<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>

int ans = 0;
vector<map<int,int>>mp(30010);
class Solution {
public:
    class DSU{
        public:
            vector<int>size;
            vector<int>par;
            DSU(int n){
                for(int i=0;i<n;i++){
                    size.push_back(1);
                    par.push_back(i);
                }
            }
            int findUpar(int u){
                if(par[u] == u)return u;
                return par[u] = findUpar(par[u]);
            }
            void merge(int u, int v, int val){
                int pu = findUpar(u);
                int pv = findUpar(v);
                if(pu==pv)return;
                
                if(size[pv]>size[pu]){
                    par[pu] = pv;
                    size[pv]+=size[pu];
                    // for(auto it: mp[pu]){
                    // if(mp[pv].find(it.first)!=mp[pv].end()){
                    //     ans = ans + (mp[pv][it.first]*it.second);
                    //     mp[pv][it.first] += it.second;
                    // }
                    // else{
                    //     mp[pv][it.first] = it.second;
                    // }
                    // }
                    if(mp[pv].find(val)!=mp[pv].end()){
                        ans = ans + (mp[pv][val]*mp[pu][val]);
                        mp[pv][val] += mp[pu][val];
                    }
                    else{
                        mp[pv][val] = mp[pu][val];
                    }
                }
                else{
                    par[pv] = pu;
                    size[pu]+=size[pv];
                    // for(auto it: mp[pv]){
                    //     if(mp[pu].find(it.first)!=mp[pu].end()){
                    //         ans = ans + (mp[pu][it.first]*it.second);
                    //         mp[pu][it.first] += it.second;
                    //     }
                    //     else{
                    //         mp[pu][it.first] = it.second;
                    //     }
                    // }
                    if(mp[pu].find(val)!=mp[pu].end()){
                        ans = ans + (mp[pv][val]*mp[pu][val]);
                        mp[pu][val] += mp[pv][val];
                    }
                    else{
                        mp[pu][val] = mp[pv][val];
                    }
                }
            }
    };
    int numberOfGoodPaths(vector<int>& vals, vector<vector<int>>& edges) {
        int n = vals.size();
        // mp.clear();
        ans = n;
        vector<pair<int,int>>v;
        vector<int>adj[n];
        for(auto it:edges){
            adj[it[0]].push_back(it[1]);
            adj[it[1]].push_back(it[0]);
        }
        
        for(int i=0;i<n;i++){
            mp[i].clear();
            v.push_back({vals[i],i});
            mp[i][vals[i]] = 1;
        }
        // for(int i=0;i<n;i++){
        //     cout<<"Hey "<<i<<endl;
        //     for(auto itr: mp[i]){
        //         cout<<itr.first<<" "<<itr.second<<endl;
        //     }
        // }
        DSU dsu(n);
        set<int>st;
        sort(v.begin(),v.end());
        for(int i=0;i<n;i++){
            int node = v[i].second;
            int val = v[i].first;
            st.insert(node);
            // cout<<node<<endl;
            for(auto it:adj[node]){
                if(st.find(it)!=st.end())
                    dsu.merge(node,it, val);
                // cout<<node<<" "<<it<<" "<<ans<<endl;
                // cout<<mp[3][2]<<" "<<mp[1][2]<<endl;
            }
        }
        return ans;
    }
};
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