I am sorry for the link if it caused any confusion. It was just a part of
the signature. Kindly disregard the link in this context.

Anurag Sharma


On Sun, Jun 6, 2010 at 7:55 AM, Minotauraus <anike...@gmail.com> wrote:

> I think you can do this in O(n) time. Feel free to correct me where
> I'm wrong.
>
> Create a 2D array with years on one side and elephant's time alive on
> the other. example:
>    2000 2001 2002 2003 2004 2005 2006 2007 2008 2009  2010  2011 2012
> E1 1      1       1     1        1     1       1      1      1
> E2                                  1     1        1      1
> 1       1       1       1        1
> E3                                                  1      1
> 1       1
>
> Now for every years add vertical indices example 2006 = 3, 2007 = 3,
> 2008 = 3 and so on. This will give you the
> year with max elephant population. The array can be init with 0 or a
> static array can be used.
>
> @Anurag: How will you approach this problem by using LCA algorithm
> that your link leads to?
>
>
>
> On Jun 5, 6:16 am, amit <amitjaspal...@gmail.com> wrote:
> > Maximum number of elephants alive
> > Hello guyz,
> >
> > Every elephant has a birth_time and a death_time. Given N Elephants
> > with birth times and death times.. How can we find
> > 1) the maximum number of elephants that can be alive at any given
> > point of time.
> > 2) what is the year in which you can have maximum number of elephants
> > alive.
> > ex: E1 - 2000-2008 E2-2004-2012 E3-2006-2009
> > So in 2006 you have 3 elephants alive (maximum)
> > PS: ignore months and all stuff .. if a elephants live in a year
> > consider it lives that complete year
> >
> > I have O(year_max-year_min) solution and O(n^2) solution , where
> > n=number of elephants .
> > Can we do better ??
> >
> > thanks
>
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