Does sequence contain 5 numbers that are each one apart solved recursively - sql

Question

This is the data:
create table #t
(ID int)
insert into #t
values
(-2)
,(-1)
-- ,(0)
,(1)
,(2)
,(3)
,(4)
,(7)
,(5)
,(21)
,(22)
,(23)
,(24)
,(25)
,(8);
We want to know if there are 5 numbers within the above sequence that are each 1 apart e.g. 21-22-23-24-25 gives a positive result. So is there an island of 5 anywhere in the list?
None recursively I've got a few possibilities but is there a simple recursive solution?
Or is there a simpler non-recursive solution?
--::::::::::::::
--:: 1. LONG-WINDED
with t as
(
select id,
U = (id+5),
L = (id-5)
from #t
)
, up as
(
select x.id,
cnt = count(*)
from t x
join t y on
(y.id > x.L and y.id <= x.id)
group by x.id
)
, down as --<<MAYBE DOWN IS NOT NEEDED
(
select x.id,
cnt = count(*)
from t x
join t y on
(y.id < x.U and y.id >= x.id)
group by x.id
)
select id from up where cnt >= 5
union all
select id from down where cnt >= 5
Following two are better:
--::::::::::::::
--::
--:: 2. PRETTY!!
SELECT *
FROM #t A
WHERE EXISTS
(
SELECT *
FROM #t B
WHERE (
(A.id + 5) > B.id
AND
A.id <= B.id
)
HAVING COUNT(*) >=5
)
--::::::::::::::
--::
--:: 3. PRETTY PRETTY!!
--::
SELECT ID
FROM #t A
CROSS APPLY
(
SELECT cnt = COUNT(*)
FROM #t B
WHERE (A.id + 5) > B.id AND A.id <= B.id
) C
WHERE C.cnt>=5
Following used this reference to Itzak article
--::::::::::::::
--::
--:: 4. One of the Windowed functions
--::
WITH x AS
(
SELECT ID,
y = LAG(ID,4) OVER(ORDER BY ID),
dif = ID - LAG(ID,4) OVER(ORDER BY ID)
FROM #t A
)
SELECT ID,y
FROM x
WHERE dif = 4

Answer 1

Yes, there is a much simpler solution. Take the difference between the numbers and an increasing sequence of numbers. If the numbers are consecutive, the difference is constant. So, you can do:
select grp, count(*) as num_in_sequence, min(id) as first_id, max(id) as last_id
from (select t.*,
(id - row_number() over (order by id)) as grp
from #t t
) t
group by grp
having count(*) >= 5;
EDIT:
I think this is the simplest of all. One window function and a comparison:
select t.*
from (select t.*, lead(id, 4) over (order by id) as id4
from #t
) t
where id4 - id = 4;
This does make the assumption that there are no duplicates in the ids, which is true of the OP data.
As I look further, this is the last solution in the OP. Kudos!