<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 {
public:
    struct DSU{
        vector<int> parent;
        vector<int> size;

        DSU(int n){
            for(int i = 0; i < n; i++){
                parent.push_back(i);
                size.push_back(1);
            }
        }

        int get_parent(int u){
            return parent[u] = u == parent[u] ? u : get_parent(parent[u]);
        }

        void merge(int u, int v){
            u = get_parent(u);
            v = get_parent(v);
            if(u == v)
                return ;
            if(size[u] < size[v])
                swap(u, v);
            size[u] += size[v];
            parent[v] = u;
        }        
    };

    int add(int v, vector<int>& vals, vector<vector<int>>& x, vector<vector<int>>& edges, DSU& g){     
        if(x[v].empty())
            return 0;   
        while(!edges.empty() && max(vals[edges.back()[0]], vals[edges.back()[1]]) <= v){
            g.merge(edges.back()[0], edges.back()[1]);
            edges.pop_back();
        }

        int res = 0;
        vector<int> t;
        for(int i : x[v])
            t.push_back(g.get_parent(i));
        sort(t.begin(), t.end());

        vector<int> vv;
        int c = 1;
        for(int i = 1; i < t.size(); i++){
            if(t[i] != t[i - 1]){
                vv.push_back(c);
                c = 1;
            }
            else
                c++;
        }
        vv.push_back(c);

        for(auto i : vv)
            res += (i * (i + 1)) >> 1;
        
        return res;
    }

    int numberOfGoodPaths(vector<int>& vals, vector<vector<int>>& edges) {
        for(int i = 0; i < edges.size(); i++)
            edges[i].push_back(max(vals[edges[i][0]], vals[edges[i][1]]));

        sort(edges.begin(), edges.end(), [&](vector<int>& a, vector<int>& b){
	    	return a[2] < b[2];
		});
        reverse(edges.begin(), edges.end());
        
        int mx = *max_element(vals.begin(), vals.end());
        vector<vector<int>> x(mx + 1);
        for(int i = 0; i < vals.size(); i++)
            x[vals[i]].push_back(i);

        DSU g(vals.size() + 5);
        int ans = 0;
        for(int i = 0; i <= mx; i++)
            ans += add(i, vals, x, edges, g);

        return ans;
    }
};
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