Given : 4 2 8 9 5 1 9
sort the array.
sorting: 1 2 4 5 8 9 9
for ( i = 0 ; i < len ; i++)
{
if( i != len-1 )
{
if (arr[i]==arr[i+1])
{
printf("\nfound repeated element\n");
break;
}
}
}
On Mon, Nov 28, 2011 at 1:24 AM, Ankur Garg <ankurga...@gmail.com> wrote:
> 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
>> >
>> > >>>> > > --
>> > >>>> > > You received this message because you are subscribed to the
>> Google
>> > >>>> Groups
>> > >>>> > > "Algorithm Geeks" group.
>> > >>>> > > To post to this group, send email to
>> algogeeks@googlegroups.com.
>> > >>>> > > To unsubscribe from this group, send email to
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>> > >>>> > > For more options, visit this group at
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>> >
>> > >>>> > --
>> > >>>> > Regards
>> > >>>> > Kumar Raja
>> > >>>> > M.Tech(SIT)
>> > >>>> > IIT Kharagpur,
>> > >>>> > 10it60...@iitkgp.ac.in
>> >
>> > >>>> --
>> > >>>> You received this message because you are subscribed to the Google
>> > >>>> Groups "Algorithm Geeks" group.
>> > >>>> To post to this group, send email to algogeeks@googlegroups.com.
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>> >
>> > >>> --
>> > >>> Regards
>> > >>> Kumar Raja
>> > >>> M.Tech(SIT)
>> > >>> IIT Kharagpur,
>> > >>> 10it60...@iitkgp.ac.in
>> >
>> > >>> --
>> > >>> You received this message because you are subscribed to the Google
>> > >>> Groups "Algorithm Geeks" group.
>> > >>> To post to this group, send email to algogeeks@googlegroups.com.
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>> > >>> For more options, visit this group at
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>> >
>> > >> --
>> > >> Anup Ghatage
>> >
>> > >> --
>> > >> You received this message because you are subscribed to the Google
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>> > >> For more options, visit this group at
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>> >
>> > > --
>> > > Regards
>> > > Kumar Raja
>> > > M.Tech(SIT)
>> > > IIT Kharagpur,
>> > > 10it60...@iitkgp.ac.in
>> >
>> > > --
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>>
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