count rows which have max value less than specified parameter - sql

Question

I want to find in my table, max value which is less than specified in parameter and get count of rows that have the same value as max value. For example in my table I have values: (4,1,3,1,4,4,10), and it is list of parameters in string "2,9,10,4". I have to split string to separate parameters. Base on this sample values I want to get something like that:
param | max value | count
2 | 1 | 2
9 | 4 | 3
10 | 4 | 3
4 | 3 | 1
And it is my sample query:
select
[param]
, max([val]) [max_value_by_param]
, max(count) [count]
from(
select
n.value as [param]
,a.val
, count(*) as [count]
from (--mock of table
select 1 as val union all
select 3 as val union all
select 4 as val union all
select 1 as val union all
select 3 as val union all
select 4 as val union all
select 4 as val union all
select 10 as val
) a
join (select [value] from string_split('2,9,10,4', ',')) n--list of params
on a.val < n.[value]
group by n.value, a.val
) tmp
group by [param]
Is it possible to do it better/easier ?

Answer 1

Here is a way to express this using apply:
select s.value as param, a.val, a.cnt
from string_split('2,9,10,4', ',') s outer apply
(select top (1) a.val, count(*) as cnt
from a
group by a.val
having a.val < s.value
order by a.val desc
) a;
Here is a db<>fiddle.
But the fastest method is probably going to be:
with av as (
select a.val, count(*) as cnt
from a
group by a.val
union all
select s.value, null as cnt
from string_split('2,9,10,4', ',') s
)
select val, a_val, a_cnt
from (select av.*,
max(case when cnt is not null then val end) over (order by val, (case when cnt is null then 1 else 2 end)) as a_val,
max(case when cnt is not null then cnt end) over (order by val, (case when cnt is null then 1 else 2 end)) as a_cnt
from av
) av
where cnt is null;
This only aggregates the data once and should return all parameters, even those with no preceding values in a.