O(N)
my @a = @ARGV;
my ($m, $j, $k, $l) = (0, 0, 0, 0);
my $len = 0;
my $curr = 0;
for (my $i = 1; $i < @a; $i++) {
if ($a[$i] >= $a[$i-1]) {
if ($m == $k) {
$j++;
$l++;
$curr++;
$len++;
}
else {
$l++;
$curr++;
}
}
else {
if ($curr > $len) {
$m = $k;
$j = $l;
$len = $curr;
}
$k = $l = $i;
$curr = 0;
}
}
if ($curr > $len) {
print "$k $l";
}
else {
print "$m $j";
}
On Sun, Mar 14, 2010 at 8:49 PM, Ralph Boland <[email protected]> wrote:
>
> On Mar 14, 7:46 am, ASHISH MISHRA <[email protected]> wrote:
> > @ankur how u can solve it in o(n)
> > i suppose u need atleast o(n lgn)
> >
> > On Sun, Sep 6, 2009 at 2:52 PM, ankur aggarwal <[email protected]
> >wrote:
> >
> > > o(n) is the best sol known to me..
> >
> > > On Sun, Sep 6, 2009 at 1:54 PM, Pramod Negi <[email protected]>
> wrote:
> >
> > >> i guess sorting will do the work.
> > >> any other constraint??
> >
> > >> On Sun, Sep 6, 2009 at 9:52 AM, p0r$h <[email protected]> wrote:
> >
> > >>> 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).
> >
>
> I don't know how to solve this in the claimed O(N) time.
> However, I have coded a data structure that,
> given j, will find k in O(log(N)) time.
> With it you can solve your problem in O(N log N) time.
> The data structure is built in O(N) time and space.
> It is part of a larger data structure that I will implement
> and release as open source in a few months.
>
> Regards,
>
> Ralph Boland
>
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