<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 {
    #define sp <<" "<<
    class DSU{

        vector<int> par;
        

    public:
        int find(int x){
            return par[x] = (x == par[x] ? x : find(par[x]));
        }

        int join(int y1, int y2){
            // component to which y1 belongs must have par find(y1)
            // we make par[find(y1)] point to par of y2.
            return par[find(y1)] = find(y2);
    }
        
        DSU(int n){
        par.resize(n + 1);
        for(int i = 1;i < n;i++){
    par[i] = i;
    }
    }
    };

public:
    int numberOfGoodPaths(vector<int>& vals, vector<vector<int>>& edges) {
        int n = vals.size();
        DSU dsu(n);
        int ans = n;
            // we need v to attach index to values in vals so that after sort, we can track their original index.
        vector<vector<int>> g(vals.size());
        for(auto &edge : edges){
            g[edge[0]].push_back(edge[1]);
            g[edge[1]].push_back(edge[0]);
        }
            map<int,vector<int>> v;
        for(int i = 0;i < vals.size();i++){
            v[vals[i]].push_back(i);
        }


        // we sort v so that we can form graph in increasing order of x
        for(auto &[x,ys] : v){
            // we will introduce new ys into our graph and join them to the existing components
            
            // introduce new ys into our graph
            for(auto y : ys){
                for(auto neighbor : g[y]){


                    // only nodes having value <= x exits in graph right now. So need to check before merging any neighbor
                    if(vals[neighbor] <= x){
                        dsu.join(neighbor,y);
                    }
                }
        }


        // to keep track of how many ys belong to same component
        map<int,int> cnt;
        for(auto y : ys){
            cnt[dsu.find(y)]++;
        }


        for(auto &[par,ycounts] : cnt){
            // if two ys belong to same component there will definitely be a good path between them. Because the component is only made up of value <= x
            // if two ys belong to different component, there can not be a path. Because they have not been joined by any node having value <= x


            // all ys belonging to same parent form a good pair

            // cout << x sp ycounts sp par sp endl;
            ans += (ycounts*(ycounts - 1)) / 2;
        }
        }


        return ans;

    }
};
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