Hey,
Given the relation
A B
---------- ----------
a b1
a b2
a b3
a b4
b b2
select a, count(*)
from help
group by a
having count(*) >= all (select count(*) from help group by a);
returns
A COUNT(*)
---------- ----------
a 4
I removed the example with 'rownum = 1', as it contained a vital flaw being it only returns row 1, thus if 2 toppings are equal in popularity then only the first one alphabetically would be returned.