<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>

#include <ranges>

class DisjointSet {
public:
    DisjointSet(int n)
        : idToParent(n, -1)
        , rootToSize(n, 1)
    {}

    int Find(int id) {
        auto parent = idToParent[id];
        if (parent == -1) {
            return id;
        }

        auto root = Find(parent);
        idToParent[id] = root; // path compression

        return root;
    }

    void Union(int a, int b) {
        auto rootA = Find(a);
        auto rootB = Find(b);

        if (rootA == rootB) {
            return;
        }

        if (rootToSize[rootA] < rootToSize[rootB]) {
            std::swap(rootA, rootB); // union by size
        }

        idToParent[rootB] = rootA;
        rootToSize[rootA] += rootToSize[rootB];
    }

private:
    std::vector<int> idToParent;
    std::vector<int> rootToSize;
};

class Solution {
public:
    // Algorithm:
    // - sort nodes by value (non-decreasing)
    // - process nodes in order
    // - use disjoin set to keep track of group of connected nodes
    // - count nodes with the same value within a group and update good paths count.
    //      n: number of nodes with same value within group
    //      count good paths within group = (n-1) * n / 2 + n 
    int numberOfGoodPaths(vector<int>& nodeToValue, vector<vector<int>>& edges) {
        auto nodeToNeighbors = MakeGraph(nodeToValue.size(), edges);

        auto nodes = std::vector<int>(nodeToValue.size());
        std::iota(nodes.begin(), nodes.end(), 0);
        std::sort(nodes.begin(), nodes.end(), [&](auto nodea, auto nodeb) {
            return nodeToValue[nodea] < nodeToValue[nodeb];
        });

        int countGoodPathsTotal = 0;
        auto sets = DisjointSet(nodeToValue.size());

        auto nodeGroups = nodes | std::views::chunk_by([&](auto nodea, auto nodeb) {
            return nodeToValue[nodea] == nodeToValue[nodeb];
        });

        for (auto group : nodeGroups) {
            for (auto node : group) {
                auto nodeVal = nodeToValue[node];
                for (auto neighbor : nodeToNeighbors[node]) {
                    auto neighborVal = nodeToValue[neighbor];
                    if (nodeVal >= neighborVal) {
                        sets.Union(node, neighbor);
                    }
                }
            }

            auto setToCountSameValue = std::unordered_map<int, int>{};
            for (auto node : group) {
                auto setId = sets.Find(node);
                ++setToCountSameValue[setId];
            }

            for (auto [setId, count] : setToCountSameValue) {
                auto countGoodPaths = (count - 1) * count / 2 + count;
                countGoodPathsTotal += countGoodPaths;
            }
        }

        return countGoodPathsTotal;
    }

private:
    std::vector<std::vector<int>> MakeGraph(int n, std::vector<std::vector<int>>& edges) {
        auto adjList = std::vector<std::vector<int>>(n);

        for (const auto& edge : edges) {
            auto a = edge[0];
            auto b = edge[1];
            adjList[a].push_back(b);
            adjList[b].push_back(a);
        }

        return adjList;
    }
};
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