A[0] = 10    A[1] = 2    A[2] =  5
A[3] =  1    A[4] = 8    A[5] = 20

Triplet  10,5,8 is triangular.

Dave, do your solution do it?

On Tue, Aug 23, 2011 at 11:55 AM, Amol Sharma <[email protected]>wrote:

> +1 for dave's solution.....i will also do the same
> --
>
>
> Amol Sharma
> Third Year Student
> Computer Science and Engineering
> MNNIT Allahabad
>  <http://gplus.to/amolsharma99> 
> <http://twitter.com/amolsharma99><http://in.linkedin.com/pub/amol-sharma/21/79b/507><http://youtube.com/amolsharma99>
>
>
>
>
>
> On Tue, Aug 23, 2011 at 11:25 AM, Dave <[email protected]> wrote:
>
>> @Saurabh: If you can use O(n) extra space, make a copy of the array
>> and sort it: O(n log n). Then, if there is a solution, there will be a
>> solution of the form (a[i], a[i+1], a[i+2]), where 0 <=  i < n-2,
>> which can be checked with a simple for loop: O(n). Thus, the
>> complexity is O(n log n).
>>
>> Dave
>>
>> On Aug 23, 12:04 am, saurabh agrawal <[email protected]> wrote:
>> > Given an array, find out whether there exists a triplet which can form
>> sides
>> > of triangle.
>> > You are not allowed to modify the array.
>> >
>> > PLease dont give o(n^3) solution
>> >
>> > there exists a solution with nlog(n) i think
>>
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Thanks and Regards,
Raghavan KL

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