Well, looks like people are not understanding when I say "run a loop in
parallel "!!!
Please look at some of the examples on Nvidia website on how computations
can be parallelised in OpenCL or CUDA.
And also some of the high level programming languages like Scala which is
also providing these parallel constructs.
If you don't understand GPUs or not familiar with parallel constructs in
Java, then my algorithm will definitely look like O ( n ^ 2 ).
Kishen
On Wed, Oct 20, 2010 at 4:25 PM, ligerdave <david.c...@gmail.com> wrote:
> @Kishen
> as long as you have one for loop in another, you wont have O(n). it
> will most likely run O(n^2)
>
> On Oct 19, 7:41 pm, Kishen Das <kishen....@gmail.com> wrote:
> > In the below code the jth and kth inner for loops can be run in parallel
> > making them O(1) and the entire thing O(n).
> >
> > for ( i=0 to i=N-1 )
> > {
> >
> > for ( j = i to j = 0 ) {
> > sum[j] += A[ i]
> > product[j] *= A [ i]
> >
> > }
> >
> > for( k=0 to k= i )
> > if ( sum[k] == S and product[k] == P ) {
> > Answer is the sub array A[k to i ]
> > break
> >
> > }
> > }
> >
> > Kishen
> >
> > On Tue, Oct 19, 2010 at 11:36 AM, abhishek singh <iiita2007...@gmail.com
> >wrote:
> >
> >
> >
> > > @ Rahul patil ofcourse array may have negative or positive integers
> >
> > > @ Kishen both O(n) and O(n logn) solutions was asked in this yahoo
> coding
> > > round question
> >
> > > On Tue, Oct 19, 2010 at 1:28 PM, Abhishek Kumar Singh <
> > > iiita2007...@gmail.com> wrote:
> >
> > >> Given an array of length N. How will you find the minimum length
> > >> contiguous sub - array of whose sum is S and whose product is P . Here
> > >> S and P will be given to you.
> >
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