@Tech: I'm not sure I understand your algorithm. Let's try it on
{1,1,2,2,3,4,5,5,6,6,7,7}. The two number occurring an odd number of
times are 3 and 4. We xor the numbers getting 7 = 111 in binary. Now
how do we divide the numbers into two groups?
Dave
On Aug 26, 11:09 am, tech coder <techcoderonw...@gmail.com> wrote:
> it can be done in O(N) by using XOR ing the elements
> 1: Xor all the elemnts since those elemnts that even freq will nullify each
> other we get number taht will tell in which the two required number differ.
> 2: divide the array in two sets on the basis of bit in which numbers
> differ
> 3:1 element will be in one set another will be in another set
> 4: XOR both the sets again we get both the elemts
> On Thu, Aug 25, 2011 at 12:50 PM, Umesh Jayas <algowithum...@gmail.com>wrote:
>
>
>
>
>
> > int main()
> > {
> > int arr[]={1,2,5,1,5,1,1,3,2,2,};
> > int elements = sizeof(arr)/sizeof(arr[0]);
> > int count=1;
> > int num;
> > sort(arr,arr+elements);
>
> > num=arr[0];
> > for(int i=1;i<elements;i++)
> > {
> > if(arr[i]==num)
> > count++;
> > else
> > {
> > if(count%2==0)
> > { num=arr[i];
> > count=1;}
> > else
> > {cout<<"\n"<<arr[i-1];
> > count=1;
> > num=arr[i];
> > }
> > }
> > }
> > getch();
> > }
>
> > complexity: O(nlogn)
>
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