<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>

//*** using the concept of disjoint set union(DSU)
class Solution {
private:
    vector<int>parent;
    vector<int>rank;
    int find_P(int node){
        if(node==parent[node]) return node;
        return parent[node]=find_P(parent[node]); //path compression
    }
    void Union(int u,int v){
         int ultP_u=find_P(u);
         int ultP_v=find_P(v);
         if(ultP_u==ultP_v) return ; //means same component

         else if(rank[ultP_u]>rank[ultP_v]){
              parent[ultP_v]=ultP_u;
         }else if(rank[ultP_u]<rank[ultP_v]){
              parent[ultP_u]=ultP_v;
         }else {
              parent[ultP_v]=ultP_u;
              rank[ultP_u]++;
         }

    }
public:
    int numberOfGoodPaths(vector<int>& vals, vector<vector<int>>& edges) {
        int n=vals.size();
        parent.resize(n);
        rank.resize(n,1);
        for(int i=0;i<n;i++){
            parent[i]=i;
        }
        unordered_map<int,vector<int>>adj;
        for(auto it:edges){
            int u=it[0];
            int v=it[1];
            adj[u].push_back(v);
            adj[v].push_back(u);
        }

        map<int,vector<int>>val_to_nodes;//for starting with smallest vals
        for(int i=0;i<n;i++){
            val_to_nodes[vals[i]].push_back(i);
        }
        int res=n;//nodes it self

        vector<int>is_active(n,0);//non active in the beginning

        for(auto &it:val_to_nodes){
            vector<int>nodes=it.second;
            for(auto u:nodes){
                for(int v:adj[u]){
                    if(is_active[v]){
                        Union(u,v);
                    }
                }
                is_active[u]=true;
            }
            vector<int>tumhare_parents;
            for(int u:nodes){
                tumhare_parents.push_back(find_P(u));
            }
            sort(tumhare_parents.begin(),tumhare_parents.end());
            int s=tumhare_parents.size();
            for(int j=0;j<s;j++){
                int cnt=0;
                int cur_P=tumhare_parents[j];
                while(j<s&&tumhare_parents[j]==cur_P){
                    cnt++;
                    j++;
                }
                j--;
                res+=(cnt*(cnt-1))/2;
            }
        }
        return res;
    }
};





// help(https://www.youtube.com/watch?v=Y02Muq_xoCg)
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