<p>You have an undirected, connected graph of <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code>. You are given an array <code>graph</code> where <code>graph[i]</code> is a list of all the nodes connected with node <code>i</code> by an edge.</p>

<p>Return <em>the length of the shortest path that visits every node</em>. You may start and stop at any node, you may revisit nodes multiple times, and you may reuse edges.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/12/shortest1-graph.jpg" style="width: 222px; height: 183px;" />
<pre>
<strong>Input:</strong> graph = [[1,2,3],[0],[0],[0]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> One possible path is [1,0,2,0,3]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/12/shortest2-graph.jpg" style="width: 382px; height: 222px;" />
<pre>
<strong>Input:</strong> graph = [[1],[0,2,4],[1,3,4],[2],[1,2]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> One possible path is [0,1,4,2,3]
</pre>

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

<ul>
	<li><code>n == graph.length</code></li>
	<li><code>1 &lt;= n &lt;= 12</code></li>
	<li><code>0 &lt;= graph[i].length &lt;&nbsp;n</code></li>
	<li><code>graph[i]</code> does not contain <code>i</code>.</li>
	<li>If <code>graph[a]</code> contains <code>b</code>, then <code>graph[b]</code> contains <code>a</code>.</li>
	<li>The input graph is always connected.</li>
</ul>

class Solution {
public:
    int f(int pr, int mask, int n, vector<vector<int>> &adj, vector<vector<int>> &dis, vector<vector<int>> &dp){
        if(mask == ((1 << n) - 1)) return 0;
        if(dp[pr + 1][mask] != -1) return dp[pr + 1][mask];

        int ans = 1e9;
        for(int i = 0; i < n; i++){
            if(mask & (1 << i)) continue;
            int cost = (pr != -1 ? dis[pr][i] : 0);
            ans = min(ans, cost + f(i, mask | (1 << i), n, adj, dis, dp));
        }
        return dp[pr + 1][mask] = ans;
    }

    int shortestPathLength(vector<vector<int>>& adj) {
        int n = adj.size();
        vector<vector<int>> dis(n, vector<int> (n, 1e9));
        for(int i = 0; i < n; i++){
            dis[i][i] = 0;
            for(auto x: adj[i]){
                dis[i][x] = 1;
                dis[x][i] = 1;
            }
        }
        
        for(int k = 0; k < n; k++){
            for(int i = 0; i < n; i++){
                for(int j = 0; j < n; j++){
                    dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
                }
            }
        }

        vector<vector<int>> dp(n + 1, vector<int> ((1 << n), -1));
        int ans = f(-1, 0, n, adj, dis, dp);
        return ans;
    }
};
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