<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 __init__(self):
        self.parent = {}
        self.rank = {}
        self.edges = {}
        
    def makeSet(self, x):
        self.parent[x] = x
        self.rank[x] = 0
    
    def findSet(self, x):
        verts = set()
        while self.parent[x] != x:
            x = self.parent[x]
            verts.add(x)
        for v in verts:
            self.parent[v] = x
        return x
    
    def union(self, x, y):
        a = self.findSet(x)
        b = self.findSet(y)
        if a != b:
            if self.rank[a] > self.rank[b]:
                self.parent[b] = a
            else:
                self.parent[a] = b
                self.rank[b] += 1
    
    def getFact(self, n):
        res = 0
        for i in range(n, 0, -1):
            res += i
        return res
    
    def numberOfGoodPaths(self, vals: list[int], edges: list[list[int]]) -> int:
        from collections import defaultdict
        n = len(vals)
        tab = defaultdict(list)
        self.edges = defaultdict(list)
        ans = n
        if not edges:
            return ans
        for x in edges:
            self.edges[x[0]].append(x[1])
            self.edges[x[1]].append(x[0])
        for i in range(n):
            tab[vals[i]].append(i)
            self.makeSet(i)
        st = sorted(enumerate(vals), key=lambda x:x[1])
        i, val = 0, st[0][1]
        while i < n:
            j = 0
            while st[i + j][1] == val:
                vertex = st[i + j][0]
                for x in self.edges[vertex]:
                    if vals[x] <= val:
                        self.union(vertex, x)
                j += 1
                if i + j >= n:
                    break
            compTab = defaultdict(list)
            for z in tab[val]:
                f = self.findSet(z)
                compTab[f].append(z)
#             print(x, compTab)
            for key in compTab.keys():
                if len(compTab[key]) > 1:
#                     print('key:', key, compTab[key],  self.getFact(len(compTab[key]) - 1))
                    ans += self.getFact(len(compTab[key]) - 1) 
            val = st[min(i + j, n-1)][1]
            i += j
        return ans
        
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