<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:
    
//     vector<int> parent ,rank;
//     int findParent(int x)
//     {
        
//         if(x==parent[x])
//             return x;
//         return parent[x]=findParent(parent[x]);
        
        
//     }
//     void Union(int x, int y)
//     {
        
//         int parentX=findParent(x);
//         int parentY=findParent(y);
        
//         if(parentX==parentY)
//             return ;
        
//         if(rank[parentY]<rank[parentX])
//         {
//             parent[parentY]=parentX;
//         }
//         else if(rank[parentY]>rank[parentX])
//         {
//             parent[parentX]=parentY;
//         }
//         else
//         {
//             parent[parentX]=parentY;
//             rank[parentY]++;
            
//         }
        
        
        
        
//     }
    
    

    
//     int numberOfGoodPaths(vector<int>& vals, vector<vector<int>>& edges) {
        
//       int n=vals.size();
        
//     int ne=edges.size();
        
//         if(ne==0)
//             return n;
        
        
        
//         parent.resize(n);
//         rank.resize(n,0);
        
        
//         map<int,vector<int>> mp;
//         vector<int> adj[n];
        
//         vector<int>  vis(n,-1);
//         for(int i=0;i<ne;i++)
//         {
            
//             adj[edges[i][0]].push_back(edges[i][1]);
//             adj[edges[i][1]].push_back(edges[i][0]);
            
//         }
//         for(int i=0;i<n;i++)
//         {
//             mp[vals[i]].push_back(i);
//             parent[i]=i;
        
            
//         }
    
//         int ans=n;
        
        
//         for(auto it: mp)
//         {
            
        
//         vector<int> nodes=it.second;  
//             int size=nodes.size();     
//             unordered_map<int ,int> freq;
//             for(int i=0;i<size;i++ )
//             {
                
//                 for(auto it: adj[nodes[i]])
//                 {
//                           if(vis[nodes[i]]==1)
//                         Union(it, nodes[i]);          
                                
//                 }    
//                 vis[nodes[i]]=1;
//             }
        
//             for(int i=0;i<size;i++)
//             {   
//                 int x=findParent(nodes[i]);
//                 cout<<nodes[i]<<parent[nodes[i]]<<endl;
//                 freq[x]++;    
//             }
            
            
            
//             for(auto it: freq)
//             {             
//                 if(it.second>=2) 
//                     ans+=((it.second)*(it.second-1)/2);
//             }   
            
//         }
        
        
        
        
        
        
        
        
        
        
        
//         return ans;
        
        
        
        
        
        
        
        
        
        
        
        
//     }
// };



class Solution {
public:
    vector<int> parent, rank;

    // Function to find the parent of a node with path compression
    int findParent(int x) {
        if (x != parent[x]) {
            parent[x] = findParent(parent[x]);  // Path compression
        }
        return parent[x];
    }

    // Function to perform union of two sets using rank
    void Union(int x, int y) {
        int parentX = findParent(x);
        int parentY = findParent(y);
        
        if (parentX != parentY) {
            if (rank[parentX] > rank[parentY]) {
                parent[parentY] = parentX;
            } else if (rank[parentX] < rank[parentY]) {
                parent[parentX] = parentY;
            } else {
                parent[parentY] = parentX;
                rank[parentX]++;
            }
        }
    }
    
    int numberOfGoodPaths(vector<int>& vals, vector<vector<int>>& edges) {
        int n = vals.size();
        
        // Initialize the parent and rank arrays
        parent.resize(n);
        rank.resize(n, 0);
        
        // Initialize the parent to be the node itself
        for (int i = 0; i < n; ++i) {
            parent[i] = i;
        }
        
        // Adjacency list representation of the graph
        vector<int> adj[n];
        for (auto& edge : edges) {
            adj[edge[0]].push_back(edge[1]);
            adj[edge[1]].push_back(edge[0]);
        }
        
        // Map to store nodes with the same values
        map<int, vector<int>> valueGroups;
        for (int i = 0; i < n; ++i) {
            valueGroups[vals[i]].push_back(i);
        }
        
        int goodPaths = n; // Each node is a good path by itself
        
        // Process nodes in the order of their values
        for (auto& [value, nodes] : valueGroups) {
            // Union nodes within the same connected component
            for (int node : nodes) {
                for (int neighbor : adj[node]) {
                    if (vals[neighbor] <= value) {
                        Union(node, neighbor);
                    }
                }
            }
            
            // Count the size of each component
            unordered_map<int, int> componentSize;
            for (int node : nodes) {
                int root = findParent(node);
                componentSize[root]++;
            }
            
            // Calculate the number of good paths
            for (auto& [root, size] : componentSize) {
                if (size > 1) {
                    goodPaths += (size * (size - 1)) / 2;
                }
            }
        }
        
        return goodPaths;
    }
};

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