I think I could do it in O(n) .
var bits=0
var @array
for each i in array
if (bits & i)!=0
return i
else
bits=bits | i;
end if
end for
I just use the pattern for powers of two
let array be 2,4,2
so first
bits=0 i=2 so bits(000) & i(010) is 0 so bits= bits | i which is
bits=010
next
bits=2 i=4 so bits(010) & i(100) is 0 so bits= bits | i which is
bits=110
next
bits=6 i=2 so bits(110) & i(010) is 2 which is !=0 so we return i.
I think that should be it.
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