<p>There is an <strong>undirected</strong> graph with <code>n</code> nodes numbered from <code>0</code> to <code>n - 1</code> (<strong>inclusive</strong>). You are given a <strong>0-indexed</strong> integer array <code>values</code> where <code>values[i]</code> is the <strong>value </strong>of the <code>i<sup>th</sup></code> node. You are also given a <strong>0-indexed</strong> 2D integer array <code>edges</code>, where each <code>edges[j] = [u<sub>j</sub>, v<sub>j</sub>, time<sub>j</sub>]</code> indicates that there is an undirected edge between the nodes <code>u<sub>j</sub></code> and <code>v<sub>j</sub></code>,<sub> </sub>and it takes <code>time<sub>j</sub></code> seconds to travel between the two nodes. Finally, you are given an integer <code>maxTime</code>.</p>

<p>A <strong>valid</strong> <strong>path</strong> in the graph is any path that starts at node <code>0</code>, ends at node <code>0</code>, and takes <strong>at most</strong> <code>maxTime</code> seconds to complete. You may visit the same node multiple times. The <strong>quality</strong> of a valid path is the <strong>sum</strong> of the values of the <strong>unique nodes</strong> visited in the path (each node&#39;s value is added <strong>at most once</strong> to the sum).</p>

<p>Return <em>the <strong>maximum</strong> quality of a valid path</em>.</p>

<p><strong>Note:</strong> There are <strong>at most four</strong> edges connected to each node.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/10/19/ex1drawio.png" style="width: 269px; height: 170px;" />
<pre>
<strong>Input:</strong> values = [0,32,10,43], edges = [[0,1,10],[1,2,15],[0,3,10]], maxTime = 49
<strong>Output:</strong> 75
<strong>Explanation:</strong>
One possible path is 0 -&gt; 1 -&gt; 0 -&gt; 3 -&gt; 0. The total time taken is 10 + 10 + 10 + 10 = 40 &lt;= 49.
The nodes visited are 0, 1, and 3, giving a maximal path quality of 0 + 32 + 43 = 75.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/10/19/ex2drawio.png" style="width: 269px; height: 170px;" />
<pre>
<strong>Input:</strong> values = [5,10,15,20], edges = [[0,1,10],[1,2,10],[0,3,10]], maxTime = 30
<strong>Output:</strong> 25
<strong>Explanation:</strong>
One possible path is 0 -&gt; 3 -&gt; 0. The total time taken is 10 + 10 = 20 &lt;= 30.
The nodes visited are 0 and 3, giving a maximal path quality of 5 + 20 = 25.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/10/19/ex31drawio.png" style="width: 236px; height: 170px;" />
<pre>
<strong>Input:</strong> values = [1,2,3,4], edges = [[0,1,10],[1,2,11],[2,3,12],[1,3,13]], maxTime = 50
<strong>Output:</strong> 7
<strong>Explanation:</strong>
One possible path is 0 -&gt; 1 -&gt; 3 -&gt; 1 -&gt; 0. The total time taken is 10 + 13 + 13 + 10 = 46 &lt;= 50.
The nodes visited are 0, 1, and 3, giving a maximal path quality of 1 + 2 + 4 = 7.
</pre>

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

<ul>
	<li><code>n == values.length</code></li>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
	<li><code>0 &lt;= values[i] &lt;= 10<sup>8</sup></code></li>
	<li><code>0 &lt;= edges.length &lt;= 2000</code></li>
	<li><code>edges[j].length == 3 </code></li>
	<li><code>0 &lt;= u<sub>j </sub>&lt; v<sub>j</sub> &lt;= n - 1</code></li>
	<li><code>10 &lt;= time<sub>j</sub>, maxTime &lt;= 100</code></li>
	<li>All the pairs <code>[u<sub>j</sub>, v<sub>j</sub>]</code> are <strong>unique</strong>.</li>
	<li>There are <strong>at most four</strong> edges connected to each node.</li>
	<li>The graph may not be connected.</li>
</ul>

class Solution:
    def maximalPathQuality(self, values: List[int], edges: List[List[int]], maxTime: int) -> int:
        
        
        graph = collections.defaultdict(set)
        time = {}
        for a, b, t in edges:
            time[(a,b)] = t
            time[(b,a)] = t
            graph[a].add(b)
            graph[b].add(a)
        ans  = 0
        def dfs(root, t, visit, s):
            if t > maxTime:
                return
            if root == 0:
                nonlocal ans
                ans = max(ans, s)
    
            visit.add(root)
            for v in graph[root]:
                if v not in visit:
                    dfs(v, t+time[(root,v)], set(visit), s+values[v])
                else:
                    dfs(v,t+time[(root,v)], set(visit), s)
        
        
        dfs(0,  0, set(), values[0])
        return ans



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