<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 defaultdict
from typing import List

class Solution:
    def countSubgraphsForEachDiameter(self, n: int, edges: List[List[int]]) -> List[int]:
        # Initialize the result array to store the count of subtrees for each diameter
        res = [0] * (n - 1)
        
        # Graph representation using adjacency list
        g = defaultdict(list)
        
        # Bitmask representation of graph connections
        gs = [0] * n
        
        # Dynamic programming array to store the maximum distance for each state
        dp = [0] * (1 << n)
        
        # Build the graph and initialize the dp array for direct edges
        for a, b in edges:
            a -= 1
            b -= 1
            g[a].append(b)
            g[b].append(a)
            gs[a] |= (1 << b)
            gs[b] |= (1 << a)
            dp[(1 << a) | (1 << b)] = 1
        
        # Distance matrix to store distances between all pairs of nodes
        dis = [[0] * n for _ in range(n)]
        
        # Depth-first search to calculate distances from each node to every other node
        def dfs(root, cur, depth, parent):
            dis[root][cur] = depth
            for nxt in g[cur]:
                if nxt == parent:
                    continue
                dfs(root, nxt, depth + 1, cur)
        
        # Calculate distances for all node pairs
        for i in range(n):
            dfs(i, i, 0, -1)
        
        # Iterate over all possible subsets of nodes
        for state in range(1 << n):
            if dp[state] > 0:
                # Increment the count for the current maximum distance
                res[dp[state] - 1] += 1
                for i in range(n):
                    # Check if adding node i to the current state forms a valid subtree
                    if dp[state | (1 << i)] == 0 and (state & gs[i]):
                        # Update the dp array with the new maximum distance
                        dp[state | (1 << i)] = max(dp[state], max(dis[i][j] for j in range(n) if state & (1 << j)))
        
        return res
16800