@senthilnathan: Indeed a nice logic.

http://codepad.org/hRbYV287
On Tue, Jun 8, 2010 at 8:18 AM, Minotauraus <anike...@gmail.com> wrote:

> Actually, this might not always be the best approach. Example:
>
>  -1 1 2 3 4
>  1 2 3 4 5
>  1 2 3 4 5
>  1 2 3 4 5
>  1 2 3 4 5
>
> 2*N = 10 steps.
>
> With my algo, you'll go:
>
> Step 1: top-left: negative, count++,
> Step2: [0][1] non negative, set limitRow=0 (or 1 depending on how you
> code)
> Step3: for([i][j] < limitRow)
>             check [1] [0]: non negative, set limitColumn = 0;
> since i=limitRow, j=limitCol, stop; count =1.
>
>
>
> > We can do it in O(n * log n) by individually binary-searching for zero on
> > each of the rows.  Once we get the index of the first position where zero
> > appears, counting the number of negative number is straight-forward.
>
>
> What if there are no zero elements at all?
>
> -Minotauraus.
>
>
> > Here is an even better O(N) algorithm which is very elegant:
> > Consider the bottom-left element of the given 2-D array.
> > If it is negative, the whole of first-column is negative.  So we can add
> > that count and ignore that column from then onwards.
> > If it is non-negative, the whole of last-row is non-negative.  So we can
> > ignore that row without changing the count.
> > Therefore, by just doing one comparison we are able to "eliminate" one
> row
> > or one column.
> > We can iteratively follow this approach and it will terminate in exactly
> 2*N
> > steps.
>
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