yes this is the only one till now, but i think we can do better also.
Hope a better solution will be posted by someone soon.
Sanju
:)
On Wed, Aug 24, 2011 at 5:22 AM, Venkat <venkataharishan...@gmail.com>wrote:
> @Sanju: For your input both above solution wont work...
>
> Do you ve any soultion for your input??
> For your input.... Xor all numbers - will give you the result....:)
> but its O(n)
>
> Anyway your input allow everyone to think little wider than Binay
> search.
>
>
> Thanks
> Venkat
>
>
> On Aug 24, 4:05 pm, Sanjay Rajpal <srn...@gmail.com> wrote:
> > @Venkat : suppose if the array were : 1 2 2 2 2 2 2 2 2 2 2, would ur
> > solution work ?
> >
> > Sanju
> > :)
> >
> > On Wed, Aug 24, 2011 at 3:58 AM, Ankit Minglani <
> ankit.mingl...@gmail.com>wrote:
> >
> >
> >
> >
> >
> >
> >
> > > How about this :
> > > We use a divide and conquer approach and since the array is sorted.
> > > We find the middle element and check its value with its immediate left
> and
> > > right element .. it must match with anyone of them ..
> >
> > > if it doesnt we have found such a element . and otherwise we divide the
> > > array again ..
> > > and then again find the middle element .. to check the same condition
> ..
> >
> > > This will take O(lg n ) time :)
> >
> > > On Wed, Aug 24, 2011 at 3:45 PM, Venkat <venkataharishan...@gmail.com
> >wrote:
> >
> > >> we can solve this with the help of binary search.
> >
> > >> we know N, which is odd(because of one pair missing)
> >
> > >> We divide it array. Let consider your input { 1,1,2,2,2,2,3,3,4,5,5}
> >
> > >> int find_culprit(int[] array, int start, int end)
> > >> {
> > >> if(end==start)
> > >> return -1;
> >
> > >> int mid=((end-start) / 2) + start;
> > >> if array[mid] == array[mid-1]
> > >> return find_culprit(mid,end)
> > >> if(array[mid] == array [mid +1]
> > >> return find_culprit(start, mid);
> > >> else
> > >> return array[mid];
> > >> }
> >
> > >> Run through:
> > >> Steps1: find_culprit(array,0,8)
> > >> mid=4
> > >> Step 2 : find_culprit(array,4,8))
> > >> mid=6
> > >> step 3 : find_culprit(array,6,8))
> > >> mid=7
> > >> return array[7]=4 (which dont have pair)
> >
> > >> Run time O(log n+1) = O(log n)
> >
> > >> Please ask if you ve any doubts.....
> >
> > >> Regards
> > >> Venkat.
> >
> > >> On Aug 24, 2:49 pm, atul purohit <gonewiththe...@gmail.com> wrote:
> > >> > Hi,
> >
> > >> > A* sorted *integer array contains elements in pairs. All the pairs
> are
> > >> > complete except one element whose pair is missing. Find that
> element.
> >
> > >> > Ex. { 1,1,2,2,2,2,3,3,4,5,5}
> > >> > result = 5
> >
> > >> > There is a standard solution which returns an XOR of all the
> elements.
> > >> But
> > >> > this needs O(n) time complexity. The person who asked me this
> question
> > >> said
> > >> > that this can be done in < O(n). Maybe we can eliminate some
> elements.
> > >> > Anyone knows how to do this?
> >
> > >> > Cheers,
> > >> > Atul
> >
> > >> --
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> >
> > > --
> > > The more you sweat in the field, the less you bleed in war."
> >
> > > Ankit Minglani
> > > NITK Surathkal
> >
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