<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: List[int], edges: List[List[int]]) -> int:
        # Most basic Union-Find template without ranking optimization.
        # Please share yours if you have a good Union-Find template with ranking optimization
        UF = {}
        def find(x):
            UF.setdefault(x,x)
            if x != UF[x]:
                UF[x] = find(UF[x])
            return UF[x]
        def union(x,y):
            UF[find(x)] = find(y)
        
        # tree保存每个点的临近点（是个图）
        # val2Nodes根据节点的值把节点分组
        tree = defaultdict(list)
        val2Nodes = defaultdict(set)
        for s,e in edges:
            tree[s].append(e)
            tree[e].append(s)
            val2Nodes[vals[s]].add(s)
            val2Nodes[vals[e]].add(e)
        
        # Include the one node path in the result, because it is not calculated when we do comb(n,r).
        res = len(vals)
        
        # 根据节点值从小到大寻找他们能够创建的图，对于每个v=k的节点
        # 在每个连通图中，可以根据值为k的节点的数目来判断连通路径的数量
        for v in sorted(val2Nodes.keys()):
            for node in val2Nodes[v]:
                # Q：为什么只判断邻居呢？邻居的邻居如果连通起来不是也应该计算进入么
                # A：注意这里面还有一个隐含的数据结构：v比较小的值建立起来的union会保存下来，因此v比较大的值创建的时候
                # 就是在已经建立起来的union上继续建立，这样就不用访问所有的节点了只要访问邻居就可以了
                for nei in tree[node]:
                    # 连同比他小的数字也加入是为了找到哪些node==v而且还通过neighbor连接起来了（因为被值更大node打断是无法连接的）
                    if vals[nei] <= v:
                        # NOTE：union的状态是不需要清空的，因为凡是联通的节点无论怎样都会联通
                        union(node,nei)
            
            # 只有连通的group才能有点到点的连线
            # 数每个group里面有多少个节点（节点数目可以帮助计算有多少组合）
            groupCount = defaultdict(int)
            for node in val2Nodes[v]:
                groupCount[find(node)] += 1
                
            for root in groupCount.keys():
                # The following two lines are doing the same thing
                # If there are n number of nodes that have value==v in the a group. 
                # The number of paths is the number of combinations for selecting 2 elements from n elements (repetitions are not allowed).
                
                # 如果一个group里等于v的值不够两个节点，那么这个地方就是0
                res += comb(groupCount[root],2)
                
                # res += groupCount[root] * (groupCount[root]-1) // 2
        
        return res
1714