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What I mean is if you start at i=Q[N-1], the max element, and go down, you must find the global maximum before reaching Q[N/2]. Correct me if I'm wrong.
Nice explanation ! Most elements there I managed to find too, so it kind of confirms my own conclusions. However, could you check the smaller possible optimisations ? I'd really like to have this be the fastest. I'm thinking like doing 1 ≤ i ≤ (N-1)/2, and y could be P[i]/A, which bypasses creating any subset. Thoughts ?
