In this video, let us solve an SQL Problem asked during the Amazon Interview. OdinSchool: hubs.la/Q02CX94v0 Download the scripts used in the video: techtfq.com/blog/sql-intervie... Thanks for watching!
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.
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
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;
@@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
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???
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;
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;
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
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
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
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;
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
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
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.
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
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;
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
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;
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 =================================================
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;
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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
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);
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.
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
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
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;
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
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!
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;
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;
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
