A deterministic FA for $0^*1^*$ is required A deterministic finite automaton without $\epsilon$ steps for the language $0^*1^*$ is required.
Any nice picture ? I have created a NFA for this language which has 2 states $Q_1,Q_2$,
both are accepting, $Q_1$ is initial, there is a move under $0$ from $Q_1$ to $Q_1$, a move under $1$ from $Q_1$ to $Q_2$ and finally a move from $Q_2$ to $Q_2$ under $1$.
To construct a DFA, note the following:


An empty string or any string containing only $0$s is accepted, so
any $0^*$ is accepted.
Zero or more $0$s followed by one or more $1$s is accepted, so any
$0^*11^*$ is accepted.
Zero or more $0$s followed by one or more $1$s followed by $0$ is
rejected, so any $0^*11^*0$ is rejected.
Any extension to a rejected string is also rejected (i.e. there is no
route back from a rejecting state to an accepting state), so any
$0^*11^*0(0+1)^*$ is rejected.


This suggests a three state DFA with two accepting states and one rejecting state ...