I believe this is what techcoder is saying:
int a[N];
// Find the bitwise xor of all the array values.
// These are the bits which are different between the two results.
int xor = 0;
for(i = 0; i < N; ++i)
xor ^= a[N];
// Find the low order bit of xor
int bit = 1;
while(!(xor & bit))
bit <<= 1;
// xor the values with "bit" set to get one result
// xor the values with "bit" unset to get the other result
int result1 = 0, result2 = 0;
for(i = 0; i < n; ++i)
{
if (a[i] & bit) result1 ^= a[i];
else result2 ^= a[i];
}
Now result1 & result2 are the values which appear an odd number of
times. It is O(n).
Don
On Aug 26, 12:13 pm, Dave <[email protected]> wrote:
> @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 <[email protected]> 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
> > <[email protected]>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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