@alexklibisz I was going to do a pr, to solve the issue that the old HitCounter had, returning documents with zero hits, I see that you still have the error,
the simplest way to solve it is

I also analyzed a bit you code, lets compute the time if a shard has D documents, H of which are a hit in any of the L hash tables.
The current version has a complexity or O(H) to fill the table and O(D) in time and space to find the top ones.
I also thought that maybe using a hash key/freq could be better, but numbers are difficult to be compared
Using a hashmap, the time to put the data is the hash is O(H) , then a an array of lists (List[] ) of size max_freq, could be created to store them by freq, the time needed to fill them, is (O(H) )
The time and space needed depends on the values of L,K
as for example a value 10 for K will select 0,1% of the documents, and if L=100 , this implies H to be <10% of the documents
I think that using a hashmap can be smaller in space and time.

let me add a further comment about L and K
K=1 implies a division of the space in 2 parts, by a random plane, so we should expect that 50% of the documents are at each side.
I we do K=1 and L=m, then the probability of a document to be a positive hit for any L is 1-(prob of being a negative one for all L) this is 1-(0,5^m) , which means that with L >10 is 99,9%
So kith k=1 and L=99 we are considering almost all the documents.
with K = 10 and L=99 the prob of being a hit in any hashtable is 0,5^10 , and being a negative one so
at the end we come with the probability of a document to be a hit (at leas in one has table) to be
1−(2k−1)l​
Here are the computed values of the expression ( 1 - \left(\frac{2^k - 1}{2^k}\right)^l ) for different values of ( k ) (from 1 to 10) and ( l ) (from 10 to 100, in steps of 10):

These values represent how the expression changes as ( k ) and ( l ) increase.

so a value of k=10 and L =100 will select 1/10 of the documents but with L = 100 and K < 6 selects most of the documents. This impacts you benchmarks here #160 (comment) , as you always use k=4 l and L =100 , so 99% of the documents are a hit!