<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:
    def maxMoves(self, kx, ky, positions):
        n_pawns = len(positions)
        positions_list = [(kx, ky)] + positions
        num_positions = n_pawns + 1

        # Precompute distances between all positions using BFS
        dist = [[0] * num_positions for _ in range(num_positions)]
        moves = [(-2, -1), (-1, -2), (1, -2), (2, -1),
                 (2, 1), (1, 2), (-1, 2), (-2, 1)]

        for i in range(num_positions):
            start = positions_list[i]
            visited = [[-1 for _ in range(50)] for __ in range(50)]
            queue = deque()
            queue.append((start[0], start[1], 0))
            visited[start[0]][start[1]] = 0

            while queue:
                x, y, d = queue.popleft()
                for dx, dy in moves:
                    nx, ny = x + dx, y + dy
                    if 0 <= nx < 50 and 0 <= ny < 50 and visited[nx][ny] == -1:
                        visited[nx][ny] = d + 1
                        queue.append((nx, ny, d + 1))

            for j in range(num_positions):
                if i == j:
                    dist[i][j] = 0
                else:
                    pos_j = positions_list[j]
                    dist[i][j] = visited[pos_j[0]][pos_j[1]]
                    if dist[i][j] == -1:
                        # If a pawn is unreachable, it's impossible to capture
                        # according to problem constraints, this shouldn't happen
                        pass

        # Initialize DP tables
        size = 1 << n_pawns
        dp_alice = [0] * (num_positions * size)
        dp_bob = [0] * (num_positions * size)

        for bitmask in range(size):
            for current_pos in range(num_positions):
                idx = current_pos * size + bitmask
                if bitmask == 0:
                    dp_alice[idx] = 0
                    dp_bob[idx] = 0
                    continue

                # Alice's turn: maximize the total sum
                max_sum = 0
                for pawn in range(n_pawns):
                    if bitmask & (1 << pawn):
                        move_count = dist[current_pos][pawn + 1]
                        next_pos = pawn + 1
                        next_bitmask = bitmask ^ (1 << pawn)
                        next_idx_bob = next_pos * size + next_bitmask
                        total = move_count + dp_bob[next_idx_bob]
                        if total > max_sum:
                            max_sum = total
                dp_alice[idx] = max_sum

                # Bob's turn: minimize the total sum
                min_sum = float('inf')
                for pawn in range(n_pawns):
                    if bitmask & (1 << pawn):
                        move_count = dist[current_pos][pawn + 1]
                        next_pos = pawn + 1
                        next_bitmask = bitmask ^ (1 << pawn)
                        next_idx_alice = next_pos * size + next_bitmask
                        total = move_count + dp_alice[next_idx_alice]
                        if total < min_sum:
                            min_sum = total
                dp_bob[idx] = min_sum

        initial_bitmask = (1 << n_pawns) - 1
        initial_idx = 0 * size + initial_bitmask
        return dp_alice[initial_idx]

25315