<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:
    /*
        Intuition:
        1. if a node is root, its degree must be n-1
        2. if node A is B's parent, then degreeA >= degreeB
        3. if degreeA = degreeB, A can be B's parent and vice versa
        4. node with higher degree need to be processed earlier
        5. if there is a tie, => based on 3, the processing order doesn't matter
    */

    int checkWays(vector<vector<int>>& pairs) {
        unordered_map<int, unordered_set<int>> adjMap;

        for (auto &p : pairs) {
            adjMap[p[0]].insert(p[1]);
            adjMap[p[1]].insert(p[0]);
        }

        bool multipleParent = false;
        int nodes = adjMap.size();
        unordered_set<int> visited;
        priority_queue<vector<int>> pq; // max heap, item = { degree, node }

        for (auto &[node, neighbors] : adjMap) {
            pq.push({ (int)neighbors.size(), node });
        }

        while (!pq.empty()) {
            int curDegree = pq.top()[0];
            int curNode = pq.top()[1];

            pq.pop();
            
            int parent = -1;
            int parentDegree = INT_MAX;

            // find possible ancestor with minimum degree => parent
            // possible ancestor must be visited, since it has higher degree
            for (int next : adjMap[curNode]) {
                if (visited.contains(next) && adjMap[next].size() >= curDegree && adjMap[next].size() < parentDegree) {
                    parent = next;
                    parentDegree = adjMap[next].size();
                }
            }

            visited.insert(curNode);

            if (parent == -1 && curDegree != nodes - 1) {
                return 0;
            } else if (parent > 0) {
                // check whethe parent's neighor contains all cur node's neighbor
                for (auto next : adjMap[curNode]) {
                    if (next == parent) continue;
                    if (!adjMap[parent].contains(next)) return 0;
                }

                if (parentDegree == curDegree) {
                    multipleParent = true;
                }
            }
        }

        return multipleParent ? 2 : 1;
    }
};
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