@Lakshman Patel RJIT ofcourse bro.
2nd part AND 1st part.
@Lakshman Patel RJIT bro this is the approach
Yes, very nice answer.
$NFA\rightarrow DFA$
$DFA\rightarrow NFA$
$NFA\equiv DFA$
So, both are equivalent in power.
$\Rightarrow$ Every $DFA$ is $NFA,$ but reverse need not be true.
not
Please correct me if I'm wrong?
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