Count number of repeats in SQL - sql

Question

I tried to solve one problem but without success.
I have two list of number
{1,2,3,4}
{5,6,7,8,9}
And I have table
ID Number
1 1
1 2
1 7
1 2
1 6
2 8
2 7
2 3
2 9
Now I need to count how many times number from second list come after number from first list but I should count only one by one id
in example table above result should be 2
three matched pars but because we have only two different IDs result is 2 instead 3
Pars:
1 2
1 7
1 2
1 6
2 3
2 9
note. I work with MSSQL
Edit. There is one more column Date which determined order
Edit2 - Solution
i write this query
SELECT * FROM table t
left JOIN table tt ON tt.ID = t.ID
AND tt.Date > t.Date
AND t.Number IN (1,2,3,4)
AND tt.Number IN (6,7,8,9)
And after this I had a plan to group by id and use only one match for each id but execution take a lot time

Answer 1

Here is a query that would do it:
select a.id, min(a.number) as a, min(b.number) as b
from mytable a
inner join mytable b
on a.id = b.id
and a.date < b.date
and b.number in (5,6,7,8,9)
where a.number in (1,2,3,4)
group by a.id
Output is:
id a b
1 1 6
2 3 9
So the two pairs are output each on one line, with the value a belonging to the first group of numbers, and the value of column b to the second group.
Here is a fiddle
Comments on attempt (edit 2 to question)
Later you added a query attempt to your question. Some comments about that attempt:
You don't need a left join because you really want to have a match for both values. inner join has in general better performance, so use that.
The condition t.Number IN (1,2,3,4) does not belong in the on clause. In combination with a left join the result will include t records that violate this condition. It should be put in the where clause.
Your concern about performance may be warranted, but can be resolved by adding a useful index on your table, i.e. on (id, number, date) or (id, date, number)