<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 DSU:
    def __init__(self):
        self.p = {}
        self.r = {}

    def init(self, n):
        for i in range(n):
            self.p[i] = i
            self.r[i] = 0

    def find(self, u):
        if u not in self.p:
            self.p[u] = u
            self.r[u] = 0
        if self.p[u] == u:
            return u
        self.p[u] = self.find(self.p[u])
        return self.p[u]

    def union(self, u, v):
        pu = self.find(u)
        pv = self.find(v)
        if pu == pv:
            return 
        if self.r[pu] > self.r[pv]:
            self.p[pv] = pu
        else:
            self.p[pu] = pv
            if self.r[pu] == self.r[pv]:
                self.r[pv] += 1

    def connected(self, u, v):
        return self.find(u) == self.find(v)

class Solution:
    def __init__(self):
        self.graph = {}
        self.nodesByVal = {}

    def putVal(self, val, node):
        self.nodesByVal[val] = self.nodesByVal.get(val, [])
        self.nodesByVal[val].append(node)

    def initGraph(self, edges, vals):
        for i in range(len(vals)):
            self.putVal(vals[i], i)
            self.graph[i] = []

        for e in edges:
            self.graph[e[0]].append(e[1])
            self.graph[e[1]].append(e[0])

    def numberOfGoodPaths(self, vals: List[int], edges: List[List[int]]) -> int:
        self.initGraph(edges, vals)

        dsu = DSU()
        dsu.init(len(vals))

        sortedVals = sorted(self.nodesByVal.keys())

        cnt = 0
        for val in sortedVals:
            nodes = self.nodesByVal[val]
            for n in nodes:
                for child in self.graph[n]:
                    if vals[child] <= vals[n]:
                        dsu.union(child, n)
            group = {}
            for n in nodes:
                root = dsu.find(n)
                if root not in group:
                    group[root] = 0
                group[root] += 1
            for size in group.values():
                cnt += size*(size+1)//2

        return cnt
1672