For up to ~9~ out of the ~15~ available marks ~(N \le 1000)~, a simple algorithm which iterates over each pair of the pieces of wood in the input suffices.

Time Complexity: ~\mathcal O(N^2)~

For the full ~15~ marks, we make the observation ~1 \le L_i \le 2000~. This means we can iterate over every (unordered) pair of lengths of pieces of wood available and add it to the corresponding total for a length of board made. Since each length of wood can be uniquely paired up with another length of wood to create a corresponding length of board, no pairs are over-counted this way. Finally, we can iterate over every possible board length to determine the length of the longest fence and the heights of all fences which have that length.

Time Complexity: ~\mathcal O(N + {L_i}^2)~