<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:
int find(vector<int>&parent,int x){
    if(parent[x]==x)return x;

    return parent[x]=find(parent,parent[x]);
}
void unionn(vector<int>&parent,vector<int>&rank,int &x,int &y){
    int xparent=find(parent,x);
    int yparent=find(parent,y);

    if(xparent!=yparent){
        if(rank[xparent]>rank[yparent]){
            parent[yparent]=xparent;
        }
       else if(rank[xparent]<rank[yparent]){
            parent[xparent]=yparent;
        }
        else {
            parent[xparent]=yparent;
            rank[yparent]++;
        }
    }
}
    int numberOfGoodPaths(vector<int>& vals, vector<vector<int>>& edges) {
        unordered_map<int,list<int>>adjlist;
        vector<int>parent(vals.size());
        for(int i=0;i<parent.size();i++)parent[i]=i;
        vector<int>rank(vals.size(),0);
        for(auto v: edges){
            adjlist[v[0]].push_back(v[1]);
            adjlist[v[1]].push_back(v[0]);
        }
      map<int,list<int>>valuenode;

      for(int i=0;i<vals.size();i++){
        valuenode[vals[i]].push_back(i);
      }

vector<bool>issafe(vals.size(),false);
int result=vals.size();
      for(auto v: valuenode){

        for(auto nbr: v.second){
            for(auto n: adjlist[nbr]){
                if(issafe[n]){
                    unionn(parent,rank,nbr,n);
                }

            }

            issafe[nbr]=true;

        }

            vector<int>parent1;
            
            for(auto nodee: v.second){
                parent1.push_back(find(parent,nodee));

            }
            sort(parent1.begin(),parent1.end());
            for(int i=0;i<parent1.size();i++){
                int count=0;
                int curr=parent1[i];
                while(i<parent1.size() && curr==parent1[i]){
                    count++;
                    i++;
                }
                i--;
               result += count * (count - 1) / 2; 
               

            }

        
      }
      return result;

    }
};
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