Is the problem like this,
Given an array of integers A[N], find the maximum value of (j-k) such
that A[k] <= A[j] & j>k." *for all i >j and i<=k, A[i]<=A[i-1] "
*
On Sat, Mar 20, 2010 at 11:19 PM, saurabh gupta <sgup...@gmail.com> 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 <rpbol...@gmail.com>wrote:
>
>>
>>
>> On Mar 15, 10:07 am, saurabh gupta <sgup...@gmail.com> 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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>
>
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-- 
Vikrant Singh

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