AMAZON SQL INTERVIEW QUESTION | Weekday puzzle

[@ragulmarley4900]

Nice bro 😊 way u teaching do some video basic to advance sql sir

[@sravankumar1767]

Superb explanation 👌 👏 👍

[@datasculptor2895]

Thank you 🙂

[@ayyappahemanth7134]

What if there are duplicates? Can I just add a row_number as id and do the joins based that id. Does that solve the problem for duplicates?

[@ragulmarley4900]

How to add this combination table do we need to create sir

[@datasculptor2895]

We need to create it. Script is there in the description

[@gsrsakhilakhil528]

bro dax meeda videos start chey bro

[@datasculptor2895]

No one is watching bro

[@ishanshubham8355]

Here i tried solving it using MYSQL
with recursive cte as (
select days,length(days) - length(replace(days,",","")) +1 as cnt
from combination
union
select days,cnt-1
from cte
where cnt >1),
cte2 as (
select *,
substring_index(substring_index(days,",",cnt),",",-1) as wrd
from cte
)
select days,
max(if(wrd="F",'TRUE',null)) as friday,
max(if(wrd="M",'TRUE',null)) as monday,
max(if(wrd="R",'TRUE',null)) as thursday,
max(if(wrd="S",'TRUE',null)) as saturday,
max(if(wrd="T",'TRUE',null)) as tuesday,
max(if(wrd="U",'TRUE',null)) as sunday,
max(if(wrd="W",'TRUE',null)) as wednesday
from cte2 as c1 join weeks as w
on c1.wrd = w.letter
group by days

[@chandramohan-bo5se]

Bro..
With out pivot..
Tell me

[@datasculptor2895]

Transform rows into columns WITHOUT PIVOT operator in SQL
ua-cam.com/video/WfQGE7FQZGI/v-deo.html