hackercupai/hackercup huggingface dataset:
Since you must eat dumplings on a plate adjacent to your opponent, after the first move, it's uniquely determined who will eat from even plates and who will eat from odd plates. Therefore, other than Alice's first move, neither player will ever eat more than one dumpling on a turn.

First, let's determine whether some board state is winning or losing for the current player. Suppose \(A[0] > A[1]\). In this case, Alice can win by eating from \(A[0]\), and will lose if she eats from \(A[1]\). Importantly, notice that if Alice eats from \(A[2...n]\), neither player will ever eat from \(A[1]\) as it will cause them to lose—their opponent can keep eating \(A[0]\), and \(A[1]\) will run out first.

Therefore (if \(A[0] > A[1]\) and the first move doesn't involve \(A[0]\)), we can replace \(A[0]\) and \(A[1]\) with \(0\)'s and it will not change the result of the game.

On the other hand if \(A[0] \leq A[1]\), then if our opponent ever eats from \(A[1]\), we should keep eating from \(A[0]\) as many times as possible to decrease \(A[1]\). We can prove this is optimal with a greedy principle: dumplings at position \(0\) can at best cause your opponent to have to eat one more dumpling from position \(1\), and at worst will never be eaten because your opponent won't give you the chance. Either way, we'll have to continue playing on pile \(2\), and it's better for us if the opponent has fewer dumplings left on pile \(1\).

Therefore (if \(A[0] \leq A[1]\) and the first move doesn't involve \(A[0]\)), then we can replace \(A[0]\) with \(0\), and \(A[1]\) with \(A[1]-A[0]\), and it will not change the result of our game.

Of course, these observations are symmetric, so they apply to \(A[N]\) and \(A[N-1]\) in the same way as \(A[0]\) and \(A[1]\).

If the first move was made at position \(A[7]\) for example, then in linear time we can repeatedly simplify each end of the array until the only elements that are left are \(A[6], A[7],\) and \(A[8]\). We either delete the first/last two elements, or delete the first/last element, and decrease the second/second last. When we're done, this leaves us with an array of size \(3\), and all that matters is the total number of dumplings in the middle vs. on the sides. If after she takes dumplings from \(A[7]\), Alice will win if \(A[7] \geq A[6]+A[8]\). This leaves us with \(\max(0, A[7]-A[6]-A[8])\) ways of winning that start with position \(7\).

This gives us an \(\mathcal{O}(N^2)\) solution since we are brute forcing the first move. But it turns out that each side is independent, so we can do a DP on each prefix and suffix to count what each prefix and suffix leaves you with, and this gives us an \(\mathcal{O}(N)\) solution.