<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 Disjoint
{
public:
    vector<int> rank;
    vector<int> parent;
    vector<int> size;
    int components;
    Disjoint(int n)
    {
        rank.resize(n + 1, 0);
        size.resize(n + 1, 1);
        parent.resize(n + 1);
        components = n;
        for (int i = 0; i <= n; i++)
            parent[i] = i;
    }
    // Highest Parent
    int FindUltimateParent(int node)
    {
        if (node == parent[node])
            return node;

        // parent[node]->used for path compression
        return parent[node] = FindUltimateParent(parent[node]);
    }

    void UnionByRank(int u, int v)
    {
        int ultimateParent_u = FindUltimateParent(u);
        int ultimateParent_v = FindUltimateParent(v);

        if (ultimateParent_u == ultimateParent_v)
            return;

        if (rank[ultimateParent_u] < rank[ultimateParent_v])
            parent[ultimateParent_u] = ultimateParent_v;
        else if (rank[ultimateParent_v] < rank[ultimateParent_u])
            parent[ultimateParent_v] = ultimateParent_u;
        else
        {
            parent[ultimateParent_u] = ultimateParent_v;
            rank[ultimateParent_u]++;
        }
    }

    bool UnionBySize(int u, int v)
    {
        int ulp_u = FindUltimateParent(u);
        int ulp_v = FindUltimateParent(v);

        if (ulp_u == ulp_v)
            return false;

        // u->smaller component and v is larger
        // we are joining smaller to the larger one and changing parent of smaller one
        if (size[ulp_u] < size[ulp_v])
        {
            parent[ulp_u] = ulp_v;
            size[ulp_v] += size[ulp_u];
        }
        else
        {
            parent[ulp_v] = ulp_u;
            size[ulp_u] += size[ulp_v];
        }
        components--;
        return true;
    }

    int NumberOfConnectedComponents(int n, vector<vector<int>> &edges)
    {
        for (auto it : edges)
        {
            int u = it[0];
            int v = it[1];

            UnionBySize(u, v);
        }

        int ans = 0;
        for (int i = 1; i <= n; i++)
            if (parent[i] == i)
                ans++;

        return ans;
    }

    bool isConnected()
    {
        return components == 1;
    }
};

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

        for (auto it : edges)
        {
            int u = it[0];
            int v = it[1];

            // we do u->v if vals[u]>=vals[v] because all the connecting path have value less than or equal
            if (vals[u] > vals[v])
                adj[u].push_back(v);
            else
                adj[v].push_back(u);
        }

        // for (auto it : adj)
        // {
        //     cout << it.first << "->";
        //     for (auto i : it.second)
        //         cout << i << " ";
        //     cout << end;
        // }

        int ans = 0;

        map<int, vector<int>> node_to_index;

        for (int i = 0; i < n; i++)
            node_to_index[vals[i]].push_back(i);

        // for (auto it : node_to_index)
        // {
        //     cout << it.first << "->";
        //     for (auto i : it.second)
        //         cout << i << " ";
        //     cout << endl;
        // }

        for (auto values : node_to_index)
        {
            unordered_map<int, int> mp;
            for (auto it : values.second)
            {
                for (auto element : adj[it])
                    graph.UnionBySize(it, element);
            }

            for (auto ele : values.second)
                mp[graph.FindUltimateParent(ele)]++;

            for (auto q : mp)
                ans += (q.second * (q.second - 1) / 2);
        }

        return ans + n;
    }
};

260143