Amazon Interview | SQL Interview Problem asked during Amazon Interview
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- Опубліковано 28 вер 2024
- In this video, let us solve an SQL Problem asked during the Amazon Interview.
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Hi , can you show one query for reverse from date ranges to single dates
Thanks for the problem and explaination!. This was my solve:
with mycte as
(
SELECT *,
rank() over(partition by employee, status order by dates) as rn,
datepart(day, dates) as theday,
(datepart(day, dates) -rank() over(partition by employee, status order by dates)) as diff
from emp_attendance
)
select employee, from_date, to_date, status
from
(
select employee, diff, status, min(dates) as from_date, max(dates) as to_date
from mycte
group by employee, diff, status
) as x
order by 1,2,3
Nice one 👍👍
Can anyone say I am fresher will I facethis type of complex queries in interview I mean is this query for freshers or experienced persons
My solution in PostgreSQL:
WITH CTE AS
(SELECT *
, dates - (ROW_NUMBER() OVER(PARTITION BY employee, status ORDER BY dates)::INT) AS grp
FROM emp_attendance)
SELECT employee, MIN(dates) AS from_date
, MAX(dates) AS end_date, status
FROM CTE
GROUP BY employee, grp, status
ORDER BY employee, from_date
In my ROW_NUMBER function, I have partitioned by employee and status, and ordered by dates. For each employee, the data is grouped by status and ordered by dates. The row number resets to 1 whenever the status changes (from present to absent or vice-versa) within each employee's partition.
I then subtracted the row number from the date to create a group identifier (grp) to identify consecutive dates within the same status for each employee.
how does this work -
dates - (ROW_NUMBER() OVER(PARTITION BY employee, status ORDER BY dates)::INT)
could you please explain with an example
Good
i did it same as you did but this solution would not work if month/year changes in the same group.
@@anshulmehta5732
When considering scenarios where the month or year changes within the same group, additional partitioning by month and year becomes necessary for the solution to work correctly.
To verify this, I included entries from Jan 29th to Feb 3rd with the status "Present" and modified the query using --- ROW_NUMBER() OVER(PARTITION BY employee, status, EXTRACT(MONTH FROM dates) ORDER BY dates)::INT
This approach created two distinct records: one from January 29th to 31st and another from February 1st to 3rd.
Similarly, we can achieve the same result for years by partitioning based on EXTRACT(YEAR FROM dates).
Full Query:
WITH CTE AS (
SELECT *
, dates - ROW_NUMBER() OVER(PARTITION BY employee, status, EXTRACT(MONTH FROM dates) ORDER BY dates)::INT AS grp
FROM emp_attendance
)
SELECT employee
, MIN(dates) AS from_date, MAX(dates) AS end_date, status
FROM CTE
GROUP BY employee, grp, status
ORDER BY 1, 2
@@anshulmehta5732 When considering scenarios where the month or year changes within the same group, additional partitioning by month and year becomes necessary for the solution to work correctly.
To verify this, I included additional entries from Jan 29th to Feb 3rd with the status "Present" and modified the query using -- ROW_NUMBER() OVER(PARTITION BY employee, status, EXTRACT(MONTH FROM dates) ORDER BY dates)::INT
This approach created two distinct records: one from January 29th to 31st and another from February 1st to 3rd.
Similarly, we can achieve the same result for years by partitioning based on EXTRACT(YEAR FROM dates).
Full Query:
WITH CTE AS (
SELECT *
, dates - ROW_NUMBER() OVER(PARTITION BY employee, status, EXTRACT(MONTH FROM dates) ORDER BY dates)::INT AS grp
FROM emp_attendance
)
SELECT employee
, MIN(dates) AS from_date, MAX(dates) AS end_date, status
FROM CTE
GROUP BY employee, grp, status
ORDER BY 1, 2
solving challenging queries from top mnc with nice explanation, great thoufiq keep it up.
with A as (select
*,
row_number() over (partition by employee,status order by dates) as rnk
from emp_attendance
),
B as (
select
*,
dates - CONCAT(rnk::text, ' day')::interval as diff
from A
)
select
employee,
min(dates) as start_date,
max(dates) as end_date,
max(status)
from
B
group by employee,diff
order by 1,2
@TFQ can we use min and max instead of first_value and last_value in the window function?
You can achieve the same differentiation between employees based on status by simply using rank() and partitioning it by employee, status (as in the first cte).
WITH rank_cte AS (
SELECT
*,
rank() OVER(partition by employee, status order by dates) as r
FROM emp_attendance
ORDER BY employee, dates
),
consec_cte AS (
SELECT
*,
r - row_number() OVER() AS consec
FROM rank_cte
)
SELECT
employee,
MIN(dates) AS start_date,
MAX(dates) AS end_date,
status
FROM consec_cte
GROUP BY employee, status, consec
ORDER BY employee, start_date;
have a question for you @techTFQ , how much time u have taken to come up for this solution ? just curious to know an approximate time
🤣🤣
this one is a very tough question, for what level role was this question asked 😰
very hard to think about this question and finish in 30 mins
Tqs For giving Valueble Infomation.
select max(amount) as thirdhighamount from orders where amount
my solution
with cte as(SELECT *
,rank() over (partition by employee,status order by dates asc) rnk
from emp_attendance
order by employee,dates),
cte2 as (
select *,(extract(day from dates) - rnk) diff
from cte)
select employee,min(dates) from_date,max(dates) to_date,status
from cte2
group by employee,status,diff
MYSQL Solution
Select employee,Min(Dates) as From_date,Max(Dates) as End_Date,Status from
(Select *,subdate(Dates,interval Row_Number() over
(Partition by Employee,Status Order by dates) Day) as Seg from
Emp_Attendance)N group by employee,Seg order by Employee, Dates;
Cte will work in Oracle db ??? Pls confirm???
Hi Taufiq ,Please confirm my solution is how optimal?
with cte as(
SELECT *,lead(status,1,null) over(partition by employee order by dates) as next_day,min(dates) over(partition by employee) as start_day FROM emp_attendance)
select employee,date_add(LAG(DATES,1,DATE_SUB(START_DAY,1)) OVER(PARTITION BY EMPLOYEE order by dates),1) AS FROM_DATE, dates as TO_DATE,status from cte where status!=next_day or next_day is null;
Where we can find the dataset?
In the description box click on script link and download that script you’ll get all queries
Lewis Anthony Perez Patricia Brown Thomas
Harris Elizabeth Garcia Sandra Wilson George
with cte as (select *,
Date - INTERVAL '1' DAY * (row_number() over(partition by Employee, Status order by Date asc)) as rnk
from EMP_ATD)
select
Employee,
min(Date),
max(Date),
Status
from cte
group by Employee, rnk, Status
order by Employee, min(date);
My solution in ms SQL server:
SELECT employee,from_date,to_date,Status
FROM(SELECT grp,employee,MIN(dates) AS from_date,MAX(dates) AS to_date,min(status) AS Status
FROM(
SELECT ROW_NUMBER() OVER(PARTITION BY employee ORDER BY dates) -ROW_NUMBER() OVER(PARTITION BY employee,status ORDER BY dates) AS grp,*
FROM emp_attendance)a
GROUP BY grp,employee)b
ORDER BY employee,from_date;
MS SQL approach
with a as (
SELECT *, ROW_NUMBER() over(partition by employee order by dates) rn
from emp_attendance)
,b as (
select *,rn - ROW_NUMBER() over(partition by employee order by dates) rn2
from a
where status like 'PRESENT')
,c as (
select *,rn - ROW_NUMBER() over(partition by employee order by dates) rn2
from a
where status not like 'PRESENT')
select employee, status, min(dates) from_date, max(dates) to_date
from b
group by rn2, employee, status
union
select employee, status, min(dates) from_date, max(dates) to_date
from c
group by rn2, employee, status
order by 1, 3
Allen Betty Moore Brenda Moore David
Harris Maria Clark Jason Moore Ronald
Brown Cynthia Wilson Michael Robinson Donna
Hall Ronald Moore Robert Robinson Robert
Bro Odin school is not a good option, i wasted my time and money , They wont provide you placements , I joined in 2022 , still i am not get a job through it, pls dont waste ur time and money
Anderson Thomas Garcia Michelle Taylor Mary
My solution:
WITH cte AS (
SELECT *,
CASE WHEN status = LAG(status, 1, status) OVER (PARTITION BY employee ORDER BY dates) THEN 0 ELSE 1 END AS flag
FROM emp_attendance
),
cte2 AS (
SELECT employee, dates, status, SUM(flag) OVER (PARTITION BY employee ORDER BY dates) AS flag_sum
FROM cte
)
SELECT employee, MIN(dates) AS from_date, MAX(dates) AS to_date, MAX(status) AS status
FROM cte2
GROUP BY employee, flag_sum
ORDER BY employee, from_date;
Sir, Is there will be any difference i use iif inplace of case Statment???
Isn't this approach more straight forward?
WITH grouped_attendance AS (
SELECT
employee,
dates,
status,
DATE_SUB(dates, INTERVAL ROW_NUMBER() OVER (PARTITION BY employee, status ORDER BY dates) DAY) AS group_date
FROM emp_attendance
)
SELECT
employee,
MIN(dates) AS from_date,
MAX(dates) AS to_date,
status
FROM grouped_attendance
GROUP BY employee, status, group_date
ORDER BY employee, from_date;
I guess you are always overcomplicating things don't know why!
with first as (
select *,lag(status,1) over(partition by employee order by dates) as prev_status
from emp_attendance
),
second as (
select b.* from (
select *,case when status = prev_status then 'SAME' else 'CHANGE' end as status_check from first
) b
where b.status_check='CHANGE'
),
final as (
select employee ,dates as from_date ,lead(dates,1) over(partition by employee order by dates)-1 as to_date,status
from second )
select employee,from_date,coalesce(to_date,from_date),status from final
order by employee,from_date;
Hi thoufiq! Here is my simple solution using SQL server:
with cte1 as(select employee, dates, dateadd(day,-1*(row_number()over(partition by employee,status order by dates)),dates) as date_grp,status
from employee)
select employee,min(dates) as from_date,max(dates) as to_date, status
from cte1
group by employee,date_grp,status
order by employee,from_date;
can you explain below part 😅😅
dateadd(day,-1*(row_number()over(partition by employee,status order by dates)),dates) as date_grp
@@mahivamsi9598 -1*(row_number()over(partition by employee,status order by dates) this value will give positive value so he decided to multiply with -1 so that it gets negative value so that difference can be created
super
Can you please start a Snowflake Bootcamp ? Will be really helpful.
WITH cte AS (
SELECT *, CASE WHEN status != prev_status THEN 1 ELSE 0 END AS flag FROM (
SELECT *, LAG(status,1,status) OVER (PARTITION BY employee ORDER BY dates) AS prev_status
from emp_attendance) t),
cte2 AS (SELECT *, SUM(flag) OVER (PARTITION BY employee ORDER BY dates) AS grp_flag
FROM cte)
SELECT employee, MIN(dates) AS from_date, MAX(dates) AS to_date, MAX(status) AS status_co
FROM cte2
GROUP BY employee, grp_flag
ORDER BY employee
Hi comments box here is my solution:
with cte as(
SELECT *,dense_rank()over( partition by employee order by employee,dates) as rn,
dense_rank() over(partition by employee,status order by employee,dates ) as rn2 from emp_attendance),
cte1 as(
select employee,dates,status,rn-rn2 as fn from cte
order by dates)
select distinct employee,first_value(dates) over(partition by employee,fn order by dates )as from_date,last_value(dates) over(partition by employee,fn) as to_date,status from cte1
order by employee,from_da
WITH cte AS(
SELECT
EMPLOYEE
,DATES
,STATUS
,rownum - SUM(CASE WHEN STATUS = 'PRESENT' THEN 1 ELSE 1 END) OVER(PARTITION BY EMPLOYEE, STATUS ORDER BY DATES) AS test
from emp_attendance
--where EMPLOYEE = 'A1'
ORDER BY EMPLOYEE, DATES
),SUMMARY AS(
SELECT
EMPLOYEE
,status
,test
,MIN(DATES) AS FROM_DATE
,MAX(DATES) AS TO_DATE
FROM cte
GROUP BY EMPLOYEE ,status,test
ORDER BY FROM_DATE
)
SELECT
EMPLOYEE
,FROM_DATE
,TO_DATE
,status
FROM summary
ORDER BY EMPLOYEE ,FROM_DATE;
Postgres solution
with base as
(
select *,ROW_NUMBER() over(PARTITION by employee order by dates asc ) as rn
from emp_attendance
)
SELECT employee,from_date,to_date,status
from
(
select employee ,status, diff,Min(dates) as from_date,max(dates) as to_date
from
(
select *,count(status) over(partition by employee , status order by employee asc, dates asc rows BETWEEN unbounded PRECEDING and CURRENT ROW) as cumulative_count,
abs(rn-count(status) over(partition by employee , status order by employee asc, dates asc rows BETWEEN unbounded PRECEDING and CURRENT ROW)) as diff
from base
)
group by 1,2,3
)
order by 1,2,3
Hernandez Mary Martinez Angela Thomas Scott
with cte as(
SELECT
*,
ROW_NUMBER()OVER(PARTITION BY employee, status ORDER BY dates, status) - ROW_NUMBER()OVER(PARTITION BY employee ORDER BY dates, status) as rnk1
FROM attendance
ORDER BY 1,2
)
SELECT
employee,
min(dates) as from_date,
max(dates) as to_date,
status
FROM cte
GROUP BY employee,status ,rnk1
ORDER BY 1, 2
Hello my solution in Sql Server:
WITH FLO AS (
SELECT *, CASE WHEN STATUS LAG(STATUS,1,'OPOUI')OVER(PARTITION BY EMPLOYEE ORDER BY DATES)THEN 1 ELSE 0 END
AS FLAG
FROM EMP_ATTENDANCE
), FLO1 AS (
SELECT * , SUM(FLAG)OVER(PARTITION BY EMPLOYEE ORDER BY DATES)AS GRP
FROM FLO
)
SELECT EMPLOYEE, MIN(DATES)AS FROM_DATE, MAX(DATES)AS TO_DATE, STATUS
FROM FLO1
GROUP BY EMPLOYEE, STATUS,GRP
ORDER BY EMPLOYEE, FROM_DATE
Hope it helps.
CTEs and window functions are new to me in learning stage, but I got this very clearly thanks for the in detail explanation
Can anyone say I am fresher will I facethis type of complex queries in interview I mean is this query for freshers or experienced persons
this is the corrected data set
(the data set in description not included A2)
drop table if exists emp_attendance;
create table emp_attendance
(
employee varchar(10),
dates date,
status varchar(20)
);
insert into emp_attendance values('A1', '2024-01-01', 'PRESENT');
insert into emp_attendance values('A1', '2024-01-02', 'PRESENT');
insert into emp_attendance values('A1', '2024-01-03', 'PRESENT');
insert into emp_attendance values('A1', '2024-01-04', 'ABSENT');
insert into emp_attendance values('A1', '2024-01-05', 'PRESENT');
insert into emp_attendance values('A1', '2024-01-06', 'PRESENT');
insert into emp_attendance values('A1', '2024-01-07', 'ABSENT');
insert into emp_attendance values('A1', '2024-01-08', 'ABSENT');
insert into emp_attendance values('A1', '2024-01-09', 'ABSENT');
insert into emp_attendance values('A1', '2024-01-10', 'PRESENT');
insert into emp_attendance values('A2', '2024-01-06', 'PRESENT');
insert into emp_attendance values('A2', '2024-01-07', 'PRESENT');
insert into emp_attendance values('A2', '2024-01-08', 'ABSENT');
insert into emp_attendance values('A2', '2024-01-09', 'PRESENT');
insert into emp_attendance values('A2', '2024-01-10', 'ABSENT');
SELECT * from emp_attendance;
Wilson Jeffrey Garcia Deborah Young Anthony
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Rodriguez Steven Hernandez Sharon Moore Patricia
How difficult sql queries are to write on real job senario? Intermediate or hard ?
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Martinez Patricia Hall Paul Robinson Jose
Solution in SQL Server
with CTE as (
select employee, dates, status, ROW_NUMBER() over(partition by employee, status order by dates)
as rn from emp_attendance), CTE2 as (
select employee, dates, status, DATEDIFF(day, rn, dates) as rn2 from CTE)
select employee, min(dates) as mindate, max(dates) as maxdates, status
from CTE2 group by employee, status, rn2 order by employee, mindate
Sir may we solve this problem using lag() window function?
Young Anna Hernandez John Miller Jason
Man you are legend....great explanation 😮
Harris Steven Jones Kimberly Hall Mark
Clark Angela Brown Kevin Young Barbara
Clark Gary Hall Robert White Elizabeth
Rodriguez Margaret Clark Margaret Brown Carol
Jones Larry Thomas Helen Moore Sharon
My solution in postgresql
WITH EMP_ID AS (
SELECT ROW_NUMBER() OVER (PARTITION BY EMPLOYEE ORDER BY EMPLOYEE,DATES) AS EMP_ID,*
FROM PRACTISE."emp_attendance"),
FLAG AS (
SELECT *,
ROW_NUMBER() OVER (PARTITION BY EMPLOYEE,STATUS ORDER BY EMPLOYEE,DATES) AS RN,
EMP_ID - ROW_NUMBER() OVER (PARTITION BY EMPLOYEE,STATUS ORDER BY EMPLOYEE,DATES) FLAG
FROM EMP_ID ORDER BY EMPLOYEE,EMP_ID,STATUS)
SELECT EMPLOYEE,MIN(DATES) AS FROM_DATE,MAX(DATES) AS TO_DATE,MIN(STATUS) AS STATUS FROM FLAG
GROUP BY EMPLOYEE,FLAG
ORDER BY EMPLOYEE,FROM_DATE
Martin Matthew Brown Mark Hernandez Karen
with cte as(SELECT *,
lag(status,1,status)over(partition by employee order by dates) as nxt
from emp_attendance),v1 as(
select *,
sum(case when status = nxt then 0 else 1 end)over(partition by employee order by dates) as grp
from cte)
select employee,min(dates) as from_date,
max(dates) as to_date,status
from v1
group by employee, grp,status;
Perfect Explanation, Thanks!
with cte as(
select employee,dates,status,DAY(dates)-ROW_NUMBER()OVER(PARTITION BY employee order by dates)rn1
from emp_attendance
where status = 'PRESENT'),
cte2 as(
select employee,dates,status,DAY(dates)- ROW_NUMBER()over(partition by employee order by dates)rn2
from emp_attendance
where status = 'ABSENT')
select employee,MIN(dates)as FROM_DATE,MAX(dates)TO_DATE,MAX(status)as status from cte
group by employee, rn1
UNION ALL
select employee,MIN(dates),MAX(dates),MAX(status) from cte2
group by employee,rn2
ORDER BY employee,FROM_DATE,TO_DATE
create table emp_attendance(employee varchar(200), Dates date, status varchar(200));
Insert into emp_attendance values
('A1','2024-01-01','PRESENT'),
('A1','2024-01-02','PRESENT'),
('A1','2024-01-03','PRESENT'),
('A1','2024-01-04','ABSENT'),
('A1','2024-01-05','PRESENT'),
('A1','2024-01-06','PRESENT'),
('A1','2024-01-07','ABSENT'),
('A1','2024-01-08','ABSENT'),
('A1','2024-01-09','ABSENT'),
('A1','2024-01-10','PRESENT'),
('A2','2024-01-06','PRESENT'),
('A2','2024-01-07','PRESENT'),
('A2','2024-01-08','ABSENT'),
('A2','2024-01-09','PRESENT'),
('A2','2024-01-10','ABSENT');
select * from emp_attendance;
Sir Website link not working?
Someone can pls solve this infosys interview question,
Text1 3
Text2 5
Text3 4
Output should be
Text1
Text1
Text1
Text2
Text2
Text2
Text2
Text2
Text3
Text3
Text3
Text3
Query should be single line query.
Thank you TFQ
Here's my take on it via MS SQL server for given dataset
=================================================
with cte as (select *,
day(dates) - row_number() over (partition by status, employee order by dates) grp
from emp_attendance)
select employee, MIN(dates) as from_date, MAX(dates) to_date, status
from cte
group by grp, employee, status
order by employee, from_date
=================================================
This is very usefull information Bro!
nice to see u again, bro the last Line of your given Data is a little 0 too much
insert into emp_attendance values('A2', '2024-01-010', 'ABSENT');
the source is alsmost the same
with cte as
(select *, row_number() over(partition by employee order by employee, dates) as rn
from emp_attendance),
cte_present as
(select *, row_number() over(partition by employee order by employee, dates) AS RN2
, rn - row_number() over(partition by employee order by employee, dates) as flag
from cte where status='PRESENT'),
cte_absent as
(select *, row_number() over(partition by employee order by employee, dates) as rn3
, rn - row_number() over(partition by employee order by employee, dates) as flag
from cte where status='ABSENT' )
select employee
, first_value(dates) over(partition by employee, flag order by employee, dates) as from_date
, last_value(dates) over(partition by employee, flag order by employee, dates
range between unbounded preceding and unbounded following) as to_date
, status
from cte_present
union
select employee
, first_value(dates) over(partition by employee, flag order by employee, dates) as from_date
, last_value(dates) over(partition by employee, flag order by employee, dates
range between unbounded preceding and unbounded following) as to_date
, status
from cte_absent
order by employee, from_date
with specification rn2, rn3 in MS SQL Server
Sir, Your videos are really awesome. can you make videos for python programming
MySql solution: with cte as (
select *, row_number() over (partition by employee, status order by dates ) as rw,
dates - row_number() over (partition by employee order by employee) as diff from emp_attendance
order by employee, dates
)
select employee, min(dates) as from_date, max(dates) as to_date, status from cte
group by employee, status, diff
🙏
Nicely explained!
If you have any time gap
Please make a video about
Frequently asking interview questions in sql for Capgemini interview...
Your videos helped me a lot in cracking my data analyst interview brother, thank you so much
Sir plz do one vd for jr data analyst interview questions and ans like pdf
I really appreciate .
As always 👍
Thank you very much clear explanation for the solution
I struggled with this. The rn - rn where status = X is a cool pattern.
Share some tips to get into a product based company
can we expect a question like this for an entry level business analyst ?
My exact thoughts! I have an interview coming up and if Im asked this I will just laugh and tell them to have a good week.
I have tried to solve this in MYSQL.
with cte as (
select *,row_number() over(partition by employee order by dates) as rn,
row_number() over(partition by employee,status order by dates) as rn1
from emp_attendance
)
select employee,min(dates) as from_date,max(dates) as to_date,status
from cte
group by employee,rn-rn1,status
order by 1,2
nice
Sir but there should be a query related to MSSQL,because there are people who are using MSSQL only.Need a Practice session on MSSQL
Copy this query and paste that in chat get type like alter this code to work in mssql
Solution Given by claude 3.5 Sonnet:
WITH grouped_attendance AS (
SELECT
*,
DATE_SUB(date, INTERVAL ROW_NUMBER() OVER (PARTITION BY employee, status ORDER BY date) DAY) AS group_date
FROM employee_attendance
)
SELECT
employee,
MIN(date) AS FROM_DATE,
MAX(date) AS TO_DATE,
status
FROM grouped_attendance
GROUP BY employee, status, group_date
ORDER BY employee, FROM_DATE;