You can take advantage of a basic property of triagle that

sum of largest side of triangle < sum of other two sides,

After sorting you could easily deside the range in which possible solution
could be found for a chosen hypotenuse

On Fri, Oct 14, 2011 at 11:10 AM, ravindra patel
<ravindra.it...@gmail.com>wrote:

> @Wladimir, yeah I have heard about that. Another way of populating
> primitive pythagoreans is, for any natural number m > 1  (m^2 - 1, 2m, m^2 +
> 1) forms a pythagorean triplet. This is useful in populating pythagorean
> tiplets but here the problem is to search such triplets from a given int
> array.
>
> @ rahul, Hash of z^2 - x^2 for each pair of z,x itself will of the size
> n*(n-1). I am not sure how it will work in O(n) time then.
>
> Thanks,
> - Ravindra
>
>
> On Fri, Oct 14, 2011 at 12:25 AM, Ankur Garg <ankurga...@gmail.com> wrote:
>
>> @rahul...How do u choose z and x for computing z^2 -x^2 ?
>>
>> On Thu, Oct 13, 2011 at 11:34 PM, rahul <rahul...@gmail.com> wrote:
>>
>>> You can create a hash with sqrt(z2-x2). This will make it o(n). The
>>> interviewer just made it lil tricky. That's all
>>>
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-- 
Regards,
Rahul Patil

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