@ross, I have posted my question in a new thread, not as a reply to any
existing thread!!!

On Sat, Jul 23, 2011 at 5:13 PM, ross <[email protected]> wrote:

> check for visited can also be implemented by using an integer variable
> and setting corresponding bits!
>
> On Jul 23, 4:39 pm, ross <[email protected]> wrote:
> > @akshata sharma:
> > Kindly post a new question as a separate thread and not as a reply to
> > an existing one so tat it would be noticed by many ppl!
> > As akash suggestd, use a bit vector called 'visited' of 26 size for
> > ASCII or of a larger size in case of Unicode ( still constant space
> > and i dont think declaring 26 variables counts as an additional DS!!),
> > if visited then , ignore the character while processing.
> > a simple algorithm,
> > int last_ptr=0;
> > for ( i = 0 - N )
> > {
> > if(visited(a[i])) continue;
> > else a[last_ptr++]=a[i];
> > visited(a[i]) = true;}
> >
> > a[last_ptr]=NULL;
> > print (%s,a) ;
> >
> > On Jul 23, 12:56 pm, Akshata Sharma <[email protected]> wrote:
> >
> >
> >
> >
> >
> >
> >
> > > better than O(n^2)..
> >
> > > On Sat, Jul 23, 2011 at 1:08 PM, Akshata Sharma
> > > <[email protected]>wrote:
> >
> > > > Given a string *Str of ASCII characters, write the pseudo code to
> remove
> > > > the duplicate elements present in them. For example, if the given
> string is
> > > > "Potato", then, the output has to be "Pota". Additional constraint
> is, the
> > > > algorithm has to be in-place( no extra data structures allowed) .
> Extend
> > > > your algorithm to remove duplicates in the string which consisted of
> UNICODE
> > > > characters.
>
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