Will not a balanced binary tree do the job ? if we will pick the root each
time for the median.


On Sat, May 14, 2011 at 9:10 PM, Dave <dave_and_da...@juno.com> wrote:

> @Ashish: The idea is to keep two heaps, a max-heap of the smallest
> half of the elements and a min-heap of the largest elements. You
> insert incoming elements into the appropriate heap. If the result is
> that the number of elements in the two heaps differs by more than 1,
> then you move the top element from the longer heap into the other one,
> thereby equalzing the number of elements. Thus, inserting an element
> is an O(log n) operation. To get the median, it is the top element of
> the longer heap, or, if the heaps are of equal length, it is the
> average of the two top elements. This is O(1).
>
> Dave
>
> On May 14, 8:34 am, Ashish Goel <ashg...@gmail.com> wrote:
> > not clear, can u elaborate..
> >
> > Best Regards
> > Ashish Goel
> > "Think positive and find fuel in failure"
> > +919985813081
> > +919966006652
> >
> > On Fri, May 13, 2011 at 7:15 PM, Bhavesh agrawal <agr.bhav...@gmail.com
> >wrote:
> >
> >
> >
> > > This problem can be solved using 2 heaps and the median can always be
> > > accessed in O(1) time ,the first node.
> >
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-- 
regards,
chinna.

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