$S_{3} is even$ ,
$S_{4k} is even+ odd + odd$ =EVEN
$S_{4k+1} is even +even + odd$ = ODD
$S_{4k+2} is odd +even + even$ =ODD
$S_{4k+3} is odd +odd + even$ = EVEN
Above is true always so pattern $S_{4},S_{5},S_{6}....$ will be $E,O,O,E,E,O,O,E....$
we can also view this as question of TOC
we have to scan last 3 symbol to determine current symbol like if last 3 symbol are E,O,O then next will be E
there are only two possible state and 4 transations