Subtask 1

Iterate through both strings simultaneously. Any excess as in $A$ must be merged with another a immediately behind to match the b correctly in $B$. If the character behind is not a or there is an extraneous b in $A$, we can evaluate the case to be impossible.

Time Complexity: $\mathcal{O}(\sum (|A|+|B|))$

Subtask 2

Similar to the first subtask, if a character in $A$ is less than the target character in $B$, we must repeatedly try to merge it with further characters in $A$ until they match. This can be implemented either recursively or iteratively using a stack.

Time Complexity: $\mathcal{O}(\sum (|A|+|B|))$

Alternately, note that if we let $o(c)$ be the order of the character $c$ in the alphabet, then the quantity $\sum 2^{o(c)}$ remains invariant over all operations. Thus, we can instead imagine repeatedly splitting the characters of $B$ back into $A$. For each split, there can be at most one candidate splitting point in $A$, which can be found by binary searching on the prefix sums of $2^{o(c)}$.

Time Complexity: $\mathcal{O}(\sum (|A|\log |A|+|B|))$