I have a scheduling problem in which there are $n$ slots and $m$ clients. I am trying to solve the problem using Q-learning so I have made the following state-action model.
A state $s_t$ is given by the current slot $t=1,2,\ldots,n$ and an action $a_t$ at slot $t$ is given by one client, $a_t\in\{1,2,\ldots,m\}$. In my situation, I do not have any reward associated with a state-action pair $(s_t,a_t)$ until the terminal state which is the last slot. In other words, for all $s_t\in\{1,2,\ldots,n-1\}$, the reward is $0$ and for $s_t=n$ I can compute the reward given $(a_1,a_2,\ldots,a_n)$.
In this situation, the Q table, $Q(s_t,a_t)$, will contain only zeros except for the last row in which it will contain the updated reward.
Can I still apply Q-learning in this situation? Why do I need a Q table if I only use the last row?