@don really cool algo man
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@ Don exactly waht u write i wanted to say
On Fri, Aug 26, 2011 at 11:52 AM, tech coder 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 binar
@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 bi
XOR all the elements in the array, the result will be the XOR of the two
numbers occuring odd number of times.
Now take any set bit of th result(u can determine the position of any bit
set in the number). Divide the array such that for the numbers for which at
this location(where the bit is set in
