<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:
class DSU {
    public : 
    vector<int> par ; 
    vector<int> rank ;
    DSU(int n) {
        for ( int i = 0 ; i<n ; i++) {
            par.push_back(i) ;
            rank.push_back(0) ;
        }

    }

    int findpar( int x) {
        while (par[x]!=x) {
            x = par[x] ;
        }
        return x ;
    }

    void unite(int x, int y) {
int first = findpar(x) ; 
int sec = findpar(y) ; 
if (first==sec) {
    return ;
}
if (rank[first]>rank[sec]) {
    par[sec] = first;
}
else if (rank[first]<rank[sec]) {
    par[first] = sec ; 
}
else {
    par[first] = sec ; 
    rank[sec]++;
}
    }

};
    int numberOfGoodPaths(vector<int>& vals, vector<vector<int>>& edges) {
        int n = vals.size() ; 
        vector<int> adj[n] ; 
        for ( auto &edge : edges) {
            int u = edge[0] ; 
            int v = edge[1] ; 
            if (vals[u]<vals[v]) {
                swap(u,v) ;
            }
            adj[u].push_back(v) ;
        }
        DSU dsu(n) ;
        map<int,vector<int>> nodes_with_val ;
        set<int> dif_values ; 
        int ans = 0 ; 
        for (int i = 0 ; i<vals.size(); i++) {
       dif_values.insert(vals[i]);
       nodes_with_val[vals[i]].push_back(i) ;

        }
        for (auto &value: dif_values) {
            for (auto &node: nodes_with_val[value]) {
                for ( auto &h:adj[node]) {
                    dsu.unite(node,h) ;
                }
            }
            map<int,int> parents ;

            for ( auto &node:nodes_with_val[value]) {
parents[dsu.findpar(node)] ++;
            }
            for ( auto &h: parents) {
                ans = ans + ((h.second)*(h.second-1))/2 ; 
                cout<<h.second<<" "<<value<<endl ;
            }
        }
        ans= ans+n;
        return ans ;
        



    }
};
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