<p>There is a <code>50 x 50</code> chessboard with <strong>one</strong> knight and some pawns on it. You are given two integers <code>kx</code> and <code>ky</code> where <code>(kx, ky)</code> denotes the position of the knight, and a 2D array <code>positions</code> where <code>positions[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> denotes the position of the pawns on the chessboard.</p>

<p>Alice and Bob play a <em>turn-based</em> game, where Alice goes first. In each player&#39;s turn:</p>

<ul>
	<li>The player <em>selects </em>a pawn that still exists on the board and captures it with the knight in the <strong>fewest</strong> possible <strong>moves</strong>. <strong>Note</strong> that the player can select <strong>any</strong> pawn, it <strong>might not</strong> be one that can be captured in the <strong>least</strong> number of moves.</li>
	<li><span>In the process of capturing the <em>selected</em> pawn, the knight <strong>may</strong> pass other pawns <strong>without</strong> capturing them</span>. <strong>Only</strong> the <em>selected</em> pawn can be captured in <em>this</em> turn.</li>
</ul>

<p>Alice is trying to <strong>maximize</strong> the <strong>sum</strong> of the number of moves made by <em>both</em> players until there are no more pawns on the board, whereas Bob tries to <strong>minimize</strong> them.</p>

<p>Return the <strong>maximum</strong> <em>total</em> number of moves made during the game that Alice can achieve, assuming both players play <strong>optimally</strong>.</p>

<p>Note that in one <strong>move, </strong>a chess knight has eight possible positions it can move to, as illustrated below. Each move is two cells in a cardinal direction, then one cell in an orthogonal direction.</p>

<p><img src="https://assets.leetcode.com/uploads/2024/08/01/chess_knight.jpg" style="width: 275px; height: 273px;" /></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">kx = 1, ky = 1, positions = [[0,0]]</span></p>

<p><strong>Output:</strong> <span class="example-io">4</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/08/16/gif3.gif" style="width: 275px; height: 275px;" /></p>

<p>The knight takes 4 moves to reach the pawn at <code>(0, 0)</code>.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">kx = 0, ky = 2, positions = [[1,1],[2,2],[3,3]]</span></p>

<p><strong>Output:</strong> <span class="example-io">8</span></p>

<p><strong>Explanation:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2024/08/16/gif4.gif" style="width: 320px; height: 320px;" /></strong></p>

<ul>
	<li>Alice picks the pawn at <code>(2, 2)</code> and captures it in two moves: <code>(0, 2) -&gt; (1, 4) -&gt; (2, 2)</code>.</li>
	<li>Bob picks the pawn at <code>(3, 3)</code> and captures it in two moves: <code>(2, 2) -&gt; (4, 1) -&gt; (3, 3)</code>.</li>
	<li>Alice picks the pawn at <code>(1, 1)</code> and captures it in four moves: <code>(3, 3) -&gt; (4, 1) -&gt; (2, 2) -&gt; (0, 3) -&gt; (1, 1)</code>.</li>
</ul>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">kx = 0, ky = 0, positions = [[1,2],[2,4]]</span></p>

<p><strong>Output:</strong> <span class="example-io">3</span></p>

<p><strong>Explanation:</strong></p>

<ul>
	<li>Alice picks the pawn at <code>(2, 4)</code> and captures it in two moves: <code>(0, 0) -&gt; (1, 2) -&gt; (2, 4)</code>. Note that the pawn at <code>(1, 2)</code> is not captured.</li>
	<li>Bob picks the pawn at <code>(1, 2)</code> and captures it in one move: <code>(2, 4) -&gt; (1, 2)</code>.</li>
</ul>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= kx, ky &lt;= 49</code></li>
	<li><code>1 &lt;= positions.length &lt;= 15</code></li>
	<li><code>positions[i].length == 2</code></li>
	<li><code>0 &lt;= positions[i][0], positions[i][1] &lt;= 49</code></li>
	<li>All <code>positions[i]</code> are unique.</li>
	<li>The input is generated such that <code>positions[i] != [kx, ky]</code> for all <code>0 &lt;= i &lt; positions.length</code>.</li>
</ul>

class Solution:
    def maxMoves(self, kx: int, ky: int, positions: List[List[int]]) -> int:
        '''
        Problem Summary:
        - Alice and Bob take turns capturing pawns on a 50x50 chessboard.
        - Alice tries to maximize the total number of moves, while Bob tries to minimize them.
        - The knight starts at position (kx, ky) and can capture pawns in the fewest moves possible.

        Approach:
        - Use dynamic programming (top-down DP) with two states: Alice's turn (turn=0) and Bob's turn (turn=1).
        - For Alice's turn, she tries to maximize the total number of moves by picking the farthest pawn.
        - For Bob's turn, he tries to minimize the total number of moves by picking the closest pawn.
        - Use a bitmask to represent the set of pawns left to capture, since the number of pawns is between 1 and 15.
        - Precompute the minimum number of moves from the knight's position to each pawn, and between every pair of pawns using BFS.
        '''

        # Define knight's possible moves (8 L-shaped moves)
        moves = [(2, 1), (2, -1), (-2, 1), (-2, -1), (1, 2), (-1, 2), (1, -2), (-1, -2)]
        INF = 10 ** 20  # Infinity for unreachable positions

        n = len(positions)  # Number of pawns

        # BFS to calculate the minimum distance from a starting point (start_x, start_y)
        def bfs(start_x, start_y):
            dist = [[INF] * 50 for _ in range(50)]  # Initialize distance grid
            queue = deque([(start_x, start_y, 0)])  # BFS queue with (x, y, distance)
            dist[start_x][start_y] = 0  # Starting point distance is 0
            while queue:
                x, y, d = queue.popleft()  # Dequeue
                for dx, dy in moves:  # Explore all 8 knight moves
                    nx, ny = x + dx, y + dy
                    if 0 <= nx < 50 and 0 <= ny < 50 and dist[nx][ny] == INF:
                        dist[nx][ny] = d + 1
                        queue.append((nx, ny, d + 1))
            return dist

        # Precompute the distance from the knight's starting position to each pawn
        distances = []
        knight_distances = bfs(kx, ky)  # Distance from knight's initial position
        for x, y in positions:
            distances.append(bfs(x, y))  # Distance from each pawn to all other positions

        # Dynamic programming function with memoization
        @cache
        def dp(mask, pos, turn):
            '''
            mask: bitmask representing which pawns are left to capture
            pos: index of the last pawn captured (or -1 if no pawns captured yet)
            turn: 0 for Alice's turn, 1 for Bob's turn
            '''
            if mask == 0:
                # If no pawns are left to capture, return 0 (no more moves)
                return 0

            if turn == 0:
                # Alice's turn (maximize total moves)
                result = 0
                for i in range(n):
                    if mask & (1 << i):  # If the i-th pawn is still available
                        if pos == -1:  # Starting from the knight's initial position
                            dist = knight_distances[positions[i][0]][positions[i][1]]
                        else:  # Moving from the last captured pawn
                            dist = distances[pos][positions[i][0]][positions[i][1]]

                        # Maximize result: Alice chooses the farthest pawn
                        result = max(result, dist + dp(mask ^ (1 << i), i, 1))
                return result

            else:
                # Bob's turn (minimize total moves)
                result = INF
                for i in range(n):
                    if mask & (1 << i):  # If the i-th pawn is still available
                        if pos == -1:  # Starting from the knight's initial position
                            dist = knight_distances[positions[i][0]][positions[i][1]]
                        else:  # Moving from the last captured pawn
                            dist = distances[pos][positions[i][0]][positions[i][1]]

                        # Minimize result: Bob chooses the closest pawn
                        result = min(result, dist + dp(mask ^ (1 << i), i, 0))
                return result

        # Initialize DP with all pawns available (mask is all 1s) and Alice's turn
        return dp((1 << n) - 1, -1, 0)

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