Hi Umesh/Priyanka,
Ur solution felt so intuitive once i understood it.
Cheers
On Jul 8, 11:21 am, umesh kewat <umesh1...@gmail.com> wrote:
> Hi Priyanka,
>
> Thanks For explaining my solution with example..
>
> --
> Thanks & Regards
> Umesh Kewat
>
> On Thu, Jul 8, 2010 at 1:32 PM, Priyanka Chatterjee
> <dona.1...@gmail.com>wrote:
>
>
>
>
>
> > I totally agree with Umesh's algo which gives O(K+1) time and an inplace
> > solution. The only point is the first K+1 numbers may get negated and the
> > array is modified. In that case after finding the duplicate we can traverse
> > the array again for the first k+1 no.s and make the negative numbers
> > positive.
>
> > code is -(considering array contains only positive numbers)
>
> > for(i=0;i<k+1;i++){
> > if(A[abs(A[i])])<0) return abs(A[i]); //1st duplicate, abs is absolute
> > function
>
> > A[abs(A[i])]*=(-1);
>
> > }
>
> > I am explaining Umesh's solution with an example
>
> > let k=6 , k<n
>
> > A= 1,6,4,5,2,6,3,..........
> > A[0]-1st element, a[k]-k+1 element in array
>
> > now A[0]=1; and A[1]=6 now A[1]= -6
> > A[1]=6 and A[6]=3 now A[6]=-3
> > A[2]=4 and A[4]=2 now A[4]=-2
> > A[3]=5 and A[5]=6 now A[5]=-6
> > A[4]=-2 and A[abs(-2)]=4 now A[2]=-4
> > A[5]=-6 and A[abs(-6]<0 therefore return 6
>
> > time complexity=O(k+1) and very much inplace solution .
>
> > --
> > Thanks & Regards,
> > Priyanka Chatterjee
> > Final Year Undergraduate Student,
>
> > Computer Science & Engineering,
> > National Institute Of Technology,Durgapur
> > India
> >http://priyanka-nit.blogspot.com/
>
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> --
> Thanks & Regards
>
> Umesh kewat
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