<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 label) : label(label), parent(nullptr), count(1) {
    }

    void* Union(DisjointSet* other) {
        DisjointSet* p1 = this->SearhAndCompress();
        DisjointSet* p2 = other->SearhAndCompress();
        if (p1 == p2) {
            return p1;
        }
        if (p1->count >= p2->count) {
            p1->count += p2->count;
            p2->parent = p1;
            return p1;
            
        }
        else {
            p2->count += p1->count;
            p1->parent = p2;
            return p2;
        }
    }

    DisjointSet* SearhAndCompress() {
       
        DisjointSet* p = this;
        while (p->parent) {
            p = p->parent;
        }

        DisjointSet* s = this;
        while (s != p) {
            DisjointSet* next_s = s->parent;
            s->parent = p;
            s = next_s;
        }
        return p;
    }


    DisjointSet* Searh() {
       
        DisjointSet* p = this;
        while (p->parent) {
            p = p->parent;
        }
        return p;
    }

     int label;
private:
    int count;
   
    DisjointSet* parent;
};


class Solution {
public:

    int numberOfGoodPaths(const vector<int>& vals, const vector<vector<int>>& edges) {
        int n = vals.size();
        int m = edges.size();
        vector<vector<int>> graph(n);
        for (const auto &e : edges) {
            int v1 = e[0];
            int v2 = e[1];
            graph[v1].push_back(v2);
            graph[v2].push_back(v1);
        }

        vector<DisjointSet*> g(n);
        for (int i = 0; i < n; ++i) {
            g[i] = new DisjointSet(i);
        }
        vector<pair<int, int>> vals_index(n);
        for (int i = 0; i < n; ++i) {
            vals_index[i] = {vals[i], i};
        }
        sort(vals_index.begin(), vals_index.end());
        int start_i = 0;
        unordered_map<int, int> counter;
        int unioned_v = 0;
        int res = n;
        for (int i = 0; i < n; ++i) {
            int v1 = vals_index[i].second;
            for (int u = 0; u < graph[v1].size(); ++u) {
                int v2 = graph[v1][u];
                if (vals[v1] >= vals[v2]) {
                    g[v1]->Union(g[v2]);
                }
            }

            if (i == n - 1 || vals_index[i].first != vals_index[i + 1].first ) {
                counter.clear();
                for (int u = unioned_v; u <= i; ++u) {
                    int v1 = vals_index[u].second;
                    int label = g[v1]->SearhAndCompress()->label;
                    ++counter[label];
                } 
                for (auto e : counter) {
                    res += e.second * (e.second - 1) / 2;
                }
                
                unioned_v = i + 1;
            }
        }
        return res;

    }
};
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