<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:
    def shortestPathLength(self, graph: List[List[int]]) -> int:
        N  = len(graph)

        INF = 10**10
        dist = [[INF] * N for _ in range(N)]

        for i, edges in enumerate(graph):
            dist[i][i] = 0
            for edge in edges:
                dist[i][edge] = 1
        
        for k in range(N):
            for i in range(N):
                for j in range(N):
                    dist[i][j]  = min(dist[i][j], dist[i][k] + dist[k][j])

        # node: 1 -- 12 -> O(N)
        # visited: O(2^N)
        @cache
        def visit(node, seen):
            if seen == (1 << N) - 1:
                return 0
            
            best = INF
            for i in range(N):
                if seen & (1 << i) == 0:
                    best = min(best, visit(i, (1 << i) | seen) + dist[node][i])
            
            return best
        
        best = INF
        for i in range(N):
            best = min(best, visit(i, 1 << i))
        
        return best




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