<p>A game on an <strong>undirected</strong> graph is played by two players, Mouse and Cat, who alternate turns.</p>

<p>The graph is given as follows: <code>graph[a]</code> is a list of all nodes <code>b</code> such that <code>ab</code> is an edge of the graph.</p>

<p>The mouse starts at node <code>1</code> and goes first, the cat starts at node <code>2</code> and goes second, and there is a hole at node <code>0</code>.</p>

<p>During each player&#39;s turn, they <strong>must</strong> travel along one&nbsp;edge of the graph that meets where they are.&nbsp; For example, if the Mouse is at node 1, it <strong>must</strong> travel to any node in <code>graph[1]</code>.</p>

<p>Additionally, it is not allowed for the Cat to travel to the Hole (node <code>0</code>).</p>

<p>Then, the game can end in three&nbsp;ways:</p>

<ul>
	<li>If ever the Cat occupies the same node as the Mouse, the Cat wins.</li>
	<li>If ever the Mouse reaches the Hole, the Mouse wins.</li>
	<li>If ever a position is repeated (i.e., the players are in the same position as a previous turn, and&nbsp;it is the same player&#39;s turn to move), the game is a draw.</li>
</ul>

<p>Given a <code>graph</code>, and assuming both players play optimally, return</p>

<ul>
	<li><code>1</code>&nbsp;if the mouse wins the game,</li>
	<li><code>2</code>&nbsp;if the cat wins the game, or</li>
	<li><code>0</code>&nbsp;if the game is a draw.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/17/cat1.jpg" style="width: 300px; height: 300px;" />
<pre>
<strong>Input:</strong> graph = [[2,5],[3],[0,4,5],[1,4,5],[2,3],[0,2,3]]
<strong>Output:</strong> 0
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/17/cat2.jpg" style="width: 200px; height: 200px;" />
<pre>
<strong>Input:</strong> graph = [[1,3],[0],[3],[0,2]]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= graph.length &lt;= 50</code></li>
	<li><code>1&nbsp;&lt;= graph[i].length &lt; graph.length</code></li>
	<li><code>0 &lt;= graph[i][j] &lt; graph.length</code></li>
	<li><code>graph[i][j] != i</code></li>
	<li><code>graph[i]</code> is unique.</li>
	<li>The mouse and the cat can always move.&nbsp;</li>
</ul>



class Solution:
    from collections import deque
    def catMouseGame(self, graph):
        n = len(graph)
        DRAW, MOUSE_WIN, CAT_WIN = 0, 1, 2

        # Initialize the 3D array: dp[mouse][cat][turn]
        dp = [[[DRAW] * 2 for _ in range(n)] for _ in range(n)]

        # Base cases
        for i in range(n):
            dp[0][i][0] = dp[0][i][1] = MOUSE_WIN  # Mouse at hole
            if i > 0:
                dp[i][i][0] = dp[i][i][1] = CAT_WIN  # Cat catches mouse

        queue = deque()

        # Enqueue base cases
        for i in range(n):
            for t in range(2):
                if dp[0][i][t] == MOUSE_WIN:
                    queue.append((0, i, t, MOUSE_WIN))
                if dp[i][i][t] == CAT_WIN and i > 0:
                    queue.append((i, i, t, CAT_WIN))

        while queue:
            mouse, cat, turn, result = queue.popleft()
            prev_turn = 1 - turn  # 0: Mouse's turn, 1: Cat's turn

            # Get predecessors
            if prev_turn == 0:
                # Mouse's predecessors
                for prev_mouse in graph[mouse]:
                    if dp[prev_mouse][cat][prev_turn] == DRAW:
                        if result == MOUSE_WIN:
                            dp[prev_mouse][cat][prev_turn] = MOUSE_WIN
                            queue.append((prev_mouse, cat, prev_turn, MOUSE_WIN))
                        elif all(dp[next_mouse][cat][turn] == CAT_WIN for next_mouse in graph[prev_mouse]):
                            dp[prev_mouse][cat][prev_turn] = CAT_WIN
                            queue.append((prev_mouse, cat, prev_turn, CAT_WIN))
            else:
                # Cat's predecessors
                for prev_cat in graph[cat]:
                    if prev_cat == 0:
                        continue  # Cat cannot move to hole
                    if dp[mouse][prev_cat][prev_turn] == DRAW:
                        if result == CAT_WIN:
                            dp[mouse][prev_cat][prev_turn] = CAT_WIN
                            queue.append((mouse, prev_cat, prev_turn, CAT_WIN))
                        elif all(dp[mouse][next_cat][turn] == MOUSE_WIN for next_cat in graph[prev_cat] if next_cat != 0):
                            dp[mouse][prev_cat][prev_turn] = MOUSE_WIN
                            queue.append((mouse, prev_cat, prev_turn, MOUSE_WIN))

        return dp[1][2][0]


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