We'll iterate over the ~N~ Pikachus while maintaining two values ~m_1~ and ~m_2~ – the largest and second-largest weights of catchable Pikachus seen so far, respectively. Initially, ~m_1 = m_2 = 0~.

When considering the ~i~-th Pikachu, we should ignore it completely if ~W_i > 100~. Otherwise, if ~W_i > m_1~, then we've found a new heaviest catchable Pikachu, demoting the previous heaviest one to now be the second-heaviest – therefore, we should first set ~m_2~ to equal ~m_1~, and then set ~m_1~ to equal ~W_i~. Otherwise, if ~W_i > m_2~, then we've found a new second-heaviest catchable Pikachu, so we should simply set ~m_2~ to be equal to ~W_i~.

After considering all ~N~ Pikachus in this fashion, ~m_1+m_2~ comes out to the sum of the two heaviest catchable Pikachus' weights, which is the answer we're looking for. Note that if there were fewer than two catchable Pikachus, ~m_1~ and/or ~m_2~ will still be equal to ~0~, which is desirable as it represents Jessie and/or James remaining empty-handed.