<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 dsu{
    public:
    vector<int>par,siz;
    dsu(int n){
        par.resize(n);
        siz.resize(n,1);
        for(int i=0;i<n;i++) par[i]=i;
    }   
    int findUpar(int u)
    {
        if(par[u]==u) return u;
        return par[u]=findUpar(par[u]);
    }
    void unio(int u,int v)
    {
        int up=findUpar(u),vp=findUpar(v);
        if(up==vp) return;
        if(siz[up]<siz[vp])
        {
            siz[vp]+=siz[up];
            par[up]=vp;
        }
        else{
            siz[up]+=siz[vp];
            par[vp]=up;
        }
    }
};
class Solution {
public:
    int numberOfGoodPaths(vector<int>& vals, vector<vector<int>>& edges) {
        vector<pair<int,int>>vc;
        int n=vals.size();
        vector<int>adj[n];
        for(int i=0;i<edges.size();i++)
        {
            adj[edges[i][0]].push_back(edges[i][1]);
            adj[edges[i][1]].push_back(edges[i][0]);
        }
        for(int i=0;i<vals.size();i++) vc.push_back({vals[i],i});
        sort(vc.begin(),vc.end(),[](const auto&v1,const auto&v2){return v1.first<v2.first;});
        int ans=0;
        dsu obj(n);
        for(int i=0;i<n;i++)
        {
            int l;
            for(l=i;l<n;l++){
                int x=vc[l].second;
                if(vc[l].first!=vc[i].first) break;
                for(int j=0;j<adj[x].size();j++)
                {
                    if(vals[adj[x][j]]<=vc[l].first){
                    obj.unio(vc[l].second,adj[x][j]);
                    }
                }
            }
            map<int,int>mp;
            for(int k=i;k<l;k++) mp[obj.findUpar(vc[k].second)]++;
            for(auto it:mp)
            {
                ans+=((it.second*(it.second-1))/2);
            }
            i=l-1;
        }
        return ans+n;
    }
};
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