<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
from functools import lru_cache

class Solution:
    # Direction vectors for knight movement
    directions = [(-2, -1), (-1, -2), (1, -2), (2, -1), 
                  (2, 1), (1, 2), (-1, 2), (-2, 1)]

    def _knight_distance(self, start_x, start_y, end_x, end_y):
        """Use BFS to compute the knight's minimum number of moves from (start_x, start_y) to (end_x, end_y)."""
        if start_x == end_x and start_y == end_y:
            return 0
        
        queue = deque([(start_x, start_y, 0)])
        visited = set([(start_x, start_y)])

        while queue:
            x, y, moves = queue.popleft()

            for dx, dy in self.directions:
                new_x, new_y = x + dx, y + dy

                if 0 <= new_x < 50 and 0 <= new_y < 50 and (new_x, new_y) not in visited:
                    if new_x == end_x and new_y == end_y:
                        return moves + 1

                    visited.add((new_x, new_y))
                    queue.append((new_x, new_y, moves + 1))

        return float('inf')  # Should never reach here due to valid board bounds

    def maxMoves(self, kx, ky, positions):
        n = len(positions)
        
        # Precompute the distances between the starting knight position and each pawn
        distances = [[0] * n for _ in range(n + 1)]
        
        # Distance from knight's current position to each pawn
        for i in range(n):
            distances[0][i] = self._knight_distance(kx, ky, positions[i][0], positions[i][1])
        
        # Distance between every pair of pawns
        for i in range(n):
            for j in range(n):
                if i != j:
                    distances[i + 1][j] = self._knight_distance(positions[i][0], positions[i][1], positions[j][0], positions[j][1])
        
        @lru_cache(None)
        def dp(mask, knight_pos):
            # mask is a bitmask indicating which pawns have been captured
            if mask == (1 << n) - 1:
                return 0  # All pawns captured
            
            turn = bin(mask).count('1')
            is_alice_turn = (turn % 2 == 0)

            result = float('-inf') if is_alice_turn else float('inf')

            for i in range(n):
                if mask & (1 << i) == 0:  # If pawn i has not been captured
                    new_mask = mask | (1 << i)
                    move_cost = distances[knight_pos][i]

                    if is_alice_turn:
                        # Alice maximizes the total moves
                        result = max(result, move_cost + dp(new_mask, i + 1))
                    else:
                        # Bob minimizes the total moves
                        result = min(result, move_cost + dp(new_mask, i + 1))
            
            return result
        
        # Start from the initial position of the knight and no pawns captured (mask = 0)
        return dp(0, 0)


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