hackercupai/hackercup huggingface dataset:
It's helpful to work through an example, so let's use the third sample case. We'll sort each player's hand in decreasing order for convenience:

- \(A1: [15, 13, 11, 6]\)
- \(B1: [16, 12, 2, 1]\)
- \(A2: [9, 8, 7, 5]\)
- \(B2: [14, 10, 4, 3]\)

Note that the order that player \(A1\) plays their cards has no effect on the strategy of other players. In fact, since all players have perfect information, it helps to imagine reorganizing the game in a slightly different way: player \(A1\) plays all of their cards at once, one each into \(N/4\) different stacks. Then player \(B1\) plays all of their cards, again one each onto each of the stacks, and so on. After all cards are played, the team with the highest card in a stack scores a point for that stack.

We'll keep track of the highest card in each stack, and which team it belongs to. In our example, after player \(A1\) plays, the stacks are: \((15, A)\), \((13, A)\), \((11, A)\), \((6, A)\).

The general strategy for the other players is:
- First priority: cover as many of the opposing team's highest cards as possible
- Second priority: cover as many of your own team's lowest cards as possible

Here "cover" means "play a card higher than". As we'll see, this second priority is only meaningful to player \(A2\).

Player \(B1\) will cover as many of player \(A1\)'s highest cards as they can, using their own highest cards. In this example, \(B1\) will cover the \(15\) with the \(16\) and the \(11\) with the \(12\). Now the stacks are: \((16, B)\), \((13, A)\), \((12, B)\), and \((6, A)\).

Player \(A2\) similarly wants to cover as many cards owned by team \(B\) as possible, which in this case is none of them. \(A2\)'s second priority is to cover their *own* team's *lowest* cards with the highest cards possible. In this case \(A2\) can cover the \(6\) with the \(9\), and now the stacks are: \((16, B)\), \((13, A)\), \((12, B)\), \((9, A)\).

Player \(B2\) now finishes up by covering as many of team \(A\)'s cards as possible: \((16, B)\), \((14, B)\), \((12, B)\), \((10, B)\). In the end, team \(B\) scores every point.

This approach can be implemented in \(\mathcal{O}(N \log N\)\) time by sorting the initial hands and then resorting the stacks after each player plays.