My idea for problem D of the MHC is as follows: On the first turn, you cannot take an A that has a B to its right, and symmetrically the same applies if the second player were playing in that same configuration. We can mark those choices, which is equivalent to marking those characters. Now, the new problem with marked characters is equivalent to removing those elements, and the resulting problem is exactly the same as the original one. Then, we can eliminate from left to right the last available A for each B found. Finally, Alice can win iff the resulting string has the form ...A.