Cat and Mouse
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0900-0999/0913.Cat%20and%20Mouse/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0900-0999/0913.Cat%20and%20Mouse/images/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>
Graph; Topological Sort; Memoization; Math; Dynamic Programming; Game Theory
class Solution {
    private int n;
    private int[][] g;
    private int[][][] ans;
    private int[][][] degree;

    private static final int HOLE = 0, MOUSE_START = 1, CAT_START = 2;
    private static final int MOUSE_TURN = 0, CAT_TURN = 1;
    private static final int MOUSE_WIN = 1, CAT_WIN = 2, TIE = 0;

    public int catMouseGame(int[][] graph) {
        n = graph.length;
        g = graph;
        ans = new int[n][n][2];
        degree = new int[n][n][2];
        for (int i = 0; i < n; ++i) {
            for (int j = 1; j < n; ++j) {
                degree[i][j][MOUSE_TURN] = g[i].length;
                degree[i][j][CAT_TURN] = g[j].length;
            }
        }
        for (int i = 0; i < n; ++i) {
            for (int j : g[HOLE]) {
                --degree[i][j][CAT_TURN];
            }
        }
        Deque<int[]> q = new ArrayDeque<>();
        for (int j = 1; j < n; ++j) {
            ans[0][j][MOUSE_TURN] = MOUSE_WIN;
            ans[0][j][CAT_TURN] = MOUSE_WIN;
            q.offer(new int[] {0, j, MOUSE_TURN});
            q.offer(new int[] {0, j, CAT_TURN});
        }
        for (int i = 1; i < n; ++i) {
            ans[i][i][MOUSE_TURN] = CAT_WIN;
            ans[i][i][CAT_TURN] = CAT_WIN;
            q.offer(new int[] {i, i, MOUSE_TURN});
            q.offer(new int[] {i, i, CAT_TURN});
        }
        while (!q.isEmpty()) {
            int[] state = q.poll();
            int t = ans[state[0]][state[1]][state[2]];
            List<int[]> prevStates = getPrevStates(state);
            for (var prevState : prevStates) {
                int pm = prevState[0], pc = prevState[1], pt = prevState[2];
                if (ans[pm][pc][pt] == TIE) {
                    boolean win
                        = (t == MOUSE_WIN && pt == MOUSE_TURN) || (t == CAT_WIN && pt == CAT_TURN);
                    if (win) {
                        ans[pm][pc][pt] = t;
                        q.offer(prevState);
                    } else {
                        if (--degree[pm][pc][pt] == 0) {
                            ans[pm][pc][pt] = t;
                            q.offer(prevState);
                        }
                    }
                }
            }
        }
        return ans[MOUSE_START][CAT_START][MOUSE_TURN];
    }

    private List<int[]> getPrevStates(int[] state) {
        List<int[]> pre = new ArrayList<>();
        int m = state[0], c = state[1], t = state[2];
        int pt = t ^ 1;
        if (pt == CAT_TURN) {
            for (int pc : g[c]) {
                if (pc != HOLE) {
                    pre.add(new int[] {m, pc, pt});
                }
            }
        } else {
            for (int pm : g[m]) {
                pre.add(new int[] {pm, c, pt});
            }
        }
        return pre;
    }
}
