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++;
On Mar 14, 7:46 am, ASHISH MISHRA 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
> wrote:
>
> > o(n) is the best sol known to me..
>
> > On Sun, Sep 6, 2009 at 1:54 PM, Pramod Negi wrote:
>
> >> i guess sorting w
@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 wrote:
> o(n) is the best sol known to me..
>
>
> On Sun, Sep 6, 2009 at 1:54 PM, Pramod Negi wrote:
>
>> i guess sorting will do the work.
>> any other constraint??
>>
>>
>> On Sun
