<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 findPar(int u, vector<int> &par) {
        if(par[u] == u) return u;
        return par[u] = findPar(par[u], par);
    }
    void Union(int u, int v, vector<int> &par) {
        int uPar = findPar(u, par);
        int vPar = findPar(v, par);
        if(uPar != vPar) {
            par[uPar] = vPar;
        }
    }
    int numberOfGoodPaths(vector<int>& vals, vector<vector<int>>& edges) {
        // The trick is to combine nodes in increasing order of the value into DSU
        // If two nodes of same value gets in the same DSU, then the path must exists
        // as all the nodes in the DSU will only contain values lesser than the current value
        int n = vals.size();
        vector<int> par(n);
        for(int i = 0; i < n; i++) par[i] = i;
        vector<int> adj[n];
        for(auto &it : edges) {
            adj[it[0]].push_back(it[1]);
            adj[it[1]].push_back(it[0]);
        }

        int pathCount = 0;
        // we know the val of a particular index
        map<int, vector<int>> mp;// val, index of node having particular val
        for(int i = 0; i < n; i++) mp[vals[i]].push_back(i);
        
        for(auto &it : mp) {// smallest value nodes to largest value nodes
            vector<int> nodesHavingVal = it.second;
            for(auto node : nodesHavingVal) {
                for(auto nbr : adj[node]) {
                    if(vals[nbr] <= vals[node]) {
                        Union(nbr, node, par);
                    }
                }
            }
            // After trying to join all the nbr for each node having particular value
            // we have to find how many joined nodes gets inserted in the same DSU
            unordered_map<int, int> DSUpathCnt;// Ultimate Parent of a DSU, cnt of paths for particular value
            for(auto node : nodesHavingVal) {
                int ultimatePar = findPar(node, par);
                // For each inserted node for a particular value, the amount of good paths
                // get incremented by count of that particular value in that DSU
                pathCount += ++DSUpathCnt[ultimatePar];
            }
        }
        return pathCount;
    }
};
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