SQL "Group" and "Count" categories - sql-server-2012

Question

Edit. This is a follow up from another question. To simplify the question. Assume a table
date | id | type
01/01 | 1 | F
02/01 | 1 | F
02/01 | 1 | F
03/01 | 1 | S
03/01 | 1 | S
04/01 | 1 | F
04/01 | 1 | S
05/01 | 1 | S
I am looking for a way to summarise the above table by combination of transaction types per day. If a person (id) has only one transaction per day it counts as a Single type. If they have more than one it counts as a Multiple one. I've done that with my original query and it works. The output from the above table would be:
date | Single | Multiple
01/01 | 1 | 0
02/01 | 0 | 1
03/01 | 0 | 1
04/01 | 0 | 1
05/01 | 1 | 0
I got that far and it works. What's I'm struggling with (ie. don't have a clue of how to start) is how set up a query to show all possible combinations of Type (SS, FF, FS) instead of just counting the multiple transactions. The desired output would be like:
date | Single | # FF | # FS | # SS
01/01 | 1 | 0 | 0 | 0
02/01 | 0 | 1 | 0 | 0
03/01 | 0 | 0 | 0 | 1
04/01 | 0 | 0 | 1 | 0
05/01 | 1 | 0 | 0 | 0
Any constructive hints or ideas will be much appreciated.

Answer 1

this is assuming that you have max 2 types per date.
You can use the CASE WHEN statement with MIN() and MAX() to check for combination of FF, FS or SS
select [date],
case when count(*) = 1 then 1 else 0 end as Single,
case when count(*) >= 2
and min([type]) = 'F'
and max([type]) = 'F'
then 1
else 0
end as [# FF],
case when count(*) >= 2
and min([type]) = 'F'
and max([type]) = 'S'
then 1
else 0
end as [# FS],
case when count(*) >= 2
and min([type]) = 'S'
and max([type]) = 'S'
then 1
else 0
end as [# SS]
from yourtable
group by [date]
EDIT :
for more then 3 types, just change the count(*) = 2 to count(*) >= 2 as long as the type are either F or S