@Sathaian
Please explain
"Then find the longest increasing sequence of yi's -> nlog(n) "
how to find this...??
Ex. (1,2) (2,13) (5,10) (7,9)(9,1)
xi are sorted what will be the algo.. next??
On Sep 29, 8:55 am, Sathaiah Dontula <[email protected]> wrote:
> 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 <
>
> [email protected]> 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
> > <[email protected]>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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