Zepto Product Analyst SQL Interview Question 

Without using a recursive CTE, I was unsure how to approach this, but after watching your video, I found it really insightful. The way you explain things is excellent. Thank you!

Really like the problem statement it was seems like easy at first but upon watching the video realised that simple cross join will not work. However, when recursive cte is not allowed than we should not use approach of create table using r_cte. So Last solution orr approach is the only approach I guess we can follow.

Thanks for bringing this on the video.. here's my attempt on SQL server: ===================================== with series as (select * from generate_series(1, (select MAX(num) from #numbers), 1)) select s1.value from series s1 cross join series s2 where s2.value

select n2.n from numbers n1 join numbers n2 on n1.n

Hello Ankit, this is my solution: with cross_data as ( select a.int_numbers as seq_numbers, b.int_numbers from_b, rank() over(partition by a.int_numbers order by b.int_numbers) as rnk from tbl_numbers a cross join tbl_numbers b ) select seq_numbers from cross_data where rnk

superb explanation...! in python df = pd.DataFrame({'num': [1, 2, 3, 4, 5,9]}) df_repeated = df.loc[df.index.repeat(df['num'])].reset_index(drop=True) df_repeated

This should work: select n from numbers cross apply generate_series(1, n)

Superb explanation Ankit 👌 👏 👍

Mysql Solution: with recursive cte as ( select max(n) as n from numbers union all select n - 1 from cte where n - 1 >= 1 ) select n2.n from numbers as n1 cross join numbers as n2 on n1.n

Ankit hi. Thanks a lot. But I think your explanation of recursive CTE is little bit wrong. 1. In the first step we get all rows from numbers, it is 1 to 5. 2. in the second step we get rows from CTE according WHERE statement, it is 2 to 5. And the important thing is that the step 3 is going to contain rows we just got in step 2. 3. in the third step we again get rows from CTE according WHERE statement, it is 3 to 5. And so on to step 5 It means in any follow step our table will contain rows from the only previous CTE. Of course, if I understand Recursive CTE correctly 🙂

Awesome

sitr how i can learn recursive cte,that the only thing in sql which find me as difficult now

Hi Ankit , I have had this doubt for many years. What is equi join ? Can inner and outer joins be equi joins ? I appreciate your reply and the clarity you will provide. Thank you

-- Sol 1 ;with cte as ( select *,1 as n1 from #numbers ) select n from cte cross apply generate_series (n1,n) order by n -- Sol 2 ; with cte as ( select min(n) as min_n , max(n) as max_n from #numbers ) , cte2 as ( select value from cte cross apply generate_series (min_n,max_n) ) select n from cte2 c left join #numbers n on c.value n_counter ) select n from r_cte order by n

Happy Teacher's Day.

WITH numbers as ( SELECT top 100 rownumber() over (order by (select null)) as num FROM sys.objects ) select i.value FROM input_table I JOIN numbers n on n.num

select b.a from xy a cross join xy b where a.a

with recursive cte as( select min(num) as n from numbers union all select n+1 from cte where n=c1.n group by n1.num order by n1.num; my approach

solution for continuous values: select n2.* from numbers n1 inner join numbers n2 on n1.n