On Fri, Sep 2, 2011 at 2:09 AM, Don <[email protected]> wrote:

> I think that the complexity might even be log(sizeA) where A is the
> smaller of the two arrays.
> Imagine that you have sizeA = 3 and sizeB = 1,000,000,000.
> You know that the median will either be in A, or it will be
> B[500,000,000+/-1].
> The binary search will start off by comparing A[1] with
> B[500,000,000]. If it does not find the median there, it will narrow
> the search down to either A[0] and B[500,000,001] or A[2] and
> B[499,999,999].
> Two steps, which is a lot less than O(3+1,000,000,000).
>
> Jay, I would suggest that you modify your code as follows:
>   Instead of initializing startB=0 and endB=sizeB
>
> startB = (sizeA/SizeB)/2;
> endB = startB + sizeA;
>
> Assuming that sizeA <= sizeB, we know that if the median is in B, it
> will be in that range.
>

+1. Sounds perfect! :-)

>
> Don
>
> On Sep 1, 2:27 pm, Jay Mahadeokar <[email protected]> wrote:
> > As you can see, each time, we are either discarding 1st half of A or
> second
> > half of A. Same for B.
> >
> > So, the total size is getting reduced by factor of 2 every time. So, the
> > time is log(N) where N = sizeA + sizeB. I hope it is clear.
> >
> > On Fri, Sep 2, 2011 at 12:47 AM, Rahul Verma <[email protected]
> >wrote:
> >
> >
> >
> > > @jay could you please explain that how you know that the complexity is
> > > log(sizeA + sizeB).
> >
> > > If you don't mind then could you please explain it in detail.
> >
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> > --
> > Regards,
> > -Jay Mahadeokar
> > MTech Final Year, CSE.
> > IIT Kanpur.
>
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-- 
Regards,
-Jay Mahadeokar

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