>Why not just change the definition of when one number is bigger than another
>and do normal sort ?
>I guess that is better and simpler.

Normal sort takes O(n log n), while Anurag's algo is O(n).


Regards,
Jagadish
http://www.cse.iitb.ac.in/~jagadish






On Jun 20, 2:18 pm, Rohit Saraf <[email protected]> wrote:
> Why not just change the definition of when one number is bigger than another
> and do normal sort ?
> I guess that is better and simpler.
> --------------------------------------------------
> Rohit Saraf
> Second Year Undergraduate,
> Dept. of Computer Science and Engineering
> IIT Bombayhttp://www.cse.iitb.ac.in/~rohitfeb14
>
> On Sun, Jun 20, 2010 at 7:52 AM, Anurag Sharma <[email protected]>wrote:
>
>
>
> > Keep 2 pointers 'start' and 'end' and make them point to start and
> > beginning of the array.
>
> > Now keep decresing *end* pointer until an odd element is found
> > Keep increasing the *start* pointer until an even element is found
> > swap the elements at start and end
> > Continue the above 3 steps till start<end
>
> > Now the start/end points to a border element which divides the array in 2
> > parts, 1st have having all odd numbers and 2nd half with all even numbers.
>
> > Now use any inplace sorting algorithm to sort in descending order the
> > portion containing all odd numbers and in increasing order the portion
> > containing all  even numbers.
> > Hope its clear.
>
> > Anurag Sharma
>
> > On Sun, Jun 20, 2010 at 2:15 AM, vijay <[email protected]> wrote:
>
> >>  There is an array of odd and even numbers. Now, sort them in such a
> >> way that the top portion of the array contains odd numbers, bottom
> >> portion contains even numbers. The odd numbers are to be sorted in
> >> descending order and the even numbers in ascending order. You are not
> >> allowed to use any extra array and it has to use a conventional
> >> sorting mechanism and should not do any pre or post processing
>
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