<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 DisjointSet {
public:
    DisjointSet(int n): parent(n, -1), size(n, 1) {}

    void add(int x) {
        if (parent[x] < 0) parent[x] = x;
    }

    int root(int x) {
        if (parent[x] == x) return x;
        return parent[x] = root(parent[x]);
    }

    void merge(int x, int y) {
        add(y);
        int px = root(x), py = root(y);
        if (px == py) return;
        if (size[px] < size[py]) swap(px, py);
        parent[py] = px;
        size[px] += size[py];
    }

private:
    vector<int> parent, size;
};

class Solution {
public:
    template<ranges::view Range>
    int process(Range& r, DisjointSet& sets) {
        unordered_map<int, int> groups;
        int last = -1;
        for (auto [x, y]: r) {
            if (x == last) continue;
            auto p = sets.root(x);
            groups[p]++;
            last = x;
        }

        return ranges::fold_left(groups, 0,
            [](int x, auto p) { return x + p.second * (p.second - 1) / 2; });
    }

    template<ranges::view Range, typename Pred>
    auto take_while(Range& r, Pred&& pred) {
        auto end = r.begin();
        while (end != r.end() && pred(*end)) ++end;
        return ranges::subrange(r.begin(), end);
    }

    int numberOfGoodPaths(const vector<int>& vals, const vector<vector<int>>& raw_edges) {
        int n = vals.size();
        vector<pair<int, int>> edges;
        DisjointSet sets(n);

        for (auto& v: raw_edges) {
            int x = v[0], y = v[1];

            if (vals[x] > vals[y]) edges.emplace_back(x, y);
            else if (vals[x] < vals[y]) edges.emplace_back(y, x);
            else {
                edges.emplace_back(x, y);
                edges.emplace_back(y, x);
            }
        }
        ranges::sort(edges,
            [&](auto e1, auto e2) {
                return pair{vals[e1.first], e1.first} < pair{vals[e2.first], e2.first};
            });

        int ans = n;
        for (auto r = ranges::subrange(edges); !r.empty(); ) {
            int v = vals[r.begin()->first];
            auto r2 = take_while(r, [v, &vals](auto e) { return vals[e.first] == v; });

            for (auto [x, y]: r2) {
                sets.add(x);
                sets.merge(x, y);
            }
            ans += process(r2, sets);

            r.advance(r2.size());
        }

        return ans;
    }
};
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