<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>

from collections import deque
from typing import List

# Constants to represent turns
MOUSE = 0
CAT = 1

# Constants to represent game outcomes
MOUSE_WIN = 1
CAT_WIN = 2
DRAW = 0

class Solution:
    def catMouseGame(self, graph: List[List[int]]) -> int:
        n = len(graph)

        # Initialize result map with terminal states (MOUSE_WIN/CAT_WIN)
        results = {}  # map (mouse_pos, cat_pos, turn) -> result
        for i in range(1, n):
            results[0, i, CAT] = results[0, i, MOUSE] = MOUSE_WIN
            results[i, i, CAT] = results[i, i, MOUSE] = CAT_WIN

        # Compute outdegree for all possible states
        degree = {}  # map (mouse_pos, cat_pos, turn) -> number of next states
        for mouse in range(1, n):
            for cat in range(1, n):
                degree[mouse, cat, MOUSE] = len(graph[mouse])
                degree[mouse, cat, CAT] = len(graph[cat]) - int(0 in graph[cat])

        # Initialize the queue with terminal states
        q = deque([state for state in results.keys()])

        # BFS to propagate results
        while q:
            mouse, cat, turn = q.popleft()
            curResult = results[mouse, cat, turn]

            # Determine previous states based on the current turn
            prevStates = []
            if turn == MOUSE:
                prevStates = [(mouse, prevCat, CAT) for prevCat in graph[cat]]
            else:
                prevStates = [(prevMouse, cat, MOUSE) for prevMouse in graph[mouse]]

            # Update previous states based on the current state's result
            for prevState in prevStates:
                if prevState in results:
                    continue
                prevMouse, prevCat, prevTurn = prevState
                if prevCat == 0:  # impossible state
                    continue

                degree[prevState] -= 1
                isMoverWinner = ((curResult == MOUSE_WIN and prevTurn == MOUSE) or
                                 (curResult == CAT_WIN and prevTurn == CAT))
                if isMoverWinner or degree[prevState] == 0:
                    results[prevState] = curResult
                    q.append(prevState)

        return results.get((1, 2, MOUSE), DRAW)
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