bitset <n> duplicates;// n- bit space..
for(int i=0;i<n;i++)
{
if(duplicates[array[i]] ==1)
print duplicate...
else duplicate[array[i]]=1;
}
there is no comparison between any 2 numbers ....O(n) time .....space is
O(n)bits ...
On Tue, Aug 30, 2011 at 5:18 PM, Dave <dave_and_da...@juno.com> wrote:
> Replying to myself... A radix sort takes O(n) extra space.
>
> Dave
>
> On Aug 30, 1:49 pm, Dave <dave_and_da...@juno.com> wrote:
> > @Kamakshii: With O(1) extra space, it can be done with O(n)
> > comparisons. Do a radix sort on the input (no comparisons), and then
> > check adjacent numbers for equality.
> >
> > Dave
> >
> > On Aug 30, 1:34 pm, Kamakshii Aggarwal <kamakshi...@gmail.com> wrote:
> >
> >
> >
> > > develop an algorithm to find duplicates in a list of numbers without
> using a
> > > binary tree..if there are n distinct numbers in the list ,how many
> times
> > > must two numbers be compared for equality in your algorithm?what if all
> > > numbers are equal?
> >
> > > --
> > > Regards,
> > > Kamakshi
> > > kamakshi...@gmail.com- Hide quoted text -
> >
> > - Show quoted text -
>
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