Let ~Z~ be the maximum possible position (~Z = 100\,000~), and let ~WM_p~ and ~WG_p~ be the ~M~ and ~G~ values of the wild Pokémon at position ~p~ (if any), with ~WM_p = WG_p = 0~ if there's no such Pokémon. Then, ~WM_1, \dots, WM_Z~ and ~WG_1, \dots, WG_Z~ may be populated in ~\mathcal O(Z+N)~ time by iterating over the ~N~ Pokémon.

We'll greedily simulate Jessie and Arbok's training process. Let ~p_1~ be the smallest position which Jessie can currently freely reach, ~p_2~ be the largest one, and ~a~ be Arbok's current level. Initially, ~p_1 = p_2 = S~ and ~a = L~.

At any point in time, if ~p_1-1 \ge 1~ and ~a \ge WM_{p_1-1}~, then Jessie is able to expand her reachable range to encompass position ~p_1-1~, and has no reason not to do so. Therefore, we can increase ~a~ by ~WG_{p_1-1}~, and then decrement ~p_1~. Similarly, if ~p_2+1 \le Z~ and ~a \ge WM_{p_2+1}~, then Jessie might as well expand her reachable range to encompass position ~p_2+1~. Therefore, we can increase ~a~ by ~WG_{p_2+1}~, and then increment ~p_2~.

We should repeat the above process of expanding Jessie's range of reachable positions left and right whenever possible until we arrive at a situation in which it cannot be expanded in either direction. At that point, nothing more can be done, and we can proceed to output the current value of ~a~.

The time complexity of the above solution is ~\mathcal O(Z+N)~. A similar algorithm running in ~\mathcal O(N \log N)~ time is also possible, if we sort and consider only positions of interest (Jessie's starting position as well all wild Pokémons' positions) instead of all ~Z~ possible positions.