Amazon Interview | SQL Interview Problem asked during Amazon Interview

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  • Опубликовано: 5 фев 2025
  • 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/bl...
    Thanks for watching!

Комментарии • 125

  • @techTFQ
    @techTFQ  7 месяцев назад +5

    Check out the upcoming Data Science bootcamp on OdinSchool: hubs.la/Q02CX94v0

    • @tptppttptppt
      @tptppttptppt 7 месяцев назад +1

      In this bootcamp you're gonna give training or it will be done by someone else?

    • @coldfirekp5059
      @coldfirekp5059 5 месяцев назад +1

      Hi , can you show one query for reverse from date ranges to single dates

    • @funofcricket_
      @funofcricket_ Месяц назад +1

      Sir, could you create a full pipeline project for us as Data Analysts using SQL, Power BI, Excel, and Python? We want it to closely resemble real-world projects in companies because we don't have a clear idea of how Data Analysts work on a daily basis.

  • @saralavasudevan5167
    @saralavasudevan5167 7 месяцев назад +4

    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

    • @TonnyPodiyan
      @TonnyPodiyan 6 месяцев назад

      Nice one 👍👍

    • @hopess-nm4ij
      @hopess-nm4ij 5 месяцев назад

      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

  • @manojdevareddy8831
    @manojdevareddy8831 7 месяцев назад +2

    CTEs and window functions are new to me in learning stage, but I got this very clearly thanks for the in detail explanation

  • @Lekhatopil
    @Lekhatopil 7 месяцев назад +25

    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.

    • @yasmeenkarachiwala9612
      @yasmeenkarachiwala9612 6 месяцев назад

      how does this work -
      dates - (ROW_NUMBER() OVER(PARTITION BY employee, status ORDER BY dates)::INT)
      could you please explain with an example

    • @abhiraj6494
      @abhiraj6494 6 месяцев назад

      Good

    • @anshulmehta5732
      @anshulmehta5732 5 месяцев назад +1

      i did it same as you did but this solution would not work if month/year changes in the same group.

    • @Lekhatopil
      @Lekhatopil 5 месяцев назад

      @@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

    • @Lekhatopil
      @Lekhatopil 5 месяцев назад

      @@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

  • @satishkumar-rp7zb
    @satishkumar-rp7zb 7 месяцев назад +2

    solving challenging queries from top mnc with nice explanation, great thoufiq keep it up.

  • @MahidharYouTube
    @MahidharYouTube 5 месяцев назад

    Your videos helped me a lot in cracking my data analyst interview brother, thank you so much

  • @dasoumya
    @dasoumya 7 месяцев назад +6

    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
      @mahivamsi9598 7 месяцев назад +2

      can you explain below part 😅😅
      dateadd(day,-1*(row_number()over(partition by employee,status order by dates)),dates) as date_grp

    • @anirbanbiswas7624
      @anirbanbiswas7624 7 месяцев назад

      @@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

    • @k.saibhargav8072
      @k.saibhargav8072 5 месяцев назад

      super

  • @adityatomar9820
    @adityatomar9820 7 месяцев назад +2

    Man you are legend....great explanation 😮

  • @sirajuddinmohamedsaleem937
    @sirajuddinmohamedsaleem937 7 месяцев назад +3

    @TFQ can we use min and max instead of first_value and last_value in the window function?

  • @balaroxx2700
    @balaroxx2700 7 месяцев назад +2

    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;

    • @333Stan
      @333Stan Месяц назад

      thanks a lot

  • @prasadreddy9754
    @prasadreddy9754 7 месяцев назад +4

    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

  • @Damon-007
    @Damon-007 7 месяцев назад +5

    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???

  • @ladhkay
    @ladhkay 6 месяцев назад +1

    Nicely explained!

  • @funofcricket_
    @funofcricket_ Месяц назад

    Sir, could you create a full pipeline project for us as Data Analysts using SQL, Power BI, Excel, and Python? We want it to closely resemble real-world projects in companies because we don't have a clear idea of how Data Analysts work on a daily basis.

  • @SASC-ot2dm
    @SASC-ot2dm 7 месяцев назад +2

    Thank you TFQ

  • @gopideveloper4375
    @gopideveloper4375 7 месяцев назад +1

    This is very usefull information Bro!

  • @akanshasaxena1138
    @akanshasaxena1138 7 месяцев назад +1

    Perfect Explanation, Thanks!

  • @sathyamoorthy2362
    @sathyamoorthy2362 7 месяцев назад +2

    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;

  • @kummithavenkatareddy2302
    @kummithavenkatareddy2302 6 месяцев назад

    Thank you very much clear explanation for the solution

  • @Jesus_777.2
    @Jesus_777.2 2 дня назад

    This is almost to one of the last problems in advanced sql 50

  • @jeromenirmal6345
    @jeromenirmal6345 2 месяца назад

    sql query:
    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),
    cte1 as (select *,sum(flag) over(partition by employee order by dates) as flagsum
    from cte)
    select distinct employee,
    min(dates)over(partition by employee,flagsum) as fromdate,
    max(dates) over(partition by employee,flagsum) as todate ,
    status from cte1

  • @manianbarasu2289
    @manianbarasu2289 5 месяцев назад

    Sir, Your videos are really awesome. can you make videos for python programming

  • @CebuProvince
    @CebuProvince 7 месяцев назад

    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

  • @KoushikT
    @KoushikT 6 месяцев назад

    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

  • @Tech_with_Srini
    @Tech_with_Srini 6 месяцев назад +6

    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

  • @andynelson2340
    @andynelson2340 7 месяцев назад

    I struggled with this. The rn - rn where status = X is a cool pattern.

  • @krishanukundu4565
    @krishanukundu4565 Месяц назад

    My solution -
    with cte as (
    SELECT
    *,
    row_number() over(partition by employee order by employee,dates) as rn1,
    row_number() over(partition by employee,status order by employee,dates ) as rn2,
    row_number() over(partition by employee order by employee,dates) - row_number() over(partition by employee,status order by employee,dates) as diff_flag
    from `stud.emp_attendance`
    order by 1, 2)
    select
    distinct employee,
    first_value(dates) over(partition by employee, diff_flag, status order by employee,dates) as from_date,
    last_value(dates) over(partition by employee, diff_flag, status order by employee,dates range between unbounded preceding and unbounded following) as to_date,
    status
    from cte
    order by 1, 2, 3

  • @shaikhanuman8012
    @shaikhanuman8012 7 месяцев назад +1

    Tqs For giving Valueble Infomation.

  • @krishnaarepalli5118
    @krishnaarepalli5118 7 месяцев назад

    If you have any time gap
    Please make a video about
    Frequently asking interview questions in sql for Capgemini interview...

  • @rakeshdebntah4738
    @rakeshdebntah4738 7 месяцев назад +1

    I really appreciate .

  • @prakash5935
    @prakash5935 7 месяцев назад

    Share some tips to get into a product based company

  • @SAURABHKUMAR-ot3sl
    @SAURABHKUMAR-ot3sl 7 месяцев назад +1

    Sir may we solve this problem using lag() window function?

  • @varunas9784
    @varunas9784 7 месяцев назад +1

    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
    =================================================

  • @andriimoskovskykh5044
    @andriimoskovskykh5044 7 месяцев назад

    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;

  • @imanelamnaoir590
    @imanelamnaoir590 6 месяцев назад

    can we expect a question like this for an entry level business analyst ?

    • @austintaylor7743
      @austintaylor7743 4 месяца назад

      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.

  • @mihirit7137
    @mihirit7137 7 месяцев назад +2

    this one is a very tough question, for what level role was this question asked 😰

    • @mihirit7137
      @mihirit7137 7 месяцев назад +3

      very hard to think about this question and finish in 30 mins

  • @fathimafarahna2633
    @fathimafarahna2633 7 месяцев назад +1

    As always 👍

  • @hopess-nm4ij
    @hopess-nm4ij 5 месяцев назад

    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

  • @sunnygoud5133
    @sunnygoud5133 7 месяцев назад +1

    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

  • @AmanRaj-p8w
    @AmanRaj-p8w 7 месяцев назад

    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

  • @amanbhattarai3273
    @amanbhattarai3273 7 месяцев назад +1

    How difficult sql queries are to write on real job senario? Intermediate or hard ?

    • @briandennehy6380
      @briandennehy6380 4 месяца назад

      Depending on the industry you work in, very hard

  • @shivaprasad-kn3kw
    @shivaprasad-kn3kw 7 месяцев назад

    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

  • @keerthis125
    @keerthis125 7 месяцев назад

    Sir plz do one vd for jr data analyst interview questions and ans like pdf

  • @SanthoshKumar-dr7gy
    @SanthoshKumar-dr7gy 6 месяцев назад

    Cte will work in Oracle db ??? Pls confirm???

  • @ishanshubham8355
    @ishanshubham8355 7 месяцев назад +2

    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

  • @raghavendrabeesa7334
    @raghavendrabeesa7334 7 месяцев назад

    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;

  • @monasanthosh9208
    @monasanthosh9208 7 месяцев назад

    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;

  • @abhinavkumar2662
    @abhinavkumar2662 7 месяцев назад +4

    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

    • @balaroxx2700
      @balaroxx2700 7 месяцев назад +1

      Copy this query and paste that in chat get type like alter this code to work in mssql

  • @Parameshi_micro
    @Parameshi_micro 7 месяцев назад

    select max(amount) as thirdhighamount from orders where amount

  • @lakshmanlee3579
    @lakshmanlee3579 6 месяцев назад

    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

  • @sahilummat8555
    @sahilummat8555 3 месяца назад

    ;with cte as (
    select *,
    lag(status,1,status)over(partition by employee order by dates) as prev_status
    from emp_attendance
    ),cte2 as (
    select * ,
    sum(case when status != prev_status then 1 else 0 end )over(partition by employee order by dates)as status_flag
    from cte
    )
    select employee,MIN(dates) as from_date,max(dates) to_dates,max(status) as status
    from cte2
    group by employee,status_flag

  • @Pathan264-f7q
    @Pathan264-f7q 5 месяцев назад

    Sir Website link not working?

  • @florincopaci6821
    @florincopaci6821 7 месяцев назад +1

    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.

  • @shubharthibhattacharyya9191
    @shubharthibhattacharyya9191 7 месяцев назад +2

    Can you please start a Snowflake Bootcamp ? Will be really helpful.

  • @arjundev4908
    @arjundev4908 7 месяцев назад

    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;

  • @prakash5935
    @prakash5935 7 месяцев назад +3

    Where we can find the dataset?

    • @VishalYadav-bj4ls
      @VishalYadav-bj4ls 7 месяцев назад

      In the description box click on script link and download that script you’ll get all queries

  • @shivinmehta7368
    @shivinmehta7368 6 месяцев назад

    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

  • @pauleedavidson9251
    @pauleedavidson9251 5 месяцев назад

    Thomas Mark Hernandez Gary Lopez William

  • @jayakrishnachanumuru
    @jayakrishnachanumuru 5 месяцев назад

    🙏

  • @alishahindia
    @alishahindia 7 месяцев назад

    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.

  • @mdmarufnawaz
    @mdmarufnawaz 5 месяцев назад

    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

  • @rajatpathak5944
    @rajatpathak5944 7 месяцев назад

    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);

  • @martinberger365
    @martinberger365 7 месяцев назад +1

    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!

  • @chiragbangera1833
    @chiragbangera1833 7 месяцев назад

    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

  • @Alexpudow
    @Alexpudow 7 месяцев назад

    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

  • @CarmenEric-o3b
    @CarmenEric-o3b 3 месяца назад

    Lewis Mary Harris Steven Garcia Matthew

  • @SallyJulius-b8d
    @SallyJulius-b8d 3 месяца назад

    2085 Athena Pass

  • @rohithr9122
    @rohithr9122 7 месяцев назад

    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

  • @boppanakishankanna6029
    @boppanakishankanna6029 6 месяцев назад

    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;

  • @SylviaFerguson-u8g
    @SylviaFerguson-u8g 5 месяцев назад

    Harris Steven Jones Kimberly Hall Mark

  • @AriesTeresa-i6n
    @AriesTeresa-i6n 5 месяцев назад

    Rodriguez Steven Hernandez Sharon Moore Patricia

  • @RoseWilson-u2t
    @RoseWilson-u2t 5 месяцев назад

    Harris Maria Clark Jason Moore Ronald

  • @Mathematica1729
    @Mathematica1729 7 месяцев назад

    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;

  • @YzkavPzlangd-t6v
    @YzkavPzlangd-t6v 5 месяцев назад

    Hall Ronald Moore Robert Robinson Robert

  • @KimberlyMyers-s3f
    @KimberlyMyers-s3f 5 месяцев назад

    Martin Matthew Brown Mark Hernandez Karen

  • @niteshnandanmahto919
    @niteshnandanmahto919 8 дней назад

    Odin school is shit , I was a student

  • @sreerag__27
    @sreerag__27 7 месяцев назад +4

    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;

  • @grzegorzko55
    @grzegorzko55 7 месяцев назад +1

    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;

  • @likinponnanna8990
    @likinponnanna8990 6 месяцев назад

    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

  • @hulhoop7197
    @hulhoop7197 4 месяца назад

    WITH CTE1 AS
    (
    SELECT
    employee,
    dates,
    status,
    EXTRACT(DAY FROM dates) -
    ROW_NUMBER() OVER(PARTITION BY employee, status ORDER BY dates) AS rn
    from emp_attendance
    ORDER BY dates)
    SELECT
    employee,
    MIN(dates) AS start_date,
    MAX(dates) AS end_date,
    status
    FROM CTE1
    GROUP BY employee, status, rn
    ORDER BY employee, start_date

  • @funofcricket_
    @funofcricket_ Месяц назад

    Sir, could you create a full pipeline project for us as Data Analysts using SQL, Power BI, Excel, and Python? We want it to closely resemble real-world projects in companies because we don't have a clear idea of how Data Analysts work on a daily basis.

  • @funofcricket_
    @funofcricket_ Месяц назад

    Sir, could you create a full pipeline project for us as Data Analysts using SQL, Power BI, Excel, and Python? We want it to closely resemble real-world projects in companies because we don't have a clear idea of how Data Analysts work on a daily basis.

  • @funofcricket_
    @funofcricket_ Месяц назад

    Sir, could you create a full pipeline project for us as Data Analysts using SQL, Power BI, Excel, and Python? We want it to closely resemble real-world projects in companies because we don't have a clear idea of how Data Analysts work on a daily basis.

  • @funofcricket_
    @funofcricket_ Месяц назад

    Sir, could you create a full pipeline project for us as Data Analysts using SQL, Power BI, Excel, and Python? We want it to closely resemble real-world projects in companies because we don't have a clear idea of how Data Analysts work on a daily basis.

  • @funofcricket_
    @funofcricket_ Месяц назад

    Sir, could you create a full pipeline project for us as Data Analysts using SQL, Power BI, Excel, and Python? We want it to closely resemble real-world projects in companies because we don't have a clear idea of how Data Analysts work on a daily basis.

  • @funofcricket_
    @funofcricket_ Месяц назад

    Sir, could you create a full pipeline project for us as Data Analysts using SQL, Power BI, Excel, and Python? We want it to closely resemble real-world projects in companies because we don't have a clear idea of how Data Analysts work on a daily basis.

  • @funofcricket_
    @funofcricket_ Месяц назад

    Sir, could you create a full pipeline project for us as Data Analysts using SQL, Power BI, Excel, and Python? We want it to closely resemble real-world projects in companies because we don't have a clear idea of how Data Analysts work on a daily basis.

  • @funofcricket_
    @funofcricket_ Месяц назад

    Sir, could you create a full pipeline project for us as Data Analysts using SQL, Power BI, Excel, and Python? We want it to closely resemble real-world projects in companies because we don't have a clear idea of how Data Analysts work on a daily basis.

  • @funofcricket_
    @funofcricket_ Месяц назад

    Sir, could you create a full pipeline project for us as Data Analysts using SQL, Power BI, Excel, and Python? We want it to closely resemble real-world projects in companies because we don't have a clear idea of how Data Analysts work on a daily basis.

  • @funofcricket_
    @funofcricket_ Месяц назад +1

    Sir, could you create a full pipeline project for us as Data Analysts using SQL, Power BI, Excel, and Python? We want it to closely resemble real-world projects in companies because we don't have a clear idea of how Data Analysts work on a daily basis.

  • @funofcricket_
    @funofcricket_ Месяц назад +1

    Sir, could you create a full pipeline project for us as Data Analysts using SQL, Power BI, Excel, and Python? We want it to closely resemble real-world projects in companies because we don't have a clear idea of how Data Analysts work on a daily basis.

  • @funofcricket_
    @funofcricket_ Месяц назад +1

    Sir, could you create a full pipeline project for us as Data Analysts using SQL, Power BI, Excel, and Python? We want it to closely resemble real-world projects in companies because we don't have a clear idea of how Data Analysts work on a daily basis.