Pair flips on a coin row

Think of each tails coin as a token. A move can slide a token one place, remove $2$ neighbouring tokens, or create $2$.

1. Each move flips $2$ coins, so tails coins disappear in pairs, and a pair of tails at distance $d$ apart can be removed in $d$ moves (slide one tail along $d-1$ places and then remove both) but not in fewer.
2. So the minimum is the cheapest way to pair up the $6$ tails by distance, and on a line the cheapest pairing joins neighbours in order.
3. The tails at $1,3$, $4,8$ and $9,12$ have distances $2$, $4$ and $3$, while any other pairing costs more (for instance $1,3$, $4,9$, $8,12$ costs $11$).
4. The minimum number of moves is $2+4+3=9$.

Answer: $9$