Getting a node that has a relationship with another node that is connected to at least X nodes


I just started with Neo4j and I can't find an answer to my question or tried to solve it the wrong way because none of it was at least close to solving my problem.

I have graph where part of it looks like this
(A)-[ x ]->(B)<-[ y ]-(C)
I need to get the A node that has connected B nodes with at least 10 C nodes connected to it and get the B node amount, so it will give a result like this.

A.key    |   B node amount
1           |   7
2           |   4
6           |   2

It's probably really simple but I have some sort of "mind blockade" and I can't get through it...

Hi @knoxrnox,

Welcome to the community !!!
could you please show us the graph pic.

However you can try like
match (a:A)-[x:X]->(b:B)-[y:Y]->(c:C) with a , b , count(y) as y where y>=10 return a ,b,y

Probably you mean (a:A)-[x:X]->(b:B)<-[y:Y]-(c:C)

oh Yes. i forgot to change that..
I made sample dataset and relationship direction as above however @knoxrnox can try with the his own dataset and relationship direction

You can try

MATCH (a:A)-[ x ]->(b:B)<--(c:C)
WHERE count(c) > 10
RETURN a.key, count(distinct b)

Than you all. That's exactly what I was trying to do and now when I see the answer I feel really stupid...
Thank you once again.

Unfortunately that's not possible, aggregations must occur in WITH or RETURN clauses, not in WHERE clauses.

Alternately you can use the size() function in a WHERE clause to get the count of paths matching the pattern:

MATCH (a:A)-[x:X]->(b:B)
WHERE size((b)<-[:Y]-(:C)) > 10
RETURN a.key as a, count(b) as bCount

Though note that we do not know that b is connected to more than 10 distinct :C nodes...if multiple :Y relationships from a :B node can connect to the same :C node, then this won't be accurate. If you need counts to distinct :C nodes, then you will have to MATCH out the whole pattern and do a series of count aggregations.