Google SQL Interview Problem | Solving SQL Interview Query

[@dataenggstudies]

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
;

[@hilarylomotey7051]

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

[@hardas81]

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;

[@GokulPonnappanPillay]

i will go with this one much simpler

[@AdinaAdina-mg5bz]

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

[@RaviKumar-pn8uw]

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

[@akiemcameron5303]

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

[@arslansohail6254]

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

[@Hemant13832]

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

[@hairavyadav6579]

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

[@fathimafarahna2633]

Awesome mate👌🏻 God bless you

[@dwaipayansaha4443]

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)

[@AbhishekSharma-vm7tr]

why count-1 not because there is one record with count 4 for this second most should be 2 but it is adam 3

[@arunadharmasena9236]

Hey man, It works great and without any problems.

[@tonysun203]

Really appreciate for your sharing. 👍

[@Sharmasurajlive]

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

[@SantiagoZuluaga]

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

[@imyours777]

how would you order them then? by default ordering is asc

[@SantiagoZuluaga]

@@imyours777 You're COUNTING, so you don't need to order them. Of course, you need to order them but in function ROW_NUMBER() so it's gonna stay as it is:
--using mysql
WITH t AS (
select U.*,
ROW_NUMBER()
OVER (PARTITION BY username ORDER BY endDate DESC) as order_activity,
COUNT(*)
OVER (PARTITION BY username) as total_activities
from useractivity AS U
)
SELECT username, activity, startDate, endDate FROM t
WHERE (order_activity = 1 AND total_activities = 1)
OR (order_activity = 2 AND total_activities >= 2);

[@imyours777]

@@SantiagoZuluaga ya your logic doesnt need to order but in his logic we need to order to use cnt-1
great 👍

[@harshavardhanachyuta2055]

@@imyours777 with cte as (
select *,
row_number() over(partition by username order by startdate) as rn,
count(*) over(partition by username) as cnt
from useractivity )
select username,activity,startdate,enddate
from cte where rn = (case when cnt > 1 then cnt-1 else 1 end)

[@sayantabarik4252]

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

[@maheshmmr9543]

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

[@adrie9522]

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

[@venkatareddykunduru1828]

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)

[@avi8016]

Thankyou very much sir for bringing this 🙏

[@Manishkumar-iw1cy]

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

[@abhipoornisnsadvocate5158]

Good to know about frame clause. Range one

[@sravankumar1767]

Superb bro 👌 👏 👍

[@dasoumya]

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

[@shadabsiddiqui28]

thanks for sharing.. love from india

[@reddaiahreddymallu]

Thank you, Thoufiq.

[@FaisalAli-ps7th]

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.

[@pavanareddy4536]

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;

[@gnataeee]

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?

[@aagastya7752]

It can be..

[@achyuthbudi9838]

yes

[@MJ-gu6uu]

Yes. And then union those rows with count=1. 👍

[@lankelajagadeesh2155]

Very nice explanation

[@SANDATA764]

Thank you thoufiq

[@techTFQ]

Thank you for alll the support Ahmed 🙏🏼
Appreciate it

[@SANDATA764]

@@techTFQ looking forward your upcoming python playlist InshaAllah

[@Krishna48784]

@@techTFQ thank you for sharing queries and explaining . Can you do a video on postgresql with json .

[@birthdaycake8438]

Fast download, thank you brother))

[@shilashm5691]

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(*)

[@vp4673]

@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

[@girishggirishg1083]

VERY INTERESTING

[@CassStevens]

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

[@rishabhgupta1029]

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)
)

[@angaddeepsingh7483]

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

[@fouzanak9162]

Awesome program

[@vikkynarepagul256]

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

[@jogerilely2828]

super! cheatwas easily installed

[@narasimhayadavmadduluri7129]

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);

[@cemanivannan27]

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);

[@mindlessscroll]

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

[@mohammadarslanjaved]

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

[@muditmishra9908]

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

[@vishalsonawane.8905]

Continue.....

[@KrishNa-vp8un]

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

[@KISHOREKUMAR-mu3xo]

@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;

[@yashkanojiya1146]

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

[@venkatgandam3573]

Thank you sir

[@AshutoshSingh-vn1kn]

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

[@srushtiOm]

Thanks much!

[@aryakurniasandi9671]

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

[@Mihiret468]

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

[@abhinasneupane2392]

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?

[@ea_naseer]

I think that you could sniff out start date collisions.

[@monasanthosh9208]

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);

[@007SAMRATROY]

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);

[@srishtijain642]

please make video on USER DEFINED FUNCTIONS in sql

[@pravaskumar4937]

Thanx

[@vishalsonawane.8905]

Please add the Data set for the practice

[@abhilasharya9435]

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

[@Mihiret468]

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

[@shibinjohn2792]

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);

[@nawalambavkar7543]

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

[@asavlogs8446]

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

[@sjsheikDawood]

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'

[@netpolun-ltd.7267]

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

[@ammar0466]

Can you make video about writing sql using github copilot

[@faizan4712]

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

[@theresepraisethelord201]

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

[@yashsaxena7754]

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)

[@karthikeyasoft]

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

[@narasimhamurty859]

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

[@hairavyadav6579]

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;

[@shubhampokar2532]

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

[@srinivasn415]

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

[@Life-d5k8g]

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

[@pavanareddy4536]

count(*) returns how many records are in the table, same for count(1), count(50), count(-1) doesn't matter the number, for count(number), it just counts total number of records in the table and assigns each record a number given in the count function.
Count(column) returns total number of values in the column which are not null. Null is not included here.

[@saraswathirajendran2395]

Using DENSE_RANK() and COUNT(), my solution is this for MYSQL:
WITH CTE AS (
SELECT username, activity, startDate, endDate,
DENSE_RANK() OVER (PARTITION BY username ORDER BY endDate DESC) AS date_rank,
COUNT(*) OVER(PARTITION BY username) AS activity_count
FROM UserActivity
)
SELECT username, activity, startDate, endDate
FROM CTE
WHERE date_rank = 2 OR activity_count = 1;

[@Leo-qo5hk]

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

[@nehalahmad5678]

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

[@sidharthadaggubati438]

No. You still need count of records per username.

[@obaiahlakkineni6065]

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?

[@jmhall1962]

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.

[@jcwynn4075]

Would this return extra records though? For example where activity_count is 1 and recency is also 1

[@jmhall1962]

@@jcwynn4075, the WHERE clause filters existing rows, it can never create addition rows.

[@jcwynn4075]

@@jmhall1962 I mean without the RANGE clause, the most recent activity for each person would have activity_count = 1, so would those get returned even if they have more than one activity?

[@jmhall1962]

@@jcwynn4075 In my query, COUNT partitions by username but does include an ORDER BY clause. This approach results in each row having activity_count equal to the total number of activities for the associated user. This is simpler than adding a RANGE clause to undo the undesired behavior caused by including ORDER BY in the COUNT function.

[@jmhall1962]

If the database supports the LEAST function, then changing the WHERE clause to "WHERE recency = LEAST(2, activity_count)" permits querying for nth recency by changing a single number.

[@pavipatil007]

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

[@ukkashs444]

Refer me any SQL related job. I am from Chennai

[@akshayvn14]

Data Engineer!

[@AbhishekSharma-vm7tr]

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

[@TargaryenGaming-u3u]

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

[@kamalmukhi3993]

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

[@chaithanyag93]

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

[@AdilShahzad-l7j]

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;

[@indergaming3053]

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*/

[@Code_With_Shami]

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;
)

[@gamingwithhemend9890]

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

[@basavapn6487]

That is correct , question is mostly second recent that means from last 2nd record needs to be fetched , if we have count=2, second recent record is first record

[@jcwynn4075]

If there are only 2 records, then the first record is the second most recent. How's it wrong?

[@gamingwithhemend9890]

Ohh okay!!!... I understand, thanks❤

[@shreymehta4795]

WITH cte AS(SELECT *,
dense_rank()over (partition by username ORDER by enddate )rn,
count(*) over (partition by username)cnt
from useractivity)
SELECT username,activity,startdate,enddate FROM cte
WHERE rn = 2 or cnt =1

[@owaowanothing1233]

man I missed this kind of tutorials lol. Great work here, thanks!!!

[@risasa1237]

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

[@aaravkumarsingh4018]

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

[@ecoprint3540]

For once, the software is actually really useful

[@raviteja8593]

i didnt understand wat is this cte

[@memesmacha61]

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

[@jacksparrow3595]

I have the same doubt,

[@jcwynn4075]

Second most RECENT is singing. Same as saying second to last

[@shivamsingla7596]

select username,activity,startdate,enddate from
(select *,row_number() over(partition by username order by startdate desc) as rnk
from useractivity) as z
where rnk=2
union
select * from useractivity where username in(select username from useractivity group by username having count(1)=1)
solution given by SQL KING nikhil anand

[@shabanashabana4603]

👍🏼