@rahul It still will be O(n^2) time complexity

On 14 October 2011 15:14, 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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>>
>>
>> --
>> Regards,
>> Rahul Patil
>>
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-- 
Bittu Sarkar
5th Year Dual Degree Student
Department of Computer Science & Engineering
Indian Institute of Technology Kharagpur

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