Suppose a bit-vector with 2^n bits. It is a balanced bit-vector if it has exactly 2^(n-1) bit 1 (so does bit 0). Given n, can I print out all 2^n-bit balanced bit-vector in O(2^n)?
Is there any way to traverse all balanced bitvector?
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You can't.
Printing single 2^n bits vector takes O(2^n) time, and printing all possible vectors takes
O(2^n*nCr(2^n, 2^(n-1)))- much more.