This does not look like a solution for the posted problem.
This fails the test case {2, 7, 6, 0, 100}, where the answer should
have been 4, for k=0 and j=4.

Manish

On Mar 20, 1:49 pm, saurabh gupta <[email protected]> wrote:
> This may not be the optimal solution to
> " Given an array of integers A[N], find the maximum value of (j-k) such
> that A[k] <= A[j] & j>k.
> I am looking for a solution with time complexity better than O(N^2)."
>
> this was in response to
> "how u can solve it in o(n)"
> and
> "how to solve this in the claimed  O(N) time."
>
>
>
> On Sat, Mar 20, 2010 at 9:14 PM, Ralph Boland <[email protected]> wrote:
>
> > On Mar 15, 10:07 am, saurabh gupta <[email protected]> wrote:
> > > while you scan the array
> > > maintain four indices plus two lengths
> > > two indices and a length mark one sub-array - start,end, length
> > > each such sub-array indicates a non-decreasing sequence,
> > > call them S1 and S2.
>
> > > while one scans, S2 keeps track of incoming elements (curr sequence)
> > > S1 keeps track of the maximum length (along with indices) so far (prev
> > > sequence)
> > > as one encounters an element which is less than the previous element,
> > > this marks the end of S2,
> > > S1 replaces S2 if len S2 > len S1 else S2 starts afresh from this new
> > > element.
>
> > > In the end if len S2 > len S1 answer = S2
> > > else answer = S1.
>
> > > PS: In the beginning and till one encounters a smaller element  than the
> > > prev one for the first time, S1 = S2
>
> > I agree that this is a nice algorthithm that solves the
> > largest non decreasing  sequence problem.
> > However, I don't agree that this solves the problem
> > originally posted.
>
> > Regards,
>
> > Ralph Boland
>
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> Says he feels all alone in a threatening world where what lies ahead is
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