# Re: Algorithm to find pairs in array [can not order]

From:
Patricia Shanahan <pats@acm.org>
Newsgroups:
comp.lang.java.programmer
Date:
Fri, 17 Nov 2006 13:56:07 GMT
Message-ID:
obaqueiro@gmail.com wrote:

Hello, so I am trying to make a better (with less complexity,
prefferably without a nested loop) to find something like pairs of
items in an ArrayList:

So far I've got this:

// shuffle Option offers
Collections.shuffle(offeredOptions);
for (int i=0;i<offeredOptions.size()-1;i++)
{
Offer of1 = offeredOptions.get(i);
// look for a matching option offer in the list
for (int j=1;j<offeredOptions.size();j++)
{
Offer of2 = offeredOptions.get(j);
/* we are looking for 2 offers for the same Option contract with
opposite
* offer types (write and hold)
*/
if (of1.getOption() == of1.getOption() &&
(of1.type == OfferType.hold && of2.type == OfferType.write) ||
(of1.type == OfferType.write && of2.type == OfferType.hold))
{
offeredOptions.remove(i);

offeredOptions.remove(j);
// do something else
...
// ...
break; // get to the
i loop

}
}
}

I know it is not optimal, but basically I need to match 2 items from
the array (which is randomized), and then remove them.
I am sure there must be a more efficient way to do this but I cant
remember the proper algorithm...

I know it is not a *proper* Java question but didn't knew which group

thanks anyway!

There is a quick halving of the work. The inner loop does not need to
begin from index 1. It can start from i+1, because any match between a
pair either of which has an index value less than i would have been
found on a previous outer loop iteration.

If you rely on direct comparison, using the unordered array, the work is
going to be O(n*n).

What is the type of a getOption() result? Can there be more than two
offers with the same contract?

Depending on those answers, there may be much more efficient solutions
for large n in which the data gets copied to a HashSet or HashMap during
the search.

Patricia

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