whoops, misunderstood the problem - I thought you meant the sum to the
end of the sequence minus the sum to the end. My bad. Please ignore
what I wrote above!
On 6 Sep, 01:21, "[EMAIL PROTECTED]" <[EMAIL PROTECTED]>
wrote:
> I know you've solved the problem, but I'm bored. Would this work? I
> was never good at analysing time constraints
>
> Given an array
> n_1
> n_2
> ...
> n_k
>
> create another array called sums:
> n_1
> n_1 + n_2
> n_1 + n_2 + n_3
> n_1 + n_2 + n_3 + ... n_k
>
> Take two indices min and max
> min = 0 max = f;
> while max < L {
> increment max until sums[max] - sums[min] >= d
> if (sums[max - 1] - sums[min]) < f record max and min
> increment min
>
> }
>
> This will give you longest possible sequences that match the
> criteria. Since every subsequence would have match the >d criterion,
> you just need to list every subsequence where j>f.
>
> How does this look? Kinda new at this, so feedback is appreciated.
>
> On 4 Sep, 03:33, Sticker <[EMAIL PROTECTED]> wrote:
>
> > I have a problem on sequences of numbers:
>
> > Given a sequence of integer numbers (could be quite long, let say, 10s
> > of thousands of numbers). Let us denote it as
> > {n_1,n_2,n_3,n_4,...,n_L}. The length of the sequence is L (meaning
> > that it contains L numbers)
>
> > >From this sequence I want to find a segment of j consecutive numbers
>
> > S={n_i,n_(i+1),n_(i+2),...,n_(i+j)} such that the result of maximum
> > number of S minus the minimum number of S is smaller than user defined
> > d. The length of the segment j has to be larger than another user
> > defined f.
>
> > If there are more than one such segment, find them all.
>
> > I wonder whether there exists some linear algorithms to handle this
> > problem. The solution is better to be quick because it is only a
> > subproblem of the complete one and this operation is repeated several
> > times.
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