SQL Server : find Cust with Continuous Enrollment - sql-server

Question

I have a task to solve well known problem in industry task to ID those CustID who have continuous activity , for given period of time and we allow little breaks between contracts.
I did first part populating matrix table like in snippet below for whole period of time and setting flag if it's active for this date, I think this is the only reliable way to do this, as contracts can have overlaps, etc..
So now I need to check if CustID is 1/0 for cont activity, I stuck into the task how to track this, let say in my example there is 3 days break which is OK, but I need to make sure that those days are one after another.
Do you have any good ideas how I can do this nicely, appreciate your help and leads. I saw some examples but they done in SAS so it's hard to understand.
declare #maxBreak int = 3 -- 3 days max allowed for continuse contract
declare #PeriodStart date = '2015-1-11', #PeriodEnd date = '2015-1-19';
;with matrix_dd as
(
select *
from
(select 111 CustID, '2015-1-11' dd, 1 Active union
select 111 CustID, '2015-1-12' dd, 0 Active union
select 111 CustID, '2015-1-13' dd, 0 Active union
select 111 CustID, '2015-1-14' dd, 0 Active union
select 111 CustID, '2015-1-15' dd, 1 Active union
select 111 CustID, '2015-1-16' dd, 1 Active union
select 111 CustID, '2015-1-17' dd, 1 Active union
select 111 CustID, '2015-1-18' dd, 1 Active union
select 111 CustID, '2015-1-19' dd, 0 Active union
select 111 CustID, '2015-1-20' dd, 0 Active) a
)
select *
from matrix_dd
Best
M

Answer 1

This solution calculates the active ranges and how long of a break it's been since the last interval ended:
declare #maxBreak int = 3 -- 3 days max allowed for continuse contract
declare #PeriodStart date = '2015-1-11', #PeriodEnd date = '2015-1-19';
with matrix_dd as
(
select * from ( values
(111, '2015-1-11', 1 ),
(111, '2015-1-12', 0 ),
(111, '2015-1-13', 0 ),
(111, '2015-1-14', 0 ),
(111, '2015-1-15', 1 ),
(111, '2015-1-16', 1 ),
(111, '2015-1-17', 1 ),
(111, '2015-1-18', 1 ),
(111, '2015-1-19', 0 ),
(111, '2015-1-20', 0 )
) as x(CustID, dd, Active)
), active_with_groups as (
select *,
row_number() over (partition by CustID order by dd) -
datediff(day, '2000-01-01', dd) as gid
from matrix_dd
where active = 1
and dd between #PeriodStart and #PeriodEnd
), islands as (
select CustId, min(dd) as islandStart, max(dd) as islandEnd
from active_with_groups
group by CustID, gid
), islands_with_gaps as (
select *,
datediff(
day,
lag(islandEnd, 1, islandStart)
over (partition by CustID order by islandStart),
islandStart
) - 1 as [break]
from islands
)
select *
from islands_with_gaps
where [break] >= #maxBreak
order by islandStart
Let's break it down. In the "active_with_groups" common table expression (CTE), all I'm doing is converting the dates into integers that have the same relationship by using datediff(). Why? Integers are easier to work with for this problem. Note that I'm also using row_number() to get a contiguous sequence and then getting the difference between that and the datediff() value. The key observation is that if the days also don't go up contiguously, that difference will be, well, different. Likewise, if the dates do go up contiguously, then the difference will be the same. Therefore, we can use this value as a group identifier for values that are in a contiguous range.
Next, we use that the group identifier to group by (bet you didn't see that coming!). This gives us the start and end of each interval. Nothing very clever is going on here.
The next step is to calculate the amount of time that's passed between when the last interval ended and the current one began. For this, we use a simple call to the lag() function. The only thing to note here is that I've chosen to have the lag() function emit a default value of islandStart in the case of the first interval. It could have just as easily been no default (which would have then caused it to emit a NULL value).
Lastly, we look for intervals with a gap over the specified threshold.

Answer 2

Similar to Ben's answer. I'm assuming that all your dates are represented in the data. So really we just need to make sure there isn't a run of zeroes longer than 3.
with inactive_runs as (
select
CustID,
row_number() over (partition by CustID order by dd)
- datediff(day, min(dd) over (partition by CustID), dd) as grp
from matrix_dd
where Active = 0
)
select distinct CustID from matrix_dd m
where 3 >= all (
select count(*) from inactive_runs ir
where ir.CustID = m.CustID
group by grp
);
http://rextester.com/AHI22250
Using all isn't particularly common. Here's an alternative:
...
with inactive_runs as (
select
CustID, dd, /* <-- had to add dd */
row_number() over (partition by CustID order by dd)
- datediff(day, min(dd) over (partition by CustID), dd) as grp
from #matrix_dd
where Active = 0
)
select distinct CustID from matrix_dd m
where not exists (
select 1 from inactive_runs ir
where ir.CustID = m.CustID
group by grp
having datediff(day, min(dd), max(dd)) > 2
);
I glanced at your comment above. I think it confirms my suspicion that you've got a single row for every date. If you've got a new version of SQL Server you can just sum over the previous three rows. Unfortunately you wouldn't be able to use a variable for the window size if the length is variable:
with cust as (
select
CustID,
case when
sum(case when Active = 0 then 1 end) over (
partition by CustID
order by dd
rows between 3 preceding and current row
) = 4 then 1
end as isBrk
from matrix_dd
)
select CustID
from cust
group by CustID
having count(isBrk) = 0;
Edit:
Based on your comment with the data in a "pre-matrix" format, yes, that's a simpler query. At that point you're just looking at the previous end date and the current row's start date.
with data as (
select * from (
values (111, 1230, '2014-12-11', '2015-01-11'),
(111, 1231, '2015-01-15', '2015-01-18'),
(111, 1232, '2015-03-22', '2015-04-01')
) as t (CustID, ContractID, StartDD, EndDD)
), gaps as (
select
CustID,
datediff(day,
lag(EndDD, 1, StartDD) over (partition by CustID order by StartDD),
StartDD
) as days
from data
)
select CustID
from gaps
group by CustID;
having max(days) <= 3;