I have a set A = {(1,2), (1,2,3), (2,3,4), (3,4), (1)}
I want to turn it into A={(1,2,3), (2,3,4)}, remove proper subsets from this set.
I'm using a HashSet to implement the set, 2 iterator to run through the set and check all pairs for proper subset condition using containsAll(c), and the remove() method to remove proper subsets.
the code looks something like this:
HashSet<Integer> hs....
Set<Integer> c=hs.values();
Iterator<Integer> it= c.iterator();
while(it.hasNext())
{
p=it.next();
Iterator<Integer> it2= c.iterator();
while(it2.hasNext())
{
q=it2.next();
if q is a subset of p
it2.remove();
else if p is a subset of q
{
it.remove();
break;
}
}
}
I get a ConcurrentModificationException the 1st time i come out of the inner while loop and do a
p=it.next();
The exception is for when modifying the Collection while iterating over it. But that's what .remove() is for.
I have used remove() when using just 1 iterator and encountered no problems there.
If the exception is because I'm removing an element from 'c' or 'hs' while iterating over it, then the exception should be thrown when it encounter the very next it 2 .next() command, but I don't see it then. I see it when it encounters the it.next() command.
I used the debugger, and the collections and iterators are in perfect order after the element has been removed. They contain and point to the proper updated set and element. it.next() contains the next element to be analyzed, it's not a deleted element.
Any ideas over how i can do what i'm trying to do without making a copy of the hashset itself and using it as an intermediate before I commit updates?
Thank you
You can't modify the collection with it2 and continue iterating it with it. Just as the exception says, it's concurrent modification, and it's not supported.
I'm afraid you're stuck with an intermediate collection.
Actually, your code doesn't seem you make sense: are you sure it's a collection of Integer and not of Set<Integer>? In your code p and q are Integers, so "if q is a subset of p" doesn't seem to make too much sense.
One obvious way to make this a little smarter: sort your sets by size first, as you go from largest to smallest, add the ones you want to keep to a new list. You only have to check each set against the keep list, not the whole original collection.
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