using properties of tiangle wont help i guess. the will give the range
of VALUES you want to restrict yourself to. not the range of INDEX's
of the array...

On Fri, Oct 14, 2011 at 3:14 PM, Ankur Garg <ankurga...@gmail.com> wrote:
> @Rahul
> Pls elaborate with an example ...
>
> On Fri, Oct 14, 2011 at 2:35 PM, rahul patil <rahul.deshmukhpa...@gmail.com>
> wrote:
>>
>> 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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>>
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>> Regards,
>> Rahul Patil
>>
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