<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:
    def checkWays(self, pairs: List[List[int]]) -> int:

        def dfs_partitions(graph):

            partitions = []
            visited = set()
            for k in graph:   
                segment = []
                if k not in visited:
                    visited.add(k)
                    segment.append(k)
                    q = collections.deque()
                    q.append(k)
                    while q:
                        node = q.popleft()
                        for v in graph[node]:
                            if v not in visited:
                                visited.add(v)
                                segment.append(v)
                                q.append(v)

                if segment:
                    if len(segment) == 1:
                        #print (f'Dropping {segment}')
                        continue
                    
                    partition = dict()
                    for k in segment:
                        partition[k] = graph[k]

                    partitions.append(partition)

            return partitions
            
        def process(graph, level):
            #print(f'level {level} {graph}')
            num_nodes = len(graph)
            #print(f'level {level} num_nodes={num_nodes}')
            if not graph:
                return 1

            if num_nodes == 2:
                return 2

            roots = {k for k, v in graph.items() if len(v) == num_nodes - 1}
            if not roots:
                return 0

            #print(f'level {level} roots {roots}')

            results = []

            if len(roots) > 1:
                # Keep it. It is possible that some of the sub-graphs 
                # would return 0, which should override 2
                results.append(2)

            #print(f'level {level} pruning graph of {num_nodes} nodes')
            for r in roots:
                graph.pop(r)
            for k in graph:
                graph[k].difference_update(roots)

            partitions = dfs_partitions(graph)  
            #print(f'level {level} found {len(partitions)} partitions')
           
            for p in partitions:
                result = process(p, level + 1)
                if result == 0:
                    return result
                results.append(result)
            
            #print(f'level {level} results {results}')
            if 2 in results:
                return 2
            return 1 

        graph = defaultdict(set)
        for v, w  in pairs:
            graph[v].add(w)
            graph[w].add(v)

        q = collections.deque()
        q.append(graph)

        return process(graph, 1)


        


        

       


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