@mahesh Gupta
Nice solution. Thank you. You have explained it well.
On Oct 4, 5:01 pm, Mukesh Gupta <mukeshgupta.2...@gmail.com> wrote:
> The problem could be solved using xor logic. First take xor of all the
> elements .Doing that we get a value which is xor of the two non repeating
> elements(as xor of similar values is 0). Now xor of two non repeating
> numbers contains bits set at those places where the two numbers differ. Now
> if we take any set bit of our result and again xor all those values of set
> where that bit is set we get first non-repeating value. Taking xor of all
> those values where that bit is not set gives the another non-repeating
> number..
> For ex
> let a[]={2,3,4,3,2,6}
>
> xor of all values=0010
> Now we need to get any set bit. We can extract the rightmost set bit of any
> number n by taking ( n & ~(n-1))
>
> Here 2nd bit is the rightmost set bit.
>
> Now when we take xor of all values where 2nd bit is set(this could be done
> as (a[i] & 0010) , we get 6
> Taking xor of all values where 2nd bit is not set yields 4.
>
> Mukesh Gupta
> Delhi College of Engineering
>
>
>
> On Mon, Oct 4, 2010 at 3:17 PM, malli <mallesh...@gmail.com> wrote:
> > I have an array. All numbers in the array repeat twice except two
> > numbers which repeat only once. All the numbers are placed randomly.
> > Goal is to find out efficiently the two numbers that have not
> > repeated with O(1) extra memory. Expecting linear solution.
>
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