Give a DFSA that accepts $L_1 = \{x : \#_{01}(x) \mod 3 = 0\}$ using as few states as possible
Hint, a correct regular expression is $1^*(00^*11^*00^*11^*00^*11^*)^*0^*$. 

There are 6 states, where $q_{0,1}$ is the initial state. Both $q_{0,1}$ and $q_{0,0}$ are accepting states.

$q_{0,1}$: the number of 01s is a multiple of 3 and it ends with 1. Or empty word.
$q_{0,0}$: the number of 01s is a multiple of 3 and it ends with 0.
$q_{1,1}$: the number of 01s is 1 plus a multiple of 3 and it ends with 1.
$q_{1,0}$: the number of 01s is 1 plus a multiple of 3 and it ends with 0.
$q_{2,1}$: the number of 01s is 2 plus a multiple of 3 and it ends with 1.
$q_{2,0}$: the number of 01s is 2 plus a multiple of 3 and it ends with 0.