Google SQL Interview Problem | Solving SQL Interview Query 

The below simplified query worked for me in mysql with cte as (select *, row_number() over (partition by username order by startdate) as rn, count(*) over (partition by username) as ct from useractivity) select * from cte where rn = case when ct = 1 then ct else ct - 1 end ;

I'm getting the same results by running the following: with cte as (select *, row_number() over (partition by username order by startdate) as rn, count(1) over (partition by username) as cnt from useractivity) select * from cte where cnt = 1 or rn = cnt - 1;

My solution with the script:- create table activity1( username varchar(20), acitivity varchar(20), startdate date, enddate date); insert into activity1 (username,acitivity,startdate,enddate) values ('Amy','Travel','2020-02-12','2020-02-20'), ('Amy','Dancing','2020-02-21','2020-02-23'), ('Amy','Travel','2020-02-24','2020-02-28'), ('Joe','Travel','2020-02-11','2020-02-18'), ('Adam','Travel','2020-02-12','2020-02-20'), ('Adam','Dancing','2020-02-21','2020-02-23'), ('Adam','Singing','2020-02-24','2020-02-28'), ('Adam','Travel','2020-03-01','2020-03-28'); with t1 as(select username,activity,startdate,enddate,row_number() over(partition by username order by enddate desc) sorted_date, count(enddate) over(partition by username) count from activity1) select username,activity,startdate,enddate from t1 where (sorted_date=2 and count >1) or (sorted_date=1 and count=1)

IT'S ALWAYS THE UNDERRATED VID THAT'S LEGIT! THANK YOU!

Genius as always. Thanks for sharing. Am sure u are overwhelmed with so many emails now. I am sure mine is missing in your inbox. Anyway love your videos. Wish you could locate my mail tho 😂😂😂. Keep it up and we love you

First Like... I'm eagerly waiting for your videos sir... Thanks 👍❤️

Same if we mention DESC in order by clause of row number & then replace 'cnt-1' to '2'

My attempt with CTE and Window functions :- sample data: create table useractivity (username text, activity text, startdate date, enddate date); insert into useractivity values ('Amy','Travel','2020-02-12','2020-02-20'); insert into useractivity values ('Amy','Dancing','2020-02-21','2020-02-23'); insert into useractivity values ('Amy','Travel','2020-02-24','2020-02-28'); insert into useractivity values ('Joe','Travel','2020-02-11','2020-02-18'); insert into useractivity values ('Adam','Travel','2020-02-12','2020-02-20'); insert into useractivity values ('Adam','Dancing','2020-02-21','2020-02-23'); insert into useractivity values ('Adam','Singing','2020-02-24','2020-02-28'); insert into useractivity values ('Adam','Travel','2020-03-01','2020-03-28'); solution : with cte as (select *, rank() over (partition by username order by startdate desc) as rnk from useractivity), cte2 as (select username, activity, startdate, enddate, rank() over (partition by username order by startdate) as rnk from cte where rnk

Thanks, TFQ and love you boss the way you explain the query and solve it's up to the mark ♥

Thanks TFQ. Very helpful I have a basic question Leaving the rows with count =1 aside for now, would it be correct to sort the partition by DESCENDING order and then select the rows with row number rn=2 instead of the cnt-1?

with cte as ( select *, row_number() over(partition by username order by enddate) as rn, count(*) over(partition by username order by username) as cnt from activity1) select username,acitivity,startdate,enddate from cte where rn= case when cnt=1 then 1 else cnt-1 end

Thank you so much for sharing! In your solution, lines 4 & 5 could be simplified using: , COUNT(*) OVER(partition by username) AS cnt

I need to work a lot of similar problems so partition and case becomes second nature. This is a great tutorial

I don't understand why do we need to have count at all when the question concerns about activity and time sorting. We could just sort the data by start_date and use nth_value() to get the second most recent activity. My solution is below: with cte as ( select *, coalesce(nth_value(activity,2) over (partition by username order by start_date desc range between unbounded preceding and unbounded following),activity) as second_most_recent_activty from activity1) select username,activity,start_date,end_date from cte where activity=second_most_recent_activty; Coalesce is used because for Joe null will be returned as it has only one row.

Really thank you for this question and concept .. Tricky question

Hello! This is my simple solution: With cte as(Select *, row_number()over(partition by username order by startdate desc) as rn,count(*)over(partition by username ) as total_count From User activity) Select username, activity, startdate,enddate From cte Where rn=2 or total_count=1

easiest solution: with cte as (select *, row_number() over(partition by username order by startdate desc) rn, count() over(partition by username) as cnt from activity1) select * from cte where (rn=1 and cnt=1) or (rn=2 and cnt>1)

Hii, your video are more useful, so please upload video about the topics like cluster, and indexex, thnq

I think simplest one select username,activity,startdate,enddate from (select *, row_number() over(partition by username order by startdate desc) as rn from Useractivity) where rn=2 union select username,activity,startdate,enddate from Useractivity group by username having count(username)=1

Thank you thoufiq

sir a kind request is that always share a data set so that we can also practice it

Awesome mate👌🏻 God bless you

Thanks T! heading to learn the frame clause next. I tried out using your logic just ordered the result by start date in descending order. with cte as (select * ,row_number() over (partition by username order by startdate desc) rn ,count(*) over (partition by username order by startdate desc range between unbounded preceding and unbounded following) cnt from useractivity order by username, startdate desc) select username, activity, startdate, enddate from cte where rn= case when rn=cnt then 1 else 2 end;

Sir my solution in MySQL - select * from (SELECT uc.*, ROW_NUMBER() OVER(PARTITION BY username ORDER BY username,startDate) as row_count,count(*) OVER(PARTITION BY username) as total_count FROM `user_activity` uc) as x where x.row_count = 2 or (x.row_count = 1 and x.total_count = 1);

have solved by : select * from ( select *, row_number()over(PARTITION by username order by endDate desc) as rn, count(*) over (PARTITION by username) as total_records from Table_1 ) where ( (rn = 1 and total_records =1) or (rn = 2 and total_records 1) )

Thanks for sharing TFQ, The question says the table does not contain Primary Key! Does that mean there could be two person with same name, there could be different records for same name, eg there could be two differrent person named Amy? Do we need to cosider this as an edge case?

Using rank instead of row_number select username, activity, startDate, endDate from (select *, case when max(rnk) over (partition by username) = 1 then 'x' when rnk = 2 then 'x' end as slct from (select *, rank() over (partition by username order by endDate desc) as rnk from UserActivity ) x ) y where y.slct is not null

I'm getting the same results by running the following: if it's wrong, please let me know with cte as(select *,row_number() over(partition by username order by startdate desc) rn from activity1 qualify rn=2) select * exclude rn from cte union all select * from activity1 where username in (select username from activity1 group by username having count(*)=1);

@TechTfq -I could see that you use window function effectively in most of the solutions. Just want to know if window function is good performancewise also... Thanks

Thank you so much this helped a lot!!!! You saved my life

I feel this is straight forward.. Kindly let me know if this approach has any drawbacks.... with intr as (select *,count(*) over (partition by username ) as cnt, rank() over (partition by username order by startDate desc ) as rnk from UserActivity), flag as (select *, case when cnt =1 then 'valid' when rnk =2 then 'valid' else 'invalid' end as flg from intr) select * from flag where flg='valid'

Hey man, It works great and without any problems.

@techtfq,, why dont we use u this way based on rowid,,, select * from (select username,activity,startsate,count(*) over (partition by username) as cnt,row_number() over (partition by username order by rowid asc) rn from useractivity) where rn=2 or cnt=1;

It's not necessary to complicate the COUNT function with RANGE: partitioning by username is adequate. The following query is simpler: WITH activity_recency AS (SELECT *, ROW_NUMBER() OVER (PARTITION BY username ORDER BY startdate DESC) AS recency, COUNT(*) OVER (PARTITION BY username) AS activity_count FROM user_activity) SELECT username, activity, startdate, enddate FROM activity_recency WHERE recency = 2 OR activity_count = 1; This query can also easily be enhanced to return any nth most recent or the most recent activity if there aren't n activities.

with t2 as(select username, activity, startdate, enddate from(select *, rank() over (partition by username order by enddate desc) as recency from useractivity u) t1 where recency=2), t3 as (select * from useractivity u where username not in (select username from t2)) select * from t2,t3

Good to know about frame clause. Range one

Really appreciate for your sharing. 👍

Thank you Tofiq, based on your explanation I can solve the problem as below, what do you think? Is it possible or? With cte_secondActivity As (Select *, row_number() over(partion by userName order by start date desc) as row_nr From user Activity) Select username, activity,startdate,start date, From cte_secondActivity Where row_nr=2

One more thing I added one row for Joe having recent activity than the given one. To make the given activity second recent.

Thank you, Thoufiq.

Easier Solution: with cte as (select *, row_number() over (partition by username order by CURDATE() - endDate) as recent_day_number from activity_table) select * from cte where recent_day_number in (2) or username in (select username from activity_table group by username having count(*)

Please add the Data set for the practice

hi , I think for windows function: row_number(), it should be-> order by end_date desc.

Thankyou very much sir for bringing this 🙏

My approach please let me know this will work or not .. with cte as(select *,row_number() over(partition by username order by startdate desc) rn,count(*) over(partition by username order by startdate range between unbounded preceding and unbounded following) as cnt from activity) select username, acitivity,startdate,enddate from cte where rn = case when cnt = 1 then rn else 2 end;

thanks for sharing.. love from india

what application do you use to create the source code database?

Sharing an alternative! with cte as (select username,activity,startDate,endDate,rank() over(partition by username order by startDate desc) as rnk from user_activity), table1 as (select username,activity,startDate,endDate,rnk,max(rnk) over(partition by username) as max_rnk from cte) select username,activity,startDate,endDate from table1 where rnk = 2 or (max_rnk = 1)

This seems like way too much work for a solution thah can be much simpler than this. Overcomplicating it into

Respected sir I hope you are well .kinldy make a video for beginner to expert which course start for DB and which DBMS use. Which DBMS have a scope in future .plsease shear your experience in video .and tell step by step which course should first then second then third etc. And how we apply for job and which compney should apply through linked-in because each company required 2-3 year experiences.But we have don't experience. Kindly tell us a plate form or you tube channel link. Thanks a lot sir

Very nice explanation

Hi, i have shared one query over mail .it is complex one can you please make a video on that one??

with narsi as (select a.*,row_number()over(partition by username order by startdate desc) rn from activity1 a) select username,acitivity,startdate,enddate from narsi where rn=2 or username in(select username from narsi group by username having max(rn)=1);

MYSQL Solutions for Freshers With CTE as (Select *,dense_rank() over (Partition by Username order by StartDate) as Rn from Activity1), CTE1 as (Select *, Case When RN=2 then 1 Else 0 end as TRn from CTE), CTE2 as (Select *,Sum(TRN) over (Partition by Username) as TRN_Sum from CTE1) Select Username,Acitivity,Startdate,enddate from CTE2 Where (Case When TRN_SUM=1 then RN=2 Else RN=1 end);

group by/having/count/limit/offset - probably 3 times shorter :D

select * from (select *,row_number()over(partition by username order by (select 0) ) as rownum ,count (*) over (partition by username order by (select 0) ) as count from useractivity)useractivity where rownum= case when count =1 then 1 else count-1 end

Hi Toufiq, How to connect with you over telegram. Thanks

i am not getting why count - 1 because if i want second most value it should be (adam-Dancing Not Adam-singing) cnt - 1 is correct for amy but not for adam

Hi.when it's 4 it has returned the third row but we still need only the second?

I think if one user has only 2 records. we need to consider another case also. cnt =2

VERY INTERESTING

Sir. Trigger, cursor aur user defined function ki video banayiye

Solution - select * from (select username, activity , row_number() over(partition by username order by startdate desc) rn, count(*) over(partition by username order by startdate range between unbounded preceding and current row ) as c from user_activity ) X where rn=2 or c=rn

I use count and row_number logic : WITH t1 AS (SELECT *, Row_number() OVER( partition BY username ORDER BY enddate ) rn, Count(username) OVER( partition BY username) cc FROM activity1 ORDER BY username, enddate) SELECT * FROM t1 WHERE cc < 2 OR rn = 2

Superb bro 👌 👏 👍

Continue.....

My solution to the problem -> 1. Get ranks based on start date partitioned by user name ordered by username and rank - store it as x, 2. select all rows from x that have have ranks < 3 and create a case when using lead() to see if the next row has the same username as current_row -> 1 if Yes, 2 if No - store it as X, 3. Select all rows from X with lead = 0

super! cheatwas easily installed

Awesome program

Fast download, thank you brother))

I think simple or clause would have done the job, like 'where rn=2 OR cnt=1'

please make video on USER DEFINED FUNCTIONS in sql

I enrolled for the classes.. I haven’t got back anything yet

with cte as ( Select a.*, rank() over (partition by username order by startdate, enddate desc) as a_rank from activity1 a ) Select username,acitivity,startdate,enddate from cte where a_rank = 2 union all select username,acitivity,startdate,enddate from cte where username not in (Select username from cte where a_rank > 1);

Thank you sir

for sql workbench users with cte as (SELECT concat(id,' ', name) as concat,ntile(4) over(order by id) as 'ntiles' FROM interviewquestion.emp) select group_concat(concat) as result from cte group by ntiles order by 1

Hello Thoufiq, Thanks for the video!! I have one question. In the problem statement it is written that a user cannot perform two activities at the same time. Suppose if table contains records with overlapping time periods, shouldn't that condition be checked as well and those records be discarded? Thanks Kamal

Thanx

Sir, Will you be able to provide solution for below question? We have two records and five columns. Actually there is Duplicate record in that two records but one column is having different name for two records. So, my question is how to write a SQL query to remove that duplicate record in the Dashboard?

If count = 2, then cnt-1 would return the wrong value !!!, so you need to write a case for that one too. Pls reply

is that query work (SELECT username, activity, startdate, enddate FROM ( SELECT username, activity, startdate, enddate, ROW_NUMBER() OVER (PARTITION BY username ORDER BY enddate DESC) AS activity_rank FROM activities_table ) AS ranked_activities WHERE activity_rank = 2; )

Hi Taufeeq, I think if we go directly with original question then we can achieve result by using only row_number () , Right?

Can you make video about writing sql using github copilot

Hi TFQ, i do have 6 years of experience in SQL ,which role is good ? DATA science or Analyst ,kindly make video on how to find remote jobs for SQL

Brother , sorting family members problem could u slove in MS SQL server.

Hi TFQ May i know What is the difference b/w count (*) and count(1) and count (column)?

more simple solution : WITH ranked_activities AS ( SELECT *, ROW_NUMBER() OVER(PARTITION BY username ORDER BY startdate desc) AS rn, COUNT(*) OVER(PARTITION BY username) AS count FROM UserActivity ) SELECT username, activity, startdate, enddate FROM ranked_activities WHERE rn=2 OR count=1

create table useractivity (username varchar(50), activity varchar(50), startdate date, enddate date); insert into useractivity values ('Amy','Travel','2020-02-12','2020-02-20'); insert into useractivity values ('Amy','Dancing','2020-02-21','2020-02-23'); insert into useractivity values ('Amy','Travel','2020-02-24','2020-02-28'); insert into useractivity values ('Joe','Travel','2020-02-11','2020-02-18'); insert into useractivity values ('Adam','Travel','2020-02-12','2020-02-20'); insert into useractivity values ('Adam','Dancing','2020-02-21','2020-02-23'); insert into useractivity values ('Adam','Singing','2020-02-24','2020-02-28'); insert into useractivity values ('Adam','Travel','2020-03-01','2020-03-28'); select Row_number() over(partition by username order by startdate asc) as row,* into #temp_table from useractivity select username,count(1) as count into #final from #temp_table group by username having count(1)>1 insert into #final select username,count(1) as count from #temp_table group by username having count(1)=1 select ff.username, ff.activity, ff.startdate, ff.enddate from #final f join #temp_table ff on f.username=ff.username and ff.row=1 and f.count=1 union all select ff.username, ff.activity, ff.startdate, ff.enddate from #final f join #temp_table ff on f.username=ff.username and ff.row=2 and f.count1

Your output has not matched. Please check it once properly.

select * from(select *, dense_rank() over(partition by username order by startDate desc) as rank, count(*) over(partition by username) as count from Google_UserActivity)x where rank=2 or count=1

please provide ddl scripts in video so that we can practise from ourself before watching your solution

Hi, Kindly review my code, improvement suggestions are highly solicited. Thank you. With master as ( Select *, Rank()over(partition by username order by endDate desc) as rank, Lead(endDate,1,0)over(partition by username order by endDate desc) as flag From UserActivity) Select Username, Activity, Startdate, Enddate From master Where rank =2 or flag = 0

i didnt understand wat is this cte

I think first time every one get confused

with cte as ( select *,LEAD(rn,1,0) over (partition by username order by (select null) ) as ld from ( select * from ( select *, row_number() over (partition by username order by (select null) ) as rn from useractivity) A)B where rn in (1,2)) select username,activity,startdate,enddate from cte where ld=0 /*please rate this query out of 10*/

WITH cte1 AS ( SELECT *, RANK() OVER(PARTITION BY username ORDER BY startdate DESC) AS rnk, COUNT(*) OVER(PARTITION BY username ORDER BY startdate DESC RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) AS total FROM activity) SELECT username, acitivity, startdate, enddate FROM cte1 WHERE rnk = 2 OR total = 1;

select username,ld as activity from( select *,ROW_NUMBER() over (partition by username order by startdate desc ) rnk FROM ( select *,lead(activity,1,activity) over (partition by username order by startdate desc) ld from [dbo].[UserActivity] )a) b where rnk =1

For once, the software is actually really useful

I think for adam 's second most activity is dancing right??

google sql query docs public