<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

# Directions for knight's possible moves
dr = [-2, -1, 1, 2, 2, 1, -1, -2]
dc = [1, 2, 2, 1, -1, -2, -2, -1]

# MinMoves stores the shortest distance between any two points on the board.
minMoves = [[None for _ in range(51)] for _ in range(51)]
dp = [[[0 for _ in range(1 << 16)] for _ in range(16)] for _ in range(2)]
done = False

def bfs(src_r, src_c):
    dist = [[float('inf')] * 50 for _ in range(50)]
    dist[src_r][src_c] = 0
    q = deque([(src_r, src_c)])
    
    while q:
        r, c = q.popleft()
        for i in range(8):
            nr, nc = r + dr[i], c + dc[i]
            if 0 <= nr < 50 and 0 <= nc < 50:
                if dist[nr][nc] > dist[r][c] + 1:
                    dist[nr][nc] = dist[r][c] + 1
                    q.append((nr, nc))
    return dist

def preprocess():
    global done
    if done:
        return
    done = True
    # Preprocess minimum distances for all board positions
    for i in range(50):
        for j in range(50):
            minMoves[i][j] = bfs(i, j)

class Solution:
    def rec(self, ind, turn, mask, positions):
        # Base case: All positions visited
        if bin(mask).count('1') == len(positions):
            return 0
        # If already computed, return result
        if dp[turn][ind][mask] != -1:
            return dp[turn][ind][mask]
        
        kx, ky = positions[ind]
        
        if turn == 0:
            # Knight's turn: Maximize result
            res = -float('inf')
            for i, (pawnPosX, pawnPosY) in enumerate(positions):
                if not mask & (1 << i):
                    dist = minMoves[kx][ky][pawnPosX][pawnPosY]
                    res = max(res, dist + self.rec(i, 1 - turn, mask | (1 << i), positions))
        else:
            # Pawn's turn: Minimize result
            res = float('inf')
            for i, (pawnPosX, pawnPosY) in enumerate(positions):
                if not mask & (1 << i):
                    dist = minMoves[kx][ky][pawnPosX][pawnPosY]
                    res = min(res, dist + self.rec(i, 1 - turn, mask | (1 << i), positions))
        
        dp[turn][ind][mask] = res
        return res
    
    def maxMoves(self, kx, ky, positions):
        preprocess()
        # Add the knight's initial position to positions list
        vec = [(kx, ky)] + positions
        pawnCount = len(vec)
        
        # Reset dp array
        for i in range(pawnCount):
            for mask in range(1 << pawnCount):
                dp[0][i][mask] = -1
                dp[1][i][mask] = -1
        
        # Start recursive calculation from knight's initial position (index 0)
        return self.rec(0, 0, 1, vec)

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