<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 __init__(self):
        self.l = [[-1, 2], [1,2], [2, 1], [2, -1], [1, -2], [-1, -2], [-2, -1], [-2, 1]]
        self.positions = None
        self.memo = defaultdict(int)
        self.graph = defaultdict(int)

    def recur(self, player: int, bit: int, prev: tuple) -> int:
        if (player, bit, prev) in self.memo: return self.memo[(player, bit, prev)]
        if player == 0:
            most = -sys.maxsize+1
            for idx in range(len(self.positions)):
                if not ((1 << idx) & bit):
                    x1, y1 = self.positions[idx]
                    dist = self.graph[( prev, (x1, y1))]
                    most = max(most, dist + self.recur(1, bit ^ (1 << idx), (x1, y1)))
            if most == -sys.maxsize+1: return 0
            self.memo[(player, bit, prev)] = most
        if player == 1:
            least = sys.maxsize
            for idx in range(len(self.positions)):
                if not ((1 << idx) & bit):
                    x1, y1 = self.positions[idx]
                    dist = self.graph[( prev, (x1, y1))]
                    least = min(least, dist + self.recur(0, bit ^ (1 << idx), (x1, y1)))
            if least == sys.maxsize: return 0
            self.memo[(player, bit, prev)] = least
        return self.memo[(player, bit, prev)]

    def bfs(self, pos1: tuple, pos2: tuple) -> int:
        (x1, y1), (x2, y2) = pos1, pos2
        deq = deque([(x1, y1, 0)])
        seen = set()
        seen.add(pos1)
        while deq:
            x,y,dist = deq.popleft()
            if x == x2 and y == y2:
                return dist
            for a,b in self.l:
                if 0 <= x + a < 50 and 0 <= b + y < 50:
                    if (x+a, b+y) not in seen:
                        seen.add((x+a, b+y))
                        deq.append((x+a, b+y, dist+1))
        return -1
    
    def createGraph(self, kx: int, ky: int) -> None:
        for x,y in self.positions:
            dist = self.bfs((kx,ky), (x,y))
            self.graph[((kx,ky), (x,y))] = dist
            self.graph[((x,y), (kx,ky))] = dist
        for i in range(len(self.positions)):
            for j in range(len(self.positions)):
                if i != j:
                    (x1,y1), (x2,y2) = self.positions[i], self.positions[j]
                    dist = self.bfs((x1,y1), (x2,y2))
                    self.graph[((x1,y1), (x2,y2))] = dist
                    self.graph[(x2,y2), (x1,y1)] = dist

    def maxMoves(self, kx: int, ky: int, positions: List[List[int]]) -> int:
        self.positions = positions
        self.createGraph(kx, ky)
        # print(self.graph)
        ans = self.recur(0, 0, (kx, ky))
        return ans
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