\begin{enumerate}[(a)] \item WLOG, assume $m \ge n$, the answer is $2n$. Construction: Alternating put $B$ and $A$ until $B$ runs out, then put the remaining $A$ in one block. There are exactly $2n$ students with 0 candy. Proof of bound: Let the set of student with 0 candy be $S_0$. Define $S_1$, $S_2$, \dots, similarly. It is easily seen that $|S_0| \ge |S_1| \ge |S_2| \ge \dots$. On the other hand, in class $B$ there are at most $n$ students with zero candies. In class $A$, there are at most $n$ students whose left immediete neighbour is from class $B$, meaning there are at most $n$ students with 0 candies. As such, $2n \ge |S_0| \ge |S_1| \ge \dots$. \item WLOG, assume $m \ge n$, the answer can be described: \begin{itemize} \item If $m > n$, the answer is $n$. \item If $m = n$ and $n$ is even, the answer is $n$. \item If $m = n$ and $n$ is odd, the answer is $n-1$. \end{itemize} Construction: Alternating 2 $B$'s and 2 $A$'s until $B$ runs out, then put all the remaining $A$'s together. This satisfies the number given above. Proof of bound: Like we've seen in part (a), we have $S_1 \ge S_2 \ge \dots $. We split the $B$'s into $t$ contiguous blocks. There are at most $\min(t, n-t)$ people in class $B$ with 1 candy. In class $A$, there are at most $\min(t, m-t)$ people with 1 candy. Indeed, when $m > n$ or $m=n$ and $n$ is even, there are at most $\min(t, n-t)+\min(t, m-t) \le n-t+t=n$ people with 1 candy. When $m=n$ and $n$ is odd, we see that the value is given by exactly $2\min(t, n-t)$, and the maximum of $\min(t, n-t)$ is $\lfloor \frac n2 \rfloor = \frac{n-1}{2}$, so there are at most $n-1$ people with $1$ candy. Thus, this value $\ge |S_1| \ge |S_2|\ge \dots$ is the desired maximum. \end{enumerate}