We model the problem as a shortest path problem in a graph in which the state is an ordered pair $(square,jumplen)$, where $square$ is the square at which Nikola is located and $jumplen$ is the length of the preceding jump.

From state $(square,jumplen)$ Nikola can move to state $(square-jumplen,jumplen)$ moving backwards, or to state $(square+jumplen+1,jumplen+1)$ moving forwards, assuming these jumps don't move him out of the playing area.

Notice that the graph is acyclic (meaning that there is no way to visit a state twice while traversing the graph), because forward jumps increase the $jumplen$ parameter. Because of this, the length of the shortest path can be calculated with dynamic programming.

Let the function $opt(square,jumplen)$ represent the smallest cost for Nikola to get from square $square$ to square $N$, if the preceding jump was of length $jumplen$.

Here's how to calculate the value of the function:

• For $square<1$ or $square>N$, $opt(square,jumplen)=\mathrm{\infty }$;
• For $square=N$, $opt(square,jumplen)=fee[N]$;
• Otherwise, $${\displaystyle opt(square,jumplen)=fee[square]+min\{opt(square-jumplen,jumplen),opt(square+jumplen+1,jumplen+1)\}}$$

The solution is then $opt(2,1)$.