<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:
        """Solution 1:"""
        res, n = 0, len(vals)
        adjacency = collections.defaultdict(list)
        for u, v in edges:
            adjacency[u].append(v)
            adjacency[v].append(u)
        
        val_to_nodes = collections.defaultdict(list)
        for i, val in enumerate(vals):
            val_to_nodes[val].append(i)

        ranks = [1] * n
        parents = list(range(n))

        def find(x, parents):
            if parents[x] == x:
                return x
            res = find(parents[x], parents)
            parents[x] = res
            return res

        def union(x, y, ranks, parents):
            if ranks[x] < ranks[y]:
                parents[find(x, parents)] = find(y, parents)
            elif ranks[x] > ranks[y]:
                parents[find(y, parents)] = find(x, parents)
            else:
                parents[find(x, parents)] = find(y, parents)
                ranks[parents[y]] += 1
        
        for val in sorted(val_to_nodes.keys()):
            for v in val_to_nodes[val]:
                for neighbor in adjacency[v]:
                    if val >= vals[neighbor]:
                        union(v, neighbor, ranks, parents)
            
            component_added_dict = collections.defaultdict(int)
            for v in val_to_nodes[val]:
                component_added_dict[find(v, parents)] += 1
            
            for component_added_parent in component_added_dict:
                component_added = component_added_dict[component_added_parent]
                res += int(component_added * (component_added - 1) / 2)
        return res + n



        

    """Solution 2:"""
    # def numberOfGoodPaths(self, vals: List[int], edges: List[List[int]]) -> int:
    #     adjacency = collections.defaultdict(list)
    #     n = len(vals)
    #     for u, v in edges:
    #         adjacency[u].append(v)
    #         adjacency[v].append(u)
        
    #     res = 0
    #     for v in range(n):
    #         total = [0]
    #         seen = set()
    #         seen.add(v)
    #         self.dfs(v, vals, adjacency, vals[v], seen, total)
    #         res += total[0]
    #     # note that res will be even because it double counts the paths, since a path
    #     # between u,v will be double counted as a path beween v,u
          # also note we add n because we want to count the paths of singleton nodes
    #     return int(res / 2) + n

    # def dfs(self, curr, vals, adjacency, initial_val, seen, total):
    #     for v in adjacency[curr]:
    #         if v not in seen:
    #             seen.add(v)
    #             if vals[v] == initial_val:
    #                 total[0] += 1
    #             if vals[v] <= initial_val:
    #                 self.dfs(v, vals, adjacency, initial_val, seen, total)
1574