Yup Gene ..rightly said and very well pointed out :) ..My Mistake :(
On Sun, Nov 27, 2011 at 12:49 AM, Gene <gene.ress...@gmail.com> wrote:
> Isn't this overkill? If you're already using a set, just check the set
> before you insert each new element, and you'll discover the
> duplicates:
>
> S = empty
> while i = input item existss
> if i in S output "i has a duplicate";
> insert i in S
> end
>
> XOR is generally useful only for detecting a single item that's
> included in a list an odd number of times rather than an even number
> of times.
>
> On Nov 24, 3:56 pm, Ankur Garg <ankurga...@gmail.com> wrote:
> > ^^+1..how matrix formed ??
> > But as Gene said we can use a set to store all the unique elements
> >
> > Now we xor all the set elements and then xor them with the elements of
> the
> > array . This wud give us the repeating element as all the elements coming
> > once will be 0(xored twice) and repeating element wud be xored twice .
> >
> > To code it as follows
> >
> > int FindSingle(int a[],int n){
> > set<int>s;
> > s.insert(a,a+n);
> > set<int>::iterator it;
> > it = s.begin();
> > int XOR= *it;
> > it++;
> > while(it!=s.end()){
> > XOR =XOR^*it;
> > it++;}
> >
> > for(int i=0;i<n;i++)
> > XOR=XOR^a[i];
> > return XOR;
> >
> > }
> >
> > On Fri, Nov 25, 2011 at 1:03 AM, kumar raja <rajkumar.cs...@gmail.com
> >wrote:
> >
> >
> >
> > > @Anup:
> > > Atleast u tell me how the M has formed???
> >
> > > On 24 November 2011 11:21, Anup Ghatage <ghat...@gmail.com> wrote:
> >
> > >> @kunzmilan
> > >> Nice idea, how do you decide the row-size or column-size of the
> matrix?
> >
> > >> On Thu, Nov 24, 2011 at 8:00 PM, kumar raja <rajkumar.cs...@gmail.com
> >wrote:
> >
> > >>> @kunzmilan :
> > >>> Can u please maintain the clarity ??
> > >>> How did u find the M
> >
> > >>> if the list is 4 2 8 9 5 1 9 how M looks like ?? please elaborate
> it...
> >
> > >>> On 24 November 2011 06:15, kunzmize an <kunzmi...@atlas.cz> wrote:
> >
> > >>>> On 24 lis, 09:09, kumar raja <rajkumar.cs...@gmail.com> wrote:
> > >>>> > @kunzmilan : i did not get u, once explain with example...
> >
> > >>>> > On 23 November 2011 23:47, kunzmilan <kunzmi...@atlas.cz> wrote:
> > >>>> Matrix M
> > >>>> 0 1 0
> > >>>> 0 1 0
> > >>>> 1 0 0
> > >>>> multiplied with M(T)
> > >>>> 0 0 1
> > >>>> 1 1 0
> > >>>> 0 0 0
> > >>>> gives
> > >>>> 1 0 0
> > >>>> 0 2 0
> > >>>> 0 0 0.
> > >>>> On its diagonal are numbers of repeated elements.
> > >>>> kunzmilan
> >
> > >>>> > > On 24 lis, 07:02, kumar raja <rajkumar.cs...@gmail.com> wrote:
> > >>>> > > > In the given array all the elements occur single time except
> one
> > >>>> element
> > >>>> > > > which occurs 2 times find it in O(n) time and O(1) space.
> >
> > >>>> > > > e.g. 2 3 4 9 3 7
> >
> > >>>> > > > output :3
> >
> > >>>> > > > If such a solution exist can we extend the logic to find "All
> the
> > >>>> > > repeated
> > >>>> > > > elements in an array in O(n) time and O(1) space"
> >
> > >>>> > > > --
> > >>>> > > > Regards
> > >>>> > > > Kumar Raja
> > >>>> > > > M.Tech(SIT)
> > >>>> > > > IIT Kharagpur,
> > >>>> > > > 10it60...@iitkgp.ac.in
> > >>>> > > > Write the list in the form of a matrix M, e.g.
> > >>>> > > > 0 1 0 0...
> > >>>> > > > 0 0 1 0...
> > >>>> > > > 0 0 0 1...
> > >>>> > > > ......etc.,
> > >>>> > > > and its quadratic form M(T)M shows, how many times each
> element
> > >>>> repeats.
> > >>>> > > kunzmilan
> >
> > >>>> > > --
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> >
> > >>>> > --
> > >>>> > Regards
> > >>>> > Kumar Raja
> > >>>> > M.Tech(SIT)
> > >>>> > IIT Kharagpur,
> > >>>> > 10it60...@iitkgp.ac.in
> >
> > >>>> --
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> >
> > >>> --
> > >>> Regards
> > >>> Kumar Raja
> > >>> M.Tech(SIT)
> > >>> IIT Kharagpur,
> > >>> 10it60...@iitkgp.ac.in
> >
> > >>> --
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> >
> > >> --
> > >> Anup Ghatage
> >
> > >> --
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> >
> > > --
> > > Regards
> > > Kumar Raja
> > > M.Tech(SIT)
> > > IIT Kharagpur,
> > > 10it60...@iitkgp.ac.in
> >
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>
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