@ Don exactly waht u write i wanted to say
On Fri, Aug 26, 2011 at 11:52 AM, tech coder <[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?
>
> see we come to know that both number differ at bit1 bit2 and bit3 we need
> only bit1
>
> set1 contain the numbers whose bit1 is set including 3
> set2 contain the numbers whose bit1 is clear including 4
>
> now xoring 1st set we get 3
> xor ing 2nd set we get 4
> (bacuase all others appear even no of time they will nullify each
> other)
>
>
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