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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