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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