<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:
    # Tookme 50mins
    # I almost figured it out on my own, I even debugged what was wrong
    # but in the end I had to look into to the Editorial to understand what was missing
    # basically my recurrence relation was wrong, I had also to consider a case of
    # exploring an option with the old mask
    def shortestPathLength(self, graph: List[List[int]]) -> int:
        n = len(graph)

        r = math.inf
        memo = {}
        original = tuple([0]*n)
        for i in range(n):
            exploration = self.dfs(i, original, set(), graph, memo)
            r = min(r, exploration-1)

        if r == math.inf:
            return 0

        return r

    def dfs(self, i, bitmask, stack, graph, memo):
        key = (i, bitmask)
        if 0 not in bitmask:
            return 0
        if key in memo:
            return memo[key]
        if key in stack:
            return math.inf
        
        new_bitmask = self.set_bit(bitmask, i)
        best = math.inf
        stack.add(key)
        for e in graph[i]:
            b = self.dfs(e, new_bitmask, stack, graph, memo) + 1
            a = self.dfs(e, bitmask, stack, graph, memo) + 1
            best = min(best, a, b)
        stack.remove(key)
    
        memo[key] = best
        return memo[key]


    def set_bit(self, bitmask, i):
        list_bitmask = list(bitmask)
        list_bitmask[i] = 1
        return tuple(list_bitmask)

# get the nodes visited as a bitmap, this will be a variable of the dp table
# the other variable of the dp will be the i of the node

# we have to keep track of DFS stack? but with what? the bitmask of visited nodes? if it doesn't change then it means we have a loop


# base case: bitmask all 1 return 0
# key is in stack: return math.inf
# key is in memo: return memo[key]
# key = (i, bitmask)

# visit i = bitmask[i] = 1
# stack.add(key)

# for in graph[i]: call recursive function with new bitmask
# we have to have a function to set_bit and return a tuple
# 
29000