Set contains other set in SQL - sql

Question

Is anybody to know how to find sets which contain all sets of other sets
table
id elem
1 A
1 C
2 B
2 D
2 A
2 C
3 A
3 E
3 F
4 F
4 F
I want to get this:
id
2
3
In that case id 2 contains (A,C) - all set of id 1
and id 3 contains (F) - all set of id 4
In my query I need to get all id which contains all set (all elements) at least one id.
I will be very grateful. Thank you.

Answer 1

I didn't have much to go off of for this one, but it was interested so I guessed. I'm using SQL Server
Follow along in Rextester
I submit the answer which has nested selects, then I follow through on the step by step logic to clarify what's going on:
Build a quick table
CREATE TABLE a (id int, t varchar(10))
INSERT INTO a (id, t)
VALUES (1,'A'),(1,'B'),(1,'B'),(1,'B'),(2,'A')
,(2,'B'),(2,'C'),(3,'C'),(3,'D'),(4,'A'),(5,'P');
Here is the solution:
select distinct p_id supersets from
(
select e.p_id, e.c_id, count(*) matches from (
select distinct c.id p_id, c.t p_t, d.id c_id, d.t c_t from a c
inner join a d on c.t = d.t and c.id <> d.id) e
group by e.p_id, e.c_id) sup
inner join
(select id, count(distinct t) reqs from a group by id) sub
on sub.id = sup.c_id and sup.matches = sub.reqs;
Here are the logical steps broken out to help explain why I'm doing what I'm doing:
--Step1
--Create a list of matches between (distinct) values where IDs are not the same
select distinct c.id p_id, c.t p_t, d.id c_id, d.t c_t from a c
inner join a d on c.t = d.t and c.id <> d.id;
--Step2
--Create a unique list of parent IDs and their child IDs and the # of distinct matches
--For example 2 has 2 matches with 1 and vice versa
select e.p_id, e.c_id, count(*) matches from (
select distinct c.id p_id, c.t p_t, d.id c_id, d.t c_t from a c
inner join a d on c.t = d.t and c.id <> d.id) e
group by e.p_id, e.c_id;
--Step2a
--Create a sub query to see how many distinct values are in each "Set"
select id, count(distinct t) reqs from a group by id;
Now we put it all together in the join (above, first) to make sure the total # of matches from Parent to Child make up 100% of the child values ie(is a super set)

Answer 2

I think the following does what you want:
select t1.id
from t t1 join
(select t.*, count(*) over (partition by id) as cnt
from t
) t2
on t1.elem = t2.elem and t1.id <> t2.id
group by t1.id, t2.id, t2.cnt
having count(*) = cnt;
This matches each id to each other id based on the elements. If the number of matches equals the count in the second set, then all match -- and you have a superset.
I notice that you have duplicate elements. Let's handle this with a CTE:
with t as (
select distinct id, elem
from t
)
select t1.id
from t t1 join
(select t.*, count(*) over (partition by id) as cnt
from t
) t2
on t1.elem = t2.elem and t1.id <> t2.id
group by t1.id, t2.id, t2.cnt
having count(*) = cnt;