<p>There are <code>n</code> cities numbered from <code>1</code> to <code>n</code>. You are given an array <code>edges</code> of size <code>n-1</code>, where <code>edges[i] = [u<sub>i</sub>, v<sub>i</sub>]</code> represents a bidirectional edge between cities <code>u<sub>i</sub></code> and <code>v<sub>i</sub></code>. There exists a unique path between each pair of cities. In other words, the cities form a <strong>tree</strong>.</p>

<p>A <strong>subtree</strong> is a subset of cities where every city is reachable from every other city in the subset, where the path between each pair passes through only the cities from the subset. Two subtrees are different if there is a city in one subtree that is not present in the other.</p>

<p>For each <code>d</code> from <code>1</code> to <code>n-1</code>, find the number of subtrees in which the <strong>maximum distance</strong> between any two cities in the subtree is equal to <code>d</code>.</p>

<p>Return <em>an array of size</em> <code>n-1</code> <em>where the </em><code>d<sup>th</sup></code><em> </em><em>element <strong>(1-indexed)</strong> is the number of subtrees in which the <strong>maximum distance</strong> between any two cities is equal to </em><code>d</code>.</p>

<p><strong>Notice</strong>&nbsp;that&nbsp;the <strong>distance</strong> between the two cities is the number of edges in the path between them.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2020/09/21/p1.png" style="width: 161px; height: 181px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 4, edges = [[1,2],[2,3],[2,4]]
<strong>Output:</strong> [3,4,0]
<strong>Explanation:
</strong>The subtrees with subsets {1,2}, {2,3} and {2,4} have a max distance of 1.
The subtrees with subsets {1,2,3}, {1,2,4}, {2,3,4} and {1,2,3,4} have a max distance of 2.
No subtree has two nodes where the max distance between them is 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 2, edges = [[1,2]]
<strong>Output:</strong> [1]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 3, edges = [[1,2],[2,3]]
<strong>Output:</strong> [2,1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 15</code></li>
	<li><code>edges.length == n-1</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>1 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n</code></li>
	<li>All pairs <code>(u<sub>i</sub>, v<sub>i</sub>)</code> are distinct.</li>
</ul>
class Solution:
    def countSubgraphsForEachDiameter(self, n: int, edges: List[List[int]]) -> List[int]:
        
        def fn(edges): 
            """ """
            graph = {} # graph as adjacency list 
            for u, v in edges:
                graph.setdefault(u-1, []).append(v-1) # 0-indexed 
                graph.setdefault(v-1, []).append(u-1)
            group = [None]*n
            dist = [0]*n
            
            def dfs(x, d=0): 
                """ """
                seen.add(x) # mark visited 
                for xx in graph.get(x, []): 
                    dist[x] = max(dist[x], d)
                    if group[xx] is None: group[xx] = group[x]
                    if xx not in seen: dfs(xx, d+1)
                
            for i in range(n): 
                seen = set()
                if group[i] is None: group[i] = i
                dfs(i)
            return group, dist 
        
        ans = {} # answer 
        for r in range(1, len(edges)+1):
            for x in combinations(edges, r): 
                temp = {}
                d = {}
                seen, dist = fn(x)
                for i in range(n): 
                    temp.setdefault(seen[i], []).append(i)
                    if seen[i] not in d: d[seen[i]] = dist[i]
                    else: d[seen[i]] = max(d[seen[i]], dist[i])
                for k, v in temp.items(): 
                    if len(v) > 1: 
                        ans.setdefault(d[k], set()).add(tuple(sorted(v)))
        return [len(ans.get(x, set())) for x in range(1, n)]
19100