<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 binary(self, num):
        return bin(num)[2:].zfill(8)
    
    def connected(self, n, subset, graph):
        size = subset.bit_count()
        firstSetBit = int(math.log2(subset & -subset))
        queue = [firstSetBit+1]
        visited = set()

        while queue:
            node = queue.pop(0)
            visited.add(node)
            for child in graph[node]:
                if (subset & (1<<(child-1))) and (child not in visited):
                    queue.append(child)

        return len(visited) == size

    def dfs(self, start, subset, dis, visited, graph):
        if (start in visited):
            return dis
        visited.add(start)
        maxDis = dis
        for v in graph[start]:
            if (subset & (1<<(v-1))) and (v not in visited):
                maxDis = max(maxDis, self.dfs(v, subset, dis+1, visited, graph))
        return maxDis

    def maxDistance(self, subset, n, graph):
        ans = 0
        for start in range(1, n+1):
            # node in set
            if subset & (1<<(start-1)):
                visited = set()
                maxDis = self.dfs(start, subset, 0, visited, graph)
                print()
                if maxDis == 0:
                    return -1
                ans = max(ans, maxDis)
        return ans

    def countSubgraphsForEachDiameter(self, n: int, edges: List[List[int]]) -> List[int]:
        ans = []
        maxDis = {}
        graph = self.buildGraph(edges)

        for subset in range(1, 1<<n):
            if not self.connected(n, subset, graph):
                continue
            maxDis[subset] = self.maxDistance(subset, n, graph)

        for d in range(1, n):
            ans.append(0)
            for subset in range(1, 1<<n):
                if (subset in maxDis) and (maxDis[subset] == d):
                    ans[-1] += 1
        return ans
    
    def buildGraph(self, edges: List[List[int]]):
        # init graph
        graph = {}
        for e in edges:
            c1 = e[0]
            c2 = e[1]
            if c1 not in graph:
                graph[c1] = []
            if c2 not in graph:
                graph[c2] = []
            graph[c1].append(c2)
            graph[c2].append(c1)
        return graph
17600