<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)
        distances = [[0 for i in range(n)] for j in range(n)]
        def bfs(node):
            visited = [0] * n
            visited[node] = 1
            q = deque()
            q.append((node, 0))
            while q:
                Node, distance = q.popleft()
                for i in graph[Node]:
                    if not visited[i]:
                        visited[i] = 1
                        distances[node][i] = distance + 1
                        q.append((i, distance + 1))
        for i in range(n): bfs(i)
        INF = float('inf')
        dp = [[INF for i in range(1 << n)] for j in range(n)]
        for i in range(n): dp[i][0] = 0
        for i in range(1, 1 << n):
            for j in range(n):
                if i & (1 << j):
                    x = i ^ (1 << j)
                    Min = INF
                    for k in range(n):
                        Min = min(Min, dp[k][x] + distances[k][j])
                    dp[j][i] = Min
        m = INF
        for i in range(n): m = min(m, dp[i][-1])
        return m

16800