<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>


from collections import deque

class Solution(object):
    def maxMoves(self, kx, ky, positions):
        """
        :type kx: int
        :type ky: int
        :type positions: List[List[int]]
        :rtype: int
        """
        knight_moves = [(-2, -1), (-2, 1), (-1, -2), (-1, 2), (1, -2), (1, 2), (2, -1), (2, 1)]
        n = len(positions)
        
        # BFS to find shortest distance between two points on the chessboard
        def bfs(sx, sy, ex, ey):
            queue = deque([(sx, sy, 0)])
            visited = [[False] * 50 for _ in range(50)]
            visited[sx][sy] = True
            while queue:
                x, y, dist = queue.popleft()
                if x == ex and y == ey:
                    return dist
                for dx, dy in knight_moves:
                    nx, ny = x + dx, y + dy
                    if 0 <= nx < 50 and 0 <= ny < 50 and not visited[nx][ny]:
                        visited[nx][ny] = True
                        queue.append((nx, ny, dist + 1))
            return float('inf')

        # Precompute distances from the knight to all pawns and between all pawns
        distances = [[0] * n for _ in range(n)]
        knight_to_pawn = []
        for i in range(n):
            px, py = positions[i]
            knight_to_pawn.append(bfs(kx, ky, px, py))
            for j in range(i + 1, n):
                px2, py2 = positions[j]
                dist = bfs(px, py, px2, py2)
                distances[i][j] = distances[j][i] = dist

        # DP with bitmasking: dp[mask][i] = max moves starting from pawn i with pawns represented by mask
        dp = [[-1] * n for _ in range(1 << n)]
        
        # Function to calculate maximum moves with dynamic programming and bitmask
        def dp_max(mask, last):
            if dp[mask][last] != -1:
                return dp[mask][last]
            
            total_pawns = bin(mask).count('1')  # Total pawns captured so far
            if total_pawns == n:  # All pawns captured
                return 0

            max_moves = 0
            for i in range(n):
                if not (mask & (1 << i)):  # If pawn i is not captured yet
                    next_mask = mask | (1 << i)  # Capture pawn i
                    move_cost = distances[last][i]
                    max_moves = max(max_moves, move_cost + dp_min(next_mask, i))  # Try to maximize Alice's move
            dp[mask][last] = max_moves
            return max_moves

        # Function to calculate minimum moves for Bob
        def dp_min(mask, last):
            if dp[mask][last] != -1:
                return dp[mask][last]
            
            total_pawns = bin(mask).count('1')  # Total pawns captured so far
            if total_pawns == n:  # All pawns captured
                return 0

            min_moves = float('inf')
            for i in range(n):
                if not (mask & (1 << i)):  # If pawn i is not captured yet
                    next_mask = mask | (1 << i)  # Capture pawn i
                    move_cost = distances[last][i]
                    min_moves = min(min_moves, move_cost + dp_max(next_mask, i))  # Bob tries to minimize moves
            dp[mask][last] = min_moves
            return min_moves

        # Start DP from the knight's initial position
        max_result = 0
        for i in range(n):
            max_result = max(max_result, knight_to_pawn[i] + dp_min(1 << i, i))

        return max_result
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