I want to build a finite automaton that accepts $a^nb^n, n \gt 0, m \ge 0$. I can't do it unless the FA has two final states, i.e.:
$delta(q0, a) = q1
delta(q1, a) = q1
delta(q1, b) = q2
delta(q2, b) = q2$
where q1 and q2 are final states. Is there a way of building the FA with just one final state and / or a minimized number of states?
Firstly, I think you mean the alphabet is $\{a,b\}$ and the language is $a^nb^m$ for $n>0$ and $m\ge0$ instead of $a^nb^n$.
Secondly, you need to specify the transitions of $q_0$ and $q_1$ under both $a$ and $b$ as $\delta$ is a map from $Q \times \sum$ to $Q$.
You should get an automaton with a fourth state which maps to itself under both letters and is mapped to by $q_0$ and $q_2$. This automaton is minimal.
You can read about Myhill Nerode Theorem and DFA Minimization which tell us how to minimize a determinitic finite automaton and when it is possible.