@atul: i didnt get your algo fully..
Can u just tell me how it works on this array.. {-3 2 1 5 -12 6 1 -2}
the aux array would become {-3 -1 0 5 -7 -1 0 -2}
then whats the next step.?
We analyze the aux for the repeated element which sets
subarray start= 2 subarray end=6
then aux contains 0 at two indices 2 and 6.this gives us the correct answer
here..
but that wont be the case everytime, right..?
everytime start of subarray cant be 0..
Can u clarify more on this..?
Coz i dont think it works on every test array..
On Sun, Sep 2, 2012 at 12:32 AM, atul anand <atul.87fri...@gmail.com> wrote:
> take aux[] array of same size and store cumulative some at
> aux[i]=sum{input[0 to i]}
> now if you find any repeated element at index i and j then,
> subarray start = i + 1;
> subarray end = j
>
> if array contain 0 at index j then,
> subarray start = 0;
> subarray end = j
>
>
>
> On 9/2/12, Puneet Gautam <puneet.nsi...@gmail.com> wrote:
> > Given an array of positive and negative integers, we need to find the
> > MAX length subarray having sum as ZERO...
> >
> > Is there a solution less than O(n^2)..?
> >
> > Please help .. i m stuck at this problem..
> >
> > Thanks
> > Puneet
> >
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