Can we do this as the quick sort does in place , consider an even element as
a pivot then keep two pointers ,one in beginning and other at the end and
then check for even numbers if even increment i and stop on reaching an odd
number now the same way with j pointer a t the end in j until even number is
encountered now swap i and j and recursively do this till you get all even
elements on left and odd elements on right .. What you say?? Does this logic
works ??

2011/7/23 яαωαт Jee <[email protected]>

> @ popli:  plz change ur email id..placement session will start soon.
>
> On Jul 22, 8:32 pm, Gaurav Popli <[email protected]> wrote:
> > an O(n) soln
> > traveres the array...as you receive odd number put that index in
> > queue....when received an even numb check if queue is empty or
> > not...if queue is empty the do nothing else swap with the head of the
> > queue....
> >
> > hope it works....it also maintains the stability of aarray...
> >
> >
> >
> >
> >
> >
> >
> > On Fri, Jul 22, 2011 at 6:39 PM, Kunal Patil <[email protected]> wrote:
> > > @Sunny: Excellent explanation (& solution) !!
> >
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-- 
Regards
Rajeev N B <http://www.opensourcemania.co.cc>

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