<p>There is a tree (i.e. a connected, undirected graph with no cycles) consisting of <code>n</code> nodes numbered from <code>0</code> to <code>n - 1</code> and exactly <code>n - 1</code> edges.</p>

<p>You are given a <strong>0-indexed</strong> integer array <code>vals</code> of length <code>n</code> where <code>vals[i]</code> denotes the value of the <code>i<sup>th</sup></code> node. You are also given a 2D integer array <code>edges</code> where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> denotes that there exists an <strong>undirected</strong> edge connecting nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code>.</p>

<p>A <strong>good path</strong> is a simple path that satisfies the following conditions:</p>

<ol>
	<li>The starting node and the ending node have the <strong>same</strong> value.</li>
	<li>All nodes between the starting node and the ending node have values <strong>less than or equal to</strong> the starting node (i.e. the starting node&#39;s value should be the maximum value along the path).</li>
</ol>

<p>Return <em>the number of distinct good paths</em>.</p>

<p>Note that a path and its reverse are counted as the <strong>same</strong> path. For example, <code>0 -&gt; 1</code> is considered to be the same as <code>1 -&gt; 0</code>. A single node is also considered as a valid path.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/08/04/f9caaac15b383af9115c5586779dec5.png" style="width: 400px; height: 333px;" />
<pre>
<strong>Input:</strong> vals = [1,3,2,1,3], edges = [[0,1],[0,2],[2,3],[2,4]]
<strong>Output:</strong> 6
<strong>Explanation:</strong> There are 5 good paths consisting of a single node.
There is 1 additional good path: 1 -&gt; 0 -&gt; 2 -&gt; 4.
(The reverse path 4 -&gt; 2 -&gt; 0 -&gt; 1 is treated as the same as 1 -&gt; 0 -&gt; 2 -&gt; 4.)
Note that 0 -&gt; 2 -&gt; 3 is not a good path because vals[2] &gt; vals[0].
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/08/04/149d3065ec165a71a1b9aec890776ff.png" style="width: 273px; height: 350px;" />
<pre>
<strong>Input:</strong> vals = [1,1,2,2,3], edges = [[0,1],[1,2],[2,3],[2,4]]
<strong>Output:</strong> 7
<strong>Explanation:</strong> There are 5 good paths consisting of a single node.
There are 2 additional good paths: 0 -&gt; 1 and 2 -&gt; 3.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/08/04/31705e22af3d9c0a557459bc7d1b62d.png" style="width: 100px; height: 88px;" />
<pre>
<strong>Input:</strong> vals = [1], edges = []
<strong>Output:</strong> 1
<strong>Explanation:</strong> The tree consists of only one node, so there is one good path.
</pre>

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

<ul>
	<li><code>n == vals.length</code></li>
	<li><code>1 &lt;= n &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>0 &lt;= vals[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>edges.length == n - 1</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li><code>edges</code> represents a valid tree.</li>
</ul>

class Solution:
    def numberOfGoodPaths(self, vals, edges):
        total = n = len(vals)
        dict1 = collections.defaultdict(int)
        freq = [collections.Counter({vals[i]:1}) for i in range(n)] #[Counter({1: 1}), Counter({1: 1}), Counter({2: 1}), Counter({2: 1}), Counter({3: 1})]
        # freq = [defaultdict(int, {vals[i]: 1}) for i in range(n)]
        edges = sorted([max(vals[i],vals[j]),i,j] for i,j in edges) #[[1, 0, 1], [2, 1, 2], [2, 2, 3], [3, 2, 4]]

        def find(x):
            if x not in dict1:
                return x
            else:
                if x != dict1[x]:
                    dict1[x] = find(dict1[x])
                return dict1[x]

        for val,i,j in edges: 
            a, b = find(i), find(j)
            total += freq[a][val]*freq[b][val]
            dict1[b] = a
            freq[a] = collections.Counter({val:freq[a][val] + freq[b][val]})
            # for key in freq[b]:
            #     freq[a][key] += freq[b][key]


        return total
2023