<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>

class Solution {
public:
    int checkWays(vector<vector<int>> const& pairs) {
        int n {}; // n is the id of max node + 1
        for (const auto& p: pairs) {
            n = std::max(n, p[1]);
        }
        vector<vector<int>> ancestors(n);
        for (const auto& p: pairs) {
            auto x = p[0]-1, y = p[1]-1;
            ancestors[x].push_back(y);
            ancestors[y].push_back(x);
        }
        int node_count {};
        for (auto& ancestor: ancestors) {
            node_count += !ancestor.empty();
            std::sort(ancestor.begin(), ancestor.end());
        }
        // a root node should be the ancestor of all other nodes
        // thus its ancestor size = n-1
        priority_queue<pair<int, int>> pq; 
        for (int node {}; node < n; ++node) {
            if (ancestors[node].empty()) continue;
            pq.emplace(static_cast<int>(ancestors[node].size()), node);
        }

        bool more_than_one {false};
        unordered_set<int> met;
        // we can find a node's parent in met
        // note that a node in met may not be an ancestor of current node
        while (!pq.empty()) {
            auto [node_size, node] = pq.top(); 
            pq.pop();

            // find parent for current node
            int parent_size = n + 1;
            int parent {-1};
            for (auto anc: ancestors[node]) {
                // note that among all ancestors of current node
                // the parent of current node has the smallest size
                if (met.count(anc) 
                    && ancestors[anc].size() < parent_size) {
                    parent_size = ancestors[anc].size();
                    parent = anc;
                }
            }

            met.insert(node);
            if (parent == -1) { // current node is root
                if (node_size != node_count-1) return 0;
                continue;
            }
            // now we know node has a parent
            // this parent must have all the ancestors that node has (except for itself)
            for (auto const& pancs = ancestors[parent];
                auto anc: ancestors[node]) {
                if (anc == parent) continue;
                auto it = std::lower_bound(pancs.begin(), pancs.end(), anc);
                if (it == pancs.end() || *it != anc) return 0;
            }
            
            more_than_one |= parent_size == node_size;
        }
        return 1 + more_than_one;
    }
};
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