@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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