Rohit's approach put into a typical c++ construct...
inline is_odd(int x)
{
return ((x & 1) == 1);
}
struct new_compare {
bool operator () (int i, int j) {
bool b_i_odd = is_odd(i);
bool b_j_odd = is_odd(j);
if ((b_i_odd && b_j_odd) || (!b_i_odd && !b_j_odd))
return (i < j);
else if (b_i_odd && !b_j_odd)
return 1;
else
return 0;
}
};
std::vector<int> array_to_sort;
std::sort( array_to_sort.begin(), array_to_sort.end(), new_compare );
Of course, everything written above can be stated as a C program, or an
algorithm.
The sort( ) function can be replaced with any in-place sorting algorithm,
the new_compare class with either a function ptr, or a simple in-place
check.
On Tue, Jun 22, 2010 at 11:10 PM, Naveen Kumar
<naveenkumarve...@gmail.com>wrote:
> Hi Rohit,
>
> Can you explain your approach a bit more?
>
> On Sun, Jun 20, 2010 at 2:48 PM, Rohit Saraf
> <rohit.kumar.sa...@gmail.com> 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 Bombay
> > http://www.cse.iitb.ac.in/~rohitfeb14
> >
> >
> > On Sun, Jun 20, 2010 at 7:52 AM, Anurag Sharma <anuragvic...@gmail.com>
> > 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 <auvija...@gmail.com> 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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> --
> Cheers
> Naveen Kumar
>
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