You can order the envelopes by area!

Wladimir Araujo Tavares
http://www.si.ufc.br/~wladimir <http://www.si.ufc.br/%7Ewladimir/>
"Fiz uma faculdade! Só não fiz a segunda porque acabaram os tijolos."




On Tue, Oct 19, 2010 at 11:23 AM, Wladimir Tavares <wladimir...@gmail.com>wrote:

> This problem is a variation of Longest Increasing Subsequence (LIS). The
> faster algorithm é O(n log n)
>
> http://en.wikipedia.org/wiki/Longest_increasing_subsequence
>
> http://www.algorithmist.com/index.php/Longest_Increasing_Subsequence
>
>
> Wladimir Araujo Tavares
> http://www.si.ufc.br/~wladimir <http://www.si.ufc.br/%7Ewladimir/>
> "Fiz uma faculdade! Só não fiz a segunda porque acabaram os tijolos."
>
>
>
>
>
> On Mon, Oct 18, 2010 at 7:53 PM, Marcelo Amorim Menegali <
> mmeneg...@gmail.com> wrote:
>
>> 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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