<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 {
    private:
    vector<int> parent;
    vector<int> rank;
    public:
    DSU(int n) {
        parent.resize(n);
        rank.resize(n);
        for(int i = 0; i<n; i++) {
            parent[i] = i;
            rank[i] = 1;
        }
    }

    int getParent(int node) {
        if(parent[node] == node) {
            return node;
        }
        return parent[node] = getParent(parent[node]);
    }

    bool isConnected(int node1, int node2) {
        return getParent(node1) == getParent(node2);
    }

    void connect(int node1, int node2) {
        int parent1 = getParent(node1);
        int parent2 = getParent(node2);
        if(parent1 == parent2) {
            return;
        }
        if(rank[parent1] >= rank[parent2]) {
            parent[parent2] = parent1;
            rank[parent1] += rank[parent2];
        }
        else {
            parent[parent1] = parent2;
            rank[parent2] += rank[parent1];
        }
    }
};

class Solution {
public:

    int numberOfGoodPaths(vector<int>& vals, vector<vector<int>>& edges) {
        int n = vals.size();
        priority_queue<pair<int,int>> edgeOrder;
        map<int, vector<int>> valMap;
        for(int i = 0; i<edges.size(); i++) {
            int u = edges[i][0];
            int v = edges[i][1];
            edgeOrder.push({-max(vals[u], vals[v]), i});
        }
        for(int i = 0; i<n; i++) {
            if(valMap.find(vals[i]) == valMap.end()) {
                valMap[vals[i]] = vector<int>();
            }
            valMap[vals[i]].push_back(i);
        }
        if(valMap.size() == 1) {
            return n * (n + 1)/2;
        }
        DSU dsu(n);
        int ans = n;
        map<int,vector<int>>::iterator it = valMap.begin();
        while(it != valMap.end()) {
            int value = (*it).first;
            vector<int> items = (*it).second;
            while(!edgeOrder.empty()) {
                pair<int,int> top = edgeOrder.top();
                if(abs(top.first) <= value) {
                    edgeOrder.pop();
                    int index = top.second;
                    dsu.connect(edges[index][0], edges[index][1]);
                }
                else {
                    break;
                }
            }
            if(items.size() > 0) {
                for(int i = 0; i<items.size(); i++) {
                    for(int j = i + 1; j<items.size(); j++) {
                        if(dsu.isConnected(items[i], items[j])) {
                            ++ans;
                            // cout<<items[i]<<":"<<items[j]<<" ";
                        }
                    }
                }
            }
            ++it;
        }
        return ans;
    }
};
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