Lazily Tying the Knot for 1 Dimensional Dynamic Programming

Question

Several years ago I took an algorithms course where we were giving the following problem (or one like it):

There is a building of n floor

Answer 1

Others have answered your direct question about dynamic programming. However, for this kind of problem I think the greedy approach works the best. It's implementation is very straightforward.

f i j :: Int -> Int -> Int
f i j = snd $ until (\(i,_) -> i == j) 
                    (\(i,x) -> (i + if i < j then 2 else (-3),x+1))
                    (i,0)