<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 {
    int n,reqFreq;
    bool f;
    vector<vector<int>> adj;
    vector<int> freq1;
    void getEdges(vector<vector<int>>& pairs) {
        unordered_set<int> hashSet;
        for( vector<int>& pair : pairs ) {
            if( freq1.size() < pair[1]+1 ) {
                freq1.resize(pair[1]+1);
            }
            hashSet.insert(pair[0]);
            hashSet.insert(pair[1]);
            freq1[pair[0]]++;
            freq1[pair[1]]++;
        }
        reqFreq = hashSet.size()-1;
        n = freq1.size();
        f = 0;
        adj = vector<vector<int>>(n);
        for( vector<int>& pair : pairs ) {
            if( freq1[pair[0]] >= freq1[pair[1]] ) {
                if( freq1[pair[0]] == freq1[pair[1]] ) {
                    f = 1;
                }
                adj[pair[0]].push_back(pair[1]);
            } else {
                adj[pair[1]].push_back(pair[0]);
            }
        }
    }
public:
    int checkWays(vector<vector<int>>& pairs) {
        getEdges(pairs);
        vector<int> freq2(n);
        for( vector<int>& childArr : adj ) {
            for( int& child : childArr ) {
                freq2[child]++;
            }
        }
        queue<int> que;
        for( int i = 0; i< n; i++ ) {
            if( freq2[i] == 0 && freq1[i] == reqFreq  ) {
                que.push(i);
            }
        }
        while( que.size() != 0 ) {
            int size = que.size();
            int countFreq = 0;
            reqFreq = 0;
            while( size-- ) {
                int parent = que.front();
                reqFreq += freq1[parent];
                que.pop();
                for( int& child : adj[parent] ) {
                    freq2[child]--;
                    freq1[child]--;
                    if( freq2[child] == 0 ) {
                        countFreq += freq1[child]+1;
                        que.push(child);
                    }
                }
            }
            if( countFreq != reqFreq ) {
                return 0;
            }
        }
        for(int& num : freq2 ) {
            if( num != 0 ) {
                return 0;
            }
        }
        return f == 1? 2 : 1;
    }
};
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