<p>You are given an array <code>pairs</code>, where <code>pairs[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>, and:</p>

<ul>
	<li>There are no duplicates.</li>
	<li><code>x<sub>i</sub> &lt; y<sub>i</sub></code></li>
</ul>

<p>Let <code>ways</code> be the number of rooted trees that satisfy the following conditions:</p>

<ul>
	<li>The tree consists of nodes whose values appeared in <code>pairs</code>.</li>
	<li>A pair <code>[x<sub>i</sub>, y<sub>i</sub>]</code> exists in <code>pairs</code> <strong>if and only if</strong> <code>x<sub>i</sub></code> is an ancestor of <code>y<sub>i</sub></code> or <code>y<sub>i</sub></code> is an ancestor of <code>x<sub>i</sub></code>.</li>
	<li><strong>Note:</strong> the tree does not have to be a binary tree.</li>
</ul>

<p>Two ways are considered to be different if there is at least one node that has different parents in both ways.</p>

<p>Return:</p>

<ul>
	<li><code>0</code> if <code>ways == 0</code></li>
	<li><code>1</code> if <code>ways == 1</code></li>
	<li><code>2</code> if <code>ways &gt; 1</code></li>
</ul>

<p>A <strong>rooted tree</strong> is a tree that has a single root node, and all edges are oriented to be outgoing from the root.</p>

<p>An <strong>ancestor</strong> of a node is any node on the path from the root to that node (excluding the node itself). The root has no ancestors.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img src="https://assets.leetcode.com/uploads/2020/12/03/trees2.png" style="width: 208px; height: 221px;" />
<pre>
<strong>Input:</strong> pairs = [[1,2],[2,3]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> There is exactly one valid rooted tree, which is shown in the above figure.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/12/03/tree.png" style="width: 234px; height: 241px;" />
<pre>
<strong>Input:</strong> pairs = [[1,2],[2,3],[1,3]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are multiple valid rooted trees. Three of them are shown in the above figures.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> pairs = [[1,2],[2,3],[2,4],[1,5]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are no valid rooted trees.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= pairs.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= x<sub>i </sub>&lt; y<sub>i</sub> &lt;= 500</code></li>
	<li>The elements in <code>pairs</code> are unique.</li>
</ul>

using namespace std;

class Solution {
public:
    int checkWays(vector<vector<int>>& pairs) {
        unordered_map<int, unordered_set<int>> graph;

        // Step 1: Build the graph
        for (const auto& pair : pairs) {
            graph[pair[0]].insert(pair[1]);
            graph[pair[1]].insert(pair[0]);
        }

        // Step 2: Sort nodes by the number of neighbors (rank)
        vector<int> nodes;
        for (const auto& entry : graph) {
            nodes.push_back(entry.first);
        }
        sort(nodes.begin(), nodes.end(), [&](int a, int b) {
            return graph[a].size() > graph[b].size();
            });

        // Step 3: Process the nodes
        unordered_set<int> ancestors;
        int ways = 1;

        for (int node : nodes) {
            const auto& neighbors = graph[node];
            int parent = -1;
            int min_rank = INT_MAX;

            // Find the immediate ancestor (parent)
            for (int neighbor : neighbors) {
                if (ancestors.count(neighbor) && graph[neighbor].size() < min_rank) {
                    parent = neighbor;
                    min_rank = graph[neighbor].size();
                }
            }

            // If there's no parent, this node must be the root
            if (parent == -1) {
                if (neighbors.size() != nodes.size() - 1) {
                    return 0;  // Not all nodes are connected
                }
            }
            else {
                // If a parent is found, check if the current node can be a valid child
                for (int neighbor : neighbors) {
                    if (neighbor != parent && !graph[parent].count(neighbor)) {
                        return 0;  // Invalid tree structure
                    }
                }

                if (neighbors.size() == graph[parent].size()) {
                    ways = 2;  // Multiple valid trees possible
                }
            }

            ancestors.insert(node);
        }

        return ways;
    }
};

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