check for visited can also be implemented by using an integer variable
and setting corresponding bits!
On Jul 23, 4:39 pm, ross <jagadish1...@gmail.com> 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 <akshatasharm...@gmail.com> wrote:
>
>
>
>
>
>
>
> > better than O(n^2)..
>
> > On Sat, Jul 23, 2011 at 1:08 PM, Akshata Sharma
> > <akshatasharm...@gmail.com>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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