<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 UnionFind {
public:
    vector<int> parent;
    vector<int> rank;

    UnionFind(int n) {
        parent = vector<int>(n);
        rank = vector<int>(n);

        for (int i = 0; i < n; i++) {
            parent[i] = i;
            rank[i] = 1;
        }
    }

    int find(int x) {
        if (x == parent[x]) {
            return x;
        }

        parent[x] = find(parent[x]);

        return parent[x];
    }

    void unionSet(int x, int y) {
        int root_x = find(x);
        int root_y = find(y);

        if (root_x != root_y) {
            if (rank[root_x] > rank[root_y]) {
                parent[root_y] = root_x;
            } else if (rank[root_x] < rank[root_y]) {
                parent[root_x] = root_y;
            } else {
                parent[root_x] = root_y;
                rank[root_y] += 1;
            }
        }
    }
};

class Solution {
public:
    int numberOfGoodPaths(vector<int>& vals, vector<vector<int>>& edges) {
        int n = vals.size();

        UnionFind uf = UnionFind(n);

        // use a vector to store (val, a list of node)
        map<int, vector<int>> val2nodes;
        for (int i = 0; i < n; i++) {
            val2nodes[vals[i]].push_back(i);
        }

        int ans = 0;

        // build a adjacency list
        vector<vector<int>> graph(n);
        for (auto& edge : edges) {
            graph[edge[0]].push_back(edge[1]);
            graph[edge[1]].push_back(edge[0]);
        }

        // use a hash set to store if a node already added into the graph
        unordered_set<int> added;
        for (auto& [val, nodes] : val2nodes) {
            cout << "==== In Building Graph ====" << endl;
            cout << val << endl;
            
            // add these nodes into the graph
            for (auto node : nodes) {
                cout << "node : " << node << endl;

                added.insert(node);

                // iterate over all the neighbor node
                for (auto neigh : graph[node]) {
                    // add the edge in the neigbor already in the graph
                    if (added.count(neigh)) {
                        uf.unionSet(node, neigh);
                    }
                }
            }

            // use a hash map to store (root, # of nodes)
            unordered_map<int, int> root2cnt;
            for (auto node : nodes) {
                int root = uf.find(node); // find the root of node
                root2cnt[root]++;
                cout << "node : " << node << ", root : " << root << endl;
            }

            // count the number of good path
            // any two nodes with same root can form a good path
            for (auto& [root, cnt] : root2cnt) {
                if (cnt != 1) {
                    ans += cnt * (cnt - 1) / 2;
                }
            }

            // a single node can also be the good path
            ans += nodes.size();
        }

        return ans;
    }
};
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