Sorry, I (and others) have explained several times what an equivalent arithmetic coding scheme would have to do to make the same assumptions (and have the same limitations) as your scheme.

It would have to assume there's _exactly_ N ones and N zeros in the bit string. Start with those counts:

remaining ones = N;

remaining zeros = N;

It would encode bit for bit. After every zero, it decrements the remaining number of zeros. After every one, it decrements the remaining number of ones. The probability of the next bit being zero or one is derived from the known remaining number of each, P(one) = (remaining ones / (remaining ones + remaining zeros)). When one of the counts reaches zero, the probability of the other becomes 100% and no more information need to be encoded.

Does the linked code do that? No.

I don't have a ready-made example because this is not an assumption about the input that is useful in the majority of cases. The linked example can encode _any sequence_, not just ones that have an equal number of ones and zeros.

If you spent time implementing this scheme you would learn a lot about this field.