<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:
        n = len(positions)
        all_positions = positions.copy()
        all_positions.append([kx, ky])  # index n is the starting position
        pos_to_index = { (x, y): i for i, (x, y) in enumerate(all_positions) }
        
        # Precompute distances between all pairs using BFS
        d = [ [0]*(n+1) for _ in range(n+1) ]
        
        # Directions for knight moves
        directions = [ (2,1), (1,2), (-1,2), (-2,1), (-2,-1), (-1,-2), (1,-2), (2,-1) ]
        
        for i in range(n+1):
            start = all_positions[i]
            queue = deque()
            queue.append( (start[0], start[1], 0) )
            visited = [ [False]*50 for _ in range(50) ]
            visited[start[0]][start[1]] = True
            while queue:
                x, y, steps = queue.popleft()
                # Check if this position is one of the target positions
                if (x, y) in pos_to_index:
                    j = pos_to_index[(x, y)]
                    if j != i:
                        if d[i][j] == 0:
                            d[i][j] = steps
                for dx, dy in directions:
                    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, steps + 1) )
        
        # Now, implement DP with memoization
        # memo[mask][last][player] where player: 0 = Alice, 1 = Bob
        # To speed up, use a flat list
        size = (1 << n) * (n+1) * 2
        memo = [ -1 ] * size
        
        def get_memo(mask, last, player):
            return mask * (n+1) * 2 + last * 2 + player
        
        def dfs(mask, last, player):
            idx = get_memo(mask, last, player)
            if memo[idx] != -1:
                return memo[idx]
            if mask == 0:
                memo[idx] = 0
                return 0
            if player == 0:  # Alice's turn
                best = -float('inf')
                p = mask
                while p:
                    lsb = p & -p
                    p_idx = (lsb).bit_length() - 1
                    p = p ^ lsb
                    # Capture pawn p_idx
                    distance = d[last][p_idx] if last != n else d[n][p_idx]
                    total = distance + dfs(mask ^ (1 << p_idx), p_idx, 1)
                    if total > best:
                        best = total
                memo[idx] = best
                return best
            else:  # Bob's turn
                best = float('inf')
                p = mask
                while p:
                    lsb = p & -p
                    p_idx = (lsb).bit_length() - 1
                    p = p ^ lsb
                    # Capture pawn p_idx
                    distance = d[last][p_idx] if last != n else d[n][p_idx]
                    total = distance + dfs(mask ^ (1 << p_idx), p_idx, 0)
                    if total < best:
                        best = total
                memo[idx] = best
                return best
        
        full_mask = (1 << n) - 1
        return dfs(full_mask, n, 0)
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