The problem statement doesn't make it clear, but I guess one can rotate the
envelopes (i.e. exchange x_i and y_i).
I mean, in the example case of {(1,2),(2,13),(5,10),(7,9),(9,1)}, isn't
(1,2),(1,9),(5,10) a valid solution (with length 3), instead of the ones
shown (with length 2)?

On Sat, Oct 16, 2010 at 5:40 PM, nishaanth <nishaant...@gmail.com> wrote:

> ya...finding the longest subsequence is the simplest method...and its
> nlogn..
>
> It works...it similar to box stacking problem
>
>
> On Fri, Oct 1, 2010 at 6:00 AM, adit.sh...@gmail.com <adit.sh...@gmail.com
> > wrote:
>
>> The problem here is more about finding the most optimal solution and not
>> just solution.
>> Rgds
>> Adi
>> -----Original Message-----
>> From: Sathaiah Dontula
>> Sent:  29/09/2010, 09:25
>> To: algogeeks@googlegroups.com
>> Subject: Re: [algogeeks] Algorithm to determine the largest number of
>> envelopes that can be nested inside one another.
>>
>>
>> I think we can do like this,
>>
>> 1. Sort all the xi's in ascending order -> nlog(n)
>> 2. Then find the longest increasing sequence of yi's -> nlog(n)
>> 3. complexity will be nlog(n).
>>
>> Thanks,
>> Sathaiah Dontula
>>
>> On Tue, Sep 28, 2010 at 11:37 PM, Prashant Kulkarni <
>> prashant.r.k...@gmail.com> wrote:
>>
>> > i think it is similar to finding max in a list  O(n) or sorting
>> algorithm
>> > O(n log n)
>> >
>> > -- Prashant Kulkarni
>> >
>> >
>> >
>> >
>> > On Tue, Sep 28, 2010 at 11:33 PM, Rahul Singal <rahulsinga...@gmail.com
>> >wrote:
>> >
>> >> A possible solution  i can think is create a directed graph where each
>> >> vertex is a envelope and edges are from a bigger envelope to smaller
>> >> envelope ( one which can fit in bigger envelope ) . Now the problem is
>> >> reduce to finding longest path in the graph .
>> >>
>> >> Regards
>> >> Rahul
>> >>
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>
>
> --
> S.Nishaanth,
> Computer Science and engineering,
> IIT Madras.
>
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