@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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> > --
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> > BTECH (INFORMATION TECHNOLOGY)
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