twoPostsBack
and onePostBack
, to represent the number of ways to paint the fence with one and two posts, respectively.twoPostsBack = k
and onePostBack = k * k
to represent the base cases.n-th
post.curr = (k - 1) * (onePostBack + twoPostsBack)
.twoPostsBack
to the value of onePostBack
, and onePostBack
to the value of curr
.onePostBack
as the final answer.