Freshworks Data Analyst SQL Interview Problem | SQL For Data Analytics

[@ankitbansal6]

Give me 1000 likes on this video and I will create a video on how to create a calendar table from scratch 😊

[@shipra959]

Bring only records with StudentMarks greater than 75. Catch is DO NOT use WHERE/GROUPBY CLAUSE Sample Input: StudentId StudentMarks 1 99 2 76 3 71 4 50 5 76 Expected Output: StudentId StudentMarks 1 99 2 76 5 76 recently i faced this qus in ey interview for data engineer with 4 yr of exp

[@naveenbhandari5097]

Hi Ankit bhai, Today I have completed all the videos from your channel. Here I just want to thank you for making such amazing videos. Your way of explaining things is really commendable, I have failed in many interviews bcos of advanced SQL concepts but this time I have gained confidence I never had. Once again thank you for making such life-changing videos. Keep doing great! may god give you all the success you wish! Thanks, Man. looking forward a great learning ahead from your channel.

[@ankitbansal6]

Glad to know that ☺️ keep rocking 💪

[@Amulya869]

This KZbin channel is more useful.Give me some more like this

[@avi8016]

Wow, this was great 💯 I guess I'll need to work on the date function Thankyou 🙏

[@girishpv8193]

Very good explanation Ankit... Initially I thought this looks simple..but the way you generalized the query is awesome.. Keep going 👏

[@ankitbansal6]

Thanks a ton🙏

[@akashgoel601]

thanks for this, posting my sol. little different approach(actually it similar to your second soln, after watching complete video i realised): with cte as ( select price_date,price, lead(DATEADD(day,-1,price_date),1,DATEADD(month,1,price_date)) OVER(order by price_date) as lead from sku ), cte_2 as ( SELECT price_date ,DATEADD(DAY, 1, EOMONTH(price_date, 0)) as frst_Date from sku ) select price_date,price from sku where DATEPART(day,price_date)=1 union all select distinct s.frst_Date,a.price from cte a join cte_2 s on s.frst_Date BETWEEN a.price_date and a.lead

[@KisaanTuber]

Hi Ankit. Thanks for posting & explaining such challenging SQL problems. Here is my stab at the problem without using calendar table: with RECURSIVE t1 as ( SELECT date_trunc('month', MIN(price_date)) as month_date from sku UNION ALL SELECT month_date+interval '1 month' as month_date from t1 where month_date=sku.price_date) SELECT month_date, month_price from t2 where price_rnk=1 ORDER by 1;

[@nidhisingh4973]

Hello Ankit, Really grateful to you for all these amazing videos.

[@MubarakAli-qs9qq]

🎉Sir mast question h

[@hairavyadav6579]

Nice explanation sir great.....

[@sandhyakumari8204]

Thanks Ankit, it will be helpful if you can create a video on making of calendar table!

[@ankitbansal6]

Okay sure

[@sachinn5503]

WITH CTE AS( SELECT *, ROW_NUMBER() OVER(PARTITION BY MONTH(PRICE_DATE)) D, LAG(PRICE) OVER(ORDER BY PRICE) S FROM SKU) SELECT PRICE_DATE,PRICE,(PRICE-S) AS DIFF FROM CTE WHERE D=1;

[@kadagaladurgesh3691]

Great explanation Thanks for the video, I have a doubt At time 13:56 to avoid duplicates we use new condition with and operator, can we achieve same result with Union instead of union all

[@gautamigaikwad4549]

Please create a video on how to create calendar table 15:10

[@grim_rreaperr]

DECLARE @StartDate DATE = CAST('2000-01-01' AS DATE); /*set start date*/ DECLARE @EndDate DATE = CAST('2024-12-31' AS DATE); /* set end date */ WITH calendar AS ( SELECT @StartDate AS cal_dates UNION ALL SELECT DATEADD(DAY, 1, cal_dates) FROM calendar WHERE cal_dates < @EndDate ) SELECT cal_dates, MONTH(cal_dates) AS cal_month, DATEPART(DAYOFYEAR, cal_dates) AS cal_year_day, DAY(cal_dates) AS cal_month_day, DATEPART(WEEK, cal_dates) AS cal_week, DATEPART(WEEKDAY, cal_dates) AS cal_week_day, DATEPART(QUARTER, cal_dates) AS cal_quarter_num FROM calendar OPTION(MAXRECURSION 0);

[@grim_rreaperr]

working as intended in ms sql server, we can other attribute columns as well like month name, day name etc

[@gautamigaikwad4549]

Thank you

[@grim_rreaperr]

@@gautamigaikwad4549 🙏

[@apna9656]

Hi Ankit, It would be helpful for us, if you can create a video on calender table

[@ankitbansal6]

Okay sure

[@SreemantaKesh]

great question

[@nash_life]

Good Explanations Sir. I failed 3 interviews in the past 3 days because of SQL. i am not sure why I am not able to build solutions. I hope to learn from your videos.

[@ankitbansal6]

Don't worry keep practicing

[@srikarrar261]

14:24 Hi Sir, may I know what will happen instead of taking UNION ALL with UNION. I think we don't need to use Subquery to filter out the price data having 1st day of month

[@addhyasumitra90]

my approach: following 1st method with lead-lag with CTE as ( select *, ROW_NUMBER() OVER(partition by sku_id, month(price_date) order by price_date desc) as rn from SKU) select sku_id, price_date, price from SKU where DATEPART(DAY, price_date)=1 UNION ALL select sku_id, datetrunc(month,isnull(LEAD(price_date) OVER(partition by sku_id order by price_date), DATEADD(month,1,price_date))) as next_month, price from CTE where rn=1

[@rajkumarrajan8059]

Ankit, Please create the calendar table from scratch !!!!!

[@grim_rreaperr]

DECLARE @StartDate DATE = CAST('2000-01-01' AS DATE); /*set start date*/ DECLARE @EndDate DATE = CAST('2024-12-31' AS DATE); /* set end date */ WITH calendar AS ( SELECT @StartDate AS cal_dates UNION ALL SELECT DATEADD(DAY, 1, cal_dates) FROM calendar WHERE cal_dates < @EndDate ) SELECT cal_dates, MONTH(cal_dates) AS cal_month, DATEPART(DAYOFYEAR, cal_dates) AS cal_year_day, DAY(cal_dates) AS cal_month_day, DATEPART(WEEK, cal_dates) AS cal_week, DATEPART(WEEKDAY, cal_dates) AS cal_week_day, DATEPART(QUARTER, cal_dates) AS cal_quarter_num FROM calendar OPTION(MAXRECURSION 0);

[@sobermenezes]

Excellent video Ankit. A query on your second method though: the inner join you used has an incomplete ON clause (on.c.cal_date). How’s that possible?

[@xiamojq621]

Good video but have you thing about procedures and functions questions there are very rare in KZbin

[@DEwithDhairy]

PySpark Version of this problem : kzbin.info/www/bejne/mWqXmI1ug8mAfqM

[@arpanscreations6954]

Hi Ankit, I think we can add the condition in the cte "where day(price_date) 1" isn't it?

[@TY-zl1vw]

Very informative, first time I heard about the DATE_TRUNC function, but it's not available for me to practice, since I'm using SQL Server 2018. Edit: Could DATEADD(DAY, 1,EOMONTH(price_date,0)) achieve the same ?

[@vishnugottipati9373]

thank you

[@vaibhavverma1340]

Here is my Attempt Sir , Please have a look. with cte as (select *, DATEFROMPARTS(year(price_date),month(price_date),'01')start_of_month ,(case when price_date > DATEFROMPARTS(year(price_date),month(price_date),'01') then lag(price,1) over (partition by sku_id order by price_date) else price end)price_start_of_month from sku) ,eliminate_duplication_months as (select sku_id, start_of_month, price_start_of_month, dense_rank() over (partition by sku_id, start_of_month order by price_date)dr from cte) select sku_id as SKU, start_of_month as [Date] , price_start_of_month as Price, price_start_of_month -lag(price_start_of_month,1,price_start_of_month) over (partition by sku_id order by price_start_of_month)Dif from eliminate_duplication_months where dr =1

[@karangupta_DE]

with recursive cte as ( select (select min(price_date) from sku) as all_dates union all select all_dates + interval '1 day' from cte where true and all_dates

[@srikarrar261]

Sir, please Create Calendar Table video

[@harshSingh-if4zb]

A little mess but getting correct output: select sku_id ,price_date, price, dr, case when month = 0 then price else lg end as final_price from (select *, lag(price,1,0) over(order by price_date) as lg from (select *, dense_rank() over(partition by month1 order by price_date desc) as dr from (select *, concat(left(price_date , 7) , "-01") as month1, datediff(price_date, concat(left(price_date , 7) , "-01")) as month from sku) a)b where dr=1)c;

[@its_anky5382]

with cte as ( Select *,row_number() over(partition by sku_id, year(price_date),month(price_date) order by sku_id,month(price_date) asc,price_date desc) rnk from sku ) select sku_id,case when day(price_date)=1 then price_date else DATETRUNC(month,price_date) end as start_date , isnull(lag(price) over(order by price_date),price) price_at_month_start from cte where rnk=1

[@Aman-lv2ee]

We can use union also instead of union all and a subquery: with cte as ( select *,dense_rank()over(partition by sku_id, extract(month from price_date), extract(year from price_date) order by price_date desc) as dk from sku ), cte2 as ( select sku_id, price_date as new_price_date, price from sku where date_part('day', price_date) = 1 union select sku_id, date(date_trunc('month',price_date+INTERVAL '1 month')) as new_price_date , price from cte where dk =1 ) select *, lag(price,1,10)over(order by extract(month from new_price_date)), price-lag(price,1,10)over(order by extract(month from new_price_date)) as difference from cte2

[@vinil9212]

smart work handling the NULL, but you could have just used COALESCE

[@LakshmiPujitha-g7b]

Hi Bhai, My answer is ;WITH CTE AS( select MIN(PRICE_DATE) dt, 1 as cnt, datepart(MONTH,max(PRICE_DATE)) CNTDT from SKU union all select DATEADD(MONTH,1, dt) dt, cnt+1 cnt, CNTDT FROM CTE where CNTDT>=CNT ), skct as( select PRICE_DATE, lead(price_date,1, DATEADD(month,1,price_date)) over(ORDER BY price_date) NXTDT, price FROM sku ) select *, price, ABS(PRICE- lag(price,1, price) over(ORDER BY price_date)) FROM cte left join skct on 1=1 and dt between price_date and nxtdt

[@Dhanushts-g7x]

with recursive cte1 as (Select min(price_date) pd from sku union all select date_add(pd,interval 1 day) pd from cte1 where pd

[@zaravind4293]

Hi Ankit I have one query how to get the alternate characters in upper case remaining in lower case like name is Rahul then output should be RaHuL. how can we achieve this in sql

[@vikasvk9174]

Hi Ankit, I have one doubt instead of 2023-01-01 we have 2023-01-10 in that case will not get first recode in our final output ryt ?

[@ritusantra8641]

with cte as (select *, cast (dateadd(mm,DATEDIFF(mm,0,price_date)+1,0) as date) as date ,rank() over(partition by year(price_date),month(price_date) order by day(price_date) desc) as rnk from sku) select sku_id, date, price from cte where rnk = 1 union all select sku_id, price_date, price from cte where day(price_date) = 1 and month(price_date) =1 order by date;

[@hariikrishnan]

How many YOE candidates can expect such questions ? Found it quite hard as a fresher (< 1YoE)

[@ankitbansal6]

It's a tough one ..4 plus YOE

[@jay_rana]

In the 1 st attempt can't we use Union instead of Union All, this will remove the duplicate record with same price date value on 1st day of month cases ??

[@apurvasaraf5828]

with cte as (select *,RANK() over(partition by sku_id,month(price_date) order by day(price_date) desc) as r from sku) select price_date,price from sku where day(price_date)=1 union all select datetrunc(MONTH,DATEADD(month,1,price_date)) as d, price from cte where r=1

[@vinil9212]

with cte as ( select sku_id,price_date,price, ROW_NUMBER() OVER(PARTITION BY month(price_date) order by price_date desc) AS rnk FROM sku), cte2 as( SELECT sku_id,DATETRUNC(month,DATEADD(month,1,price_date)) AS next_month,price FROM cte WHERE rnk=1 UNION ALL SELECT * FROM sku WHERE DATEPART(day,price_date)=1 ) SELECT *,coalesce(price-LAG(price) OVER ( ORDER BY next_month),0) AS price_diff FROM cte2 ORDER BY next_month;

[@rohitsharma-mg7hd]

my simple solution: with cte1 as (SELECT month::date FROM generate_series('2023-01-01', '2024-01-01', INTERVAL '1 Month') month), cte2 as (select *,lead(price_date,1,'2023-05-01') over() as prev_date from sku) select * from cte1 c1 join cte2 c2 on c1.month between c2.price_date and c2.prev_date

[@amanpratapsingh8631]

Here is my solution in MySQL: with cte as( select *,row_number() over(order by sku_id) as m, year(price_date) as y from sku), cte2 as( select *,concat(y,"-",m,"-","01") as concat_date from cte), cte3 as( select sku_id,price_date,price,STR_TO_DATE(concat_date,"%Y-%m-%d") AS converted_date from cte2 order by converted_date), cte4 as( select *, lag(price) over(order by price) as lag_price, datediff(converted_date,price_date) as dd from cte3) select sku_id,converted_date, case when dd>=0 then price when dd

[@madhurimadas6260]

I did like this with cte as(SELECT *, DATEADD(month, DATEDIFF(month, 0, price_date) + 1, 0) AS first_day_of_month, row_number()over(partition by year(price_date),month(price_date) order by price_date)as rnk, lead(price) over(partition by year(price_date),month(price_date) order by price_date)as pre FROM sku) select cte.sku_id,case when pre is null then price else pre end as price,first_day_of_month from cte where cte.rnk=1 union all select sku_id,price,price_date from sku where datepart(day,price_date)=1 order by first_day_of_month

[@saketarora]

Doing unions all and then not in?? Could have just done union?

[@ankitbansal6]

The price can be different ..

[@june17you]

Hi can anyone help me on what is the equivalent function of datetrunc in mysql

[@ankitbansal6]

You need to use the extract function

[@sammail96]

For oracle sql also extract function works same way as datetrunc@@ankitbansal6

[@srinivasareddybandi983]

with cte1 as ( select *,row_number() over(partition by sku_id,extract(month from price_date) order by price_date desc) rn from sku ) ,cte2 as ( select sku_id,price_date,price,cast(date_trunc('month',price_date) as date) as dt,lag(price,1,price) over() as lp from cte1 where rn = 1 union all select sku_id,cast(price_date+interval '1 month' as date) as price_date,price, cast(date_trunc('month',price_date+interval '1 month') as date) as dt,price as lp from sku where price_date = (select max(price_date) from sku) ) select dt,new_price,new_price-lag(new_price,1,new_price) over () as diff from ( select dt,price_date, case when dt=price_date then price when dt

[@atulsharma2789]

Ankit this solution is work or not ?? with cte as( select *, ROW_NUMBER() over(partition by sku_id,year(price_date),month(price_date) order by price_date desc) skudate from sku),cte2 as( select sku_id,price_date,DATEADD(Month,DATEDIFF(Month,1,price_date),0)nextofmonth,price from cte where skudate=1), cte3 as( select sku_id,price_date,price,isnull(lead(price_date) over(order by sku_id),'2023-05-01') nextmonth1 from cte2 ) select sku_id,price_date,price from sku where datepart(day,price_date)=1 union all select sku_id,DATEADD(Month,DATEDIFF(Month,1,nextmonth1),0),price from cte3

[@ykirankumar4985]

with cte as ( select price_date,price, dateadd(day,1,EOMONTH(price_date,-1)) as mo_start, dateadd(day,1,EOMONTH(price_date)) as next_mo_start, first_value(price) over(partition by month(price_date) order by price_date desc) as mo_end_price from sku), prev_month as ( select mo_start,lag(mo_end_price,1,mo_end_price) over (order by mo_start) as prev from cte ),next_month as ( select next_mo_start,mo_end_price from cte ), final as ( select distinct coalesce(mo_start,next_mo_start) as [Date], coalesce(prev,mo_end_price) as Price from prev_month full join next_month on prev_month.mo_start=next_month.next_mo_start) select [Date], Price,Price-lag(Price,1,Price) over (order by [Date]) as diff from final;

[@sonalijain9024]

I USED THIS APPROACH SO IT WILL WORK FINE ALL POSIBILITIES. IT WILL BE A GENERIC SOLUTION .PLZ TELL ME with cte as ( select *, ROW_NUMBER() over(partition by sku_id,year(price_date),month(price_date) order by price_date desc) rnk, datetrunc(MONTH,DATEADD(month,1,price_date)) next_month from sku ) select sku_id,next_month,price from cte where rnk =1 union all select sku_id,price_date,price from sku where DATEPART(day,price_date) = 1 order by next_month asc

[@AmanRaj-uf7wx]

MYSQL: with cte as ( select *, row_number() over (order by sku_id) as mth, str_to_date((concat(year(price_date), "-", row_number() over (order by sku_id), "-", "01")),"%Y-%m-%d") as updated_date from sku ) ,cte2 as ( select sku_id, price_date, price,updated_date, lag(price,1,0) over (order by price ) as lag_price, datediff(updated_date, price_date) as dd from cte order by updated_date) select sku_id, updated_date, case when dd >= 0 then price when dd

[@MITHUNKUMAR-nx8rd]

Thanks

[@sourabhpatel3834]

WITH CTE1 AS (SELECT price_date, price, ROW_NUMBER() OVER (PARTITION BY MONTH(PRICE_DATE) ORDER BY PRICE_DATE DESC) RNK, DATEADD(DAY, 1, EOMONTH(price_date)) next_month FROM SKU), CTE2 AS (SELECT price_date, price, price_date AS next_month FROM sku WHERE DAY(price_date) = 1 UNION ALL SELECT price_date, price, next_month FROM CTE1 WHERE RNK = 1) SELECT *, price - LAG(price, 1, price) OVER (ORDER BY next_month) diff FROM CTE2;

[@skkholiya]

with cte as( select *, lead(price_date) over(order by price_date) next_date, lag(price) over(order by price_date) pre_price, row_number() over(partition by sku_id) rn from sku ), cte_month as( select *,date(concat(year(price_date),"-",rn,"-","1")) month_date from cte ) select *, coalesce(price-lag(price) over(order by month_date),0) dif from ( select sku_id, month_date,if(price_date

[@ManishChauhan-fb7uz]

select sku_id as SKU ,price_date as Date ,price ,prev_price-lag(prev_price,1,prev_price) over(order by price_date) as Diff from ( select * ,lag(price, 1, price) over(order by date_part('month',price_date)) as prev_price ,rank() over(partition by date_part('month', price_date) order by date_part('day',price_date)) as rnk from sku ) as t where rnk = 1

[@story_teller_Is]

sir aap tottle hn kya

[@ankitbansal6]

Haan m totla hoon. Mere papa bhi totle hain..Mera pura khandaan totla hai. Hum sab TA ko TA bolte hain

[@story_teller_Is]

@@ankitbansal6 😃lagta h aap bhavuk hogye😁

[@ankitbansal6]

@@story_teller_Is haha just kidding 😂

[@PraveenSinghRathore-df3td]

MYSQL solution: with recursive cte as ( select min(price_date) as first_date, '2023-12-31' as end_date from sku union all select date_add(first_date,interval 1 day) as first_date, end_date from cte where first_date < end_date ), cte2 as (select first_date, day(first_date) as cal_day, month(first_date) as cal_month from cte) select s.sku_id,c.first_date,s.price from cte2 c join (select sku_id,price_date,date_add(lead(price_date,1,date_add(price_date,interval 1 month)) over(partition by sku_id order by price_date),interval -1 day) as valid_till,price from sku) s on c.first_date between s.price_date and s.valid_till where cal_day = 1 order by first_date;

[@sauravverma8355]

@ankit bansal select *,ifnull(price-lag(price,1) over(partition by sku_id),0) as Diff from (WITH cte AS ( SELECT *, ROW_NUMBER() OVER (PARTITION BY sku_id, YEAR(price_date), MONTH(price_date) ORDER BY price_date DESC) AS rn FROM sku) SELECT sku_id, STR_TO_DATE(DATE_FORMAT(price_date + INTERVAL 1 MONTH, '%Y-%m-01'), '%Y-%m-%d') AS 'date', price FROM cte WHERE rn = 1 UNION SELECT *FROM sku WHERE DAY(price_date) = 1 order by month(date) asc) as a; this one is simpler soln as compared to u

[@ManpreetSingh-tv3rw]

Could this have been done by the recursive CTE ? Like expanding the rows from Jan 1 to Jan 30 , then feb 1 to feb 29 ? I am trying this approach not sure if it will work. @ankitbansal6

[@sabesanj5509]

WITH cte AS ( SELECT SKU, DATE, PRICE ROW_NUMBER OVER (PARTITION BY SKU ORDER BY DATE) AS rnk FROM prices ) SELECT SKU, DATEFORMAT(DATE, '%y%M-01') AS start_of_month, PRICE FROM cte WHERE rnk = 1;