Consider the following language {A,B,C} and the following regex (A|B)+C. I'm a little in doubt about which of my two examples is more correct. Or are they both equally correct? e.g.,
Is it allowed to have multiple letters on a transition in a DFA?
Sorry for the paint quality.
For a regular language $L$, there are an infinite number of finite automata that recognize $L$ (just take a DFA that recognizes $L$ and add another state that transitions only to itself, for instance). However, there is a unique (up to labeling) DFA with a minimum number of states, and various algorithms for state minimization. Ceteris paribus, a DFA with fewer states would be preferable, due to being easier to understand and taking less space to store.
In this case, we have the regular expression over the alphabet $\Sigma=\{a,b,c\}$ $$R=(a\mid b)^*c, $$ which recognizes the language $$L = \{w\in\Sigma: w \text{ has exactly one } c \text{ and } w \text{ ends with a } c\}. $$ A DFA with one state must either accept the empty language or the set of all strings over $\Sigma$, i.e. $\Sigma^*$, so our automaton must have at least two states. Your second idea is on the right track. Formally, we would describe this DFA as $M=(Q, \Sigma, \delta, q_0, F)$ where $Q=\{S_1, S_2\}$ is the set of states, $\Sigma=\{a,b,c\}$ is the input alphabet, $\delta:Q\times\Sigma\to Q$ is the transition function, given by $$ \delta(q, \alpha) = \begin{cases} S_1,& q=S_1, \alpha\ne c\\ S_2,& q=S_1, \alpha = c \text{ or } q=S_2,\alpha\in\Sigma, \end{cases} $$ $q_0=S_1$ is the starting state, and $F=\{S_2\}$ is the set of final (or "accepting" states). Equivalently, we may define $M$ by a transition diagram:
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(I used the jFAST software to produce the above image.)