<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>>& pairs) {
        unordered_map<int, int> numsIdx;
        int idx = 0;//give each unique number an increasing index
        for (auto& p : pairs) {
            if (numsIdx.count(p[0]) == 0) {
                numsIdx[p[0]] = idx++;
            }
            if (numsIdx.count(p[1]) == 0) {
                numsIdx[p[1]] = idx++;
            }
        }
        const int n = numsIdx.size();
        
        vector<bitset<501>> allsubsets(n, bitset<501>());
        //function for getting index of a number
        auto subsets = [&numsIdx, &allsubsets](int num) -> auto& {
            return allsubsets[numsIdx[num]];
        };
        
        for (auto& p : pairs) {
            int f = p[0], t = p[1];//from to
            subsets(f)[t] = 1;
            subsets(f)[f] = 1;//subset includes the node itself
            
            subsets(t)[f] = 1;
            subsets(t)[t] = 1;
        }
        
        bool hasRoot = false;
        for (auto& s : allsubsets) {
            if (s.count() == n) {
                hasRoot = true;
                break;
            }
        }
        if (!hasRoot) {
            return 0;
        }
        
        int ways = 0;
        for (auto& p : pairs) {
            int a = p[0], b = p[1];//traverse each edge a<->b
            
            bool acbpob = (subsets(a) | subsets(b)) == subsets(a); //a can be parent of b <=> their subsets equal
            bool bcbpoa = (subsets(b) | subsets(a)) == subsets(b); //b can be parent of a <=> their subsets equal
            
            if (!acbpob && !bcbpoa) {
                return 0;//if neither node of this edge can be parent, no way to form this tree
            }
                
            if (acbpob && bcbpoa) {
                ways = max(ways, 2); //if either node can be parent, we have at least 2 ways to form this tree.
            }
            
            ways = max(ways, 1);
        }
        return ways;
    }
};
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