<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>
from collections import deque

class Solution:
    def countSubgraphsForEachDiameter(self, n: int, edges: List[List[int]]) -> List[int]:
        # Step 1: Build the graph
        graph = [[] for _ in range(n + 1)]
        for u, v in edges:
            graph[u].append(v)
            graph[v].append(u)
        
        # Function to check if a subset forms a subtree
        def is_subtree(subset):
            if len(subset) <= 1:
                return False
            visited = set()
            queue = deque([subset[0]])
            while queue:
                node = queue.popleft()
                if node in visited:
                    continue
                visited.add(node)
                for neighbor in graph[node]:
                    if neighbor in subset and neighbor not in visited:
                        queue.append(neighbor)
            return len(visited) == len(subset)
        
        # Function to calculate the diameter of a subtree
        def calculate_diameter(subset):
            if not is_subtree(subset):
                return 0
            n = len(subset)
            if n == 2:
                return 1
            max_dist = 0
            for i in range(n):
                for j in range(i + 1, n):
                    dist = bfs(subset[i], subset[j], subset)
                    max_dist = max(max_dist, dist)
            return max_dist
        
        # BFS to find the distance between two nodes in a subset
        def bfs(start, end, subset):
            visited = set()
            queue = deque([(start, 0)])
            while queue:
                node, dist = queue.popleft()
                if node == end:
                    return dist
                if node in visited:
                    continue
                visited.add(node)
                for neighbor in graph[node]:
                    if neighbor in subset and neighbor not in visited:
                        queue.append((neighbor, dist + 1))
            return 0
        
        # Generate all subsets of cities
        subsets = []
        for i in range(1, n + 1):
            subsets.extend(combinations(range(1, n + 1), i))
        
        # Count subtrees for each diameter
        diameter_count = [0] * (n - 1)
        for subset in subsets:
            subset = list(subset)
            diameter = calculate_diameter(subset)
            if diameter > 0:
                diameter_count[diameter - 1] += 1
        
        return diameter_count

# Example usage:
solution = Solution()
n = 4
edges = [[1,2],[2,3],[2,4]]
print(solution.countSubgraphsForEachDiameter(n, edges))  # Output: [3, 4, 0]
20400