@ dave- commplexity of radix sort is >= O(n log n). so better use heap sort.
On Wed, Aug 31, 2011 at 4:07 PM, Dave <dave_and_da...@juno.com> wrote:
> @Bharatkumar: You've tacitly assumed that the data values are in the
> range 0 to n-1. That's not given in the problem statement.
>
> Dave
>
> On Aug 31, 1:16 am, bharatkumar bagana <bagana.bharatku...@gmail.com>
> wrote:
> > 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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