<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;
    vector<int>rank;
    int findParent(int node){
        if(parent[node]==node) return node;
        return parent[node]=findParent(parent[node]);
    }
    void Union(int x,int y){
        int xParent=findParent(x);
        int yParent=findParent(y);

        if(xParent==yParent) return;
        if(rank[xParent]>rank[yParent]){
            parent[yParent]=xParent;
        }else if(rank[xParent]<rank[yParent]){
            parent[xParent]=yParent;
        }else{
            parent[xParent]=yParent;
            rank[yParent]+=1;
        }
    }
    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 edge:edges){
            int u=edge[0];
            int v=edge[1];

            adj[u].push_back(v);
            adj[v].push_back(u);
        }
        //map of values to nodes, use map instead of unordered because we need it in sorted order
        map<int,vector<int>>valToNodes;

        for(int i=0;i<n;i++){
            int value=vals[i];
            valToNodes[value].push_back(i);
        }
        int result=n;

        vector<bool>isActive(n,false);
        //now we iterate over our map in order to, starting from value 1, check which nodes
        // are have these values, make them active, check their neighbours, and connect them if they 
        // are active
        for(auto it:valToNodes){
            vector<int>nodes=it.second;
            for(int u:nodes){
                for(auto v:adj[u]){
                    if(isActive[v]){
                        Union(u,v);
                    }
                }
                isActive[u]=true;
            }
            //the nodes are now conected. we need to check for the componets
            vector<int>tumhareParents;
            for(auto u:nodes){
                int parentKaun = findParent(u);
                tumhareParents.push_back(parentKaun);
            }
            sort(tumhareParents.begin(),tumhareParents.end());
            int size=tumhareParents.size();
            for(int j=0;j<size;j++){
                long long count=0;
                int currParent=tumhareParents[j];
                while(j<size && tumhareParents[j]==currParent){
                    count++;
                    j++;
                }
                j--;//doubt
                int formulaAns = count*(count-1)/2;
                result+=formulaAns;
            }
        }
        return result;
    }
};
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