<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 {
public: // assuming maximum number of cities as 20 for feasible bitmasking

vector<int> adj[15];

// Function to build the graph from the edges
void buildGraph(int n, const vector<vector<int>>& edges) {
    for (const auto& edge : edges) {
        int u = edge[0] - 1, v = edge[1] - 1;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
}

// Recursive DFS function to check connectivity
void dfs(int node, int mask, vector<bool>& visited) {
    visited[node] = true;
    for (int neighbor : adj[node]) {
        if ((mask & (1 << neighbor)) && !visited[neighbor]) {
            dfs(neighbor, mask, visited);
        }
    }
}

// Function to check if the subset of nodes represented by 'mask' is connected
bool isConnected(int mask, int n) {
    vector<int> nodes;
    for (int i = 0; i < n; ++i) {
        if (mask & (1 << i)) {
            nodes.push_back(i);
        }
    }
    if (nodes.empty()) return false;

    vector<bool> visited(n, false);
    dfs(nodes[0], mask, visited);

    for (int node : nodes) {
        if (!visited[node]) {
            return false;
        }
    }
    return true;
}

// Recursive DFS function to find the farthest node and maximum distance
void dfsDistance(int node, int mask, vector<bool>& visited, pair<int, int>& result, int dist) {
    visited[node] = true;
    if (dist > result.second) {
        result = {node, dist};
    }
    for (int neighbor : adj[node]) {
        if ((mask & (1 << neighbor)) && !visited[neighbor]) {
            dfsDistance(neighbor, mask, visited, result, dist + 1);
        }
    }
}

// Function to find the maximum distance (diameter) in the subset represented by 'mask'
int maxDistance(int mask, int n) {
    vector<int> nodes;
    for (int i = 0; i < n; ++i) {
        if (mask & (1 << i)) {
            nodes.push_back(i);
        }
    }
    if (nodes.empty()) return 0;

    vector<bool> visited(n, false);
    pair<int, int> farthest = {nodes[0], 0};
    dfsDistance(nodes[0], mask, visited, farthest, 0);  // First DFS to find farthest node

    fill(visited.begin(), visited.end(), false);
    pair<int, int> maxDist = {farthest.first, 0};
    dfsDistance(farthest.first, mask, visited, maxDist, 0);  // Second DFS from the farthest node to find the diameter
    return maxDist.second;
}

    vector<int> countSubgraphsForEachDiameter(int n,vector<vector<int>>& edges) {
        buildGraph(n, edges);

        vector<int> count(n - 1, 0);
        for (int mask = 1; mask < (1 << n); ++mask) {
            if (isConnected(mask, n)) {
                int d = maxDistance(mask, n);
                if (1 <= d && d < n) {
                    count[d - 1]++;
                }
            }
        }
        return count;
    }
};
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