<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 {
public:
    // 0 -> mouse
    // 1 -> cat
    int dp[50][50][2];
    int hope[50][50][2];

    int catMouseGame(vector<vector<int>>& graph) {
        memset(dp, -1, sizeof dp);
        int n = graph.size();
        for(int i = 0; i < n; i++){
            for(int j = 1; j < n; j++){
                hope[i][j][0] = graph[i].size();
                for(int &nbr: graph[j]){
                    if(nbr){
                        hope[i][j][1]++;
                    }
                }
            }
        }
        queue<vector<int>> q;
        // mouse aur cat same position pe hain -> cat wins
        for(int i = 1; i < n; i++){
            dp[i][i][0] = 0;
            dp[i][i][1] = 1;
            q.push({i, i, 0});
            q.push({i, i, 1});
        }
        // mouse hole mein hai -> mouse wins
        for(int i = 1; i < n; i++){
            dp[0][i][0] = 1;
            dp[0][i][1] = 0;
            q.push({0, i, 0});
            q.push({0, i, 1});
        }
        while(!q.empty()){
            int i = q.front()[0], j = q.front()[1], p = q.front()[2];
            q.pop();
            if(!p){ // mouse ka chance hai
                for(int &nbr: graph[j]){
                    if(!nbr){ // cat 0 mein nhi ja skta
                        continue;
                    }
                    if(dp[i][nbr][!p] != -1){ // dp already assigned
                        continue;
                    }
                    if(dp[i][j][p]){ // winning state
                        hope[i][nbr][!p]--;
                        if(!hope[i][nbr][!p]){ // hope lost
                            dp[i][nbr][!p] = 0; // losing state for sure
                            q.push({i, nbr, !p});
                        }
                    }
                    else{ // losing state
                        dp[i][nbr][!p] = 1; // winning state for sure
                        q.push({i, nbr, !p});
                    }
                }
            }
            else{ // cat ka chance hai
                for(int &nbr: graph[i]){
                    if(dp[nbr][j][!p] != -1){ // result already assigned
                        continue;
                    }
                    if(dp[i][j][p]){ // winning state
                        hope[nbr][j][!p]--;
                        if(!hope[nbr][j][!p]){ // hope lost
                            dp[nbr][j][!p] = 0; // losing state for sure
                            q.push({nbr, j, !p});
                        }
                    }
                    else{ // losing state
                        dp[nbr][j][!p] = 1; // winning state for sure
                        q.push({nbr, j, !p});
                    }
                }
            }
        }
        if(dp[1][2][0] == 1){ // mouse wins
            return 1;
        }
        else if(dp[1][2][0] == 0){ // cat wins
            return 2;
        }
        else{ // draw
            return 0;
        }
    }
};

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