐭𝐢𝐨𝐧 𝐋𝐢𝐧𝐤 (𝐂𝐨𝐮𝐫𝐬𝐞 𝐀𝐜𝐜𝐞𝐬𝐬 𝐟𝐫𝐨𝐦 𝐈𝐧𝐝𝐢𝐚) : rzp.io/l/SQLINR
𝐑𝐞𝐠𝐢𝐬𝐭𝐫𝐚𝐭𝐢𝐨𝐧 𝐋𝐢𝐧𝐤 (𝐂𝐨𝐮𝐫𝐬𝐞 𝐀𝐜𝐜𝐞𝐬𝐬 𝐟𝐫𝐨𝐦 𝐨𝐮𝐭𝐬𝐢𝐝𝐞 𝐈𝐧𝐝𝐢𝐚) : rzp.io/l/SQLUSD
𝐖𝐚𝐧𝐭 𝐭𝐨 𝐥𝐞𝐚𝐫𝐧 𝐁𝐢𝐠 𝐃𝐚𝐭𝐚 𝐛𝐲 𝐒𝐮𝐦𝐢𝐭 𝐒𝐢𝐫?
𝐜𝐡𝐞𝐜𝐤𝐨𝐮𝐭 𝐭𝐡𝐞 𝐛𝐢𝐠 𝐝𝐚𝐭𝐚 𝐜𝐨𝐮𝐫𝐬𝐞 𝐝𝐞𝐭𝐚𝐢𝐥𝐬
𝐖𝐞𝐛𝐬𝐢𝐭𝐞 : trendytech.in/...
I have trained over 20,000+ professionals in the field of Data Engineering in the last 5 years.
Leetcode SQL Interview Questions | Practise SQL questions
182. Duplicate Emails
183. Customers Who Never Order
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In this video, we will solve SQL question on LeetCode. LeetCode is an excellent platform for practicing SQL Queries. The SQL Questions on LeetCode are framed pretty similar to how they are asked on SQL Interviews hence solving SQL questions on LeetCode can give you a good hands on experience on solving real world SQL questions.
#SQL #RDBMS #database #bigdata #dataengineering

[@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;

[@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.

[@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"

[@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.

[@abhisheksuman6592]

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

[@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!

[@mallikarjunpatil7553]

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

[@arunsundar3739]

able to solve sql problems on my own now :)

[@manoharmanu9354]

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

[@shananwar3199]

Thanks a lot, waiting for next session.

[@sumitmittal07]

Very soon :)

[@sayantansarbadhikary6917]

🎉

[@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?