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|))~