Hello Saurabh,

can you explain the algo??

On Sun, Mar 14, 2010 at 9:55 PM, saurabh gupta <[email protected]> wrote:

> 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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>
>
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