C++ code with detailed explanation
While reading the question, the first intuition is that we are going to find out the distance between every pair of cities. Here, a subtree is essentially a con
While reading the question, the first intuition is that we are going to find out the distance between every pair of cities. Here, a subtree is essentially a connected graph (that may or may not contain all n cities but the cities that **are** included, are connected). We\'ll check the constraints, and we find that ```n<=15```. This tells us that we can go ahead and apply Floyd Warshall\'s Algorithm (since it works in O(n^3)).\n\n```\nvector<int> countSubgraphsForEachDiameter(int n, vector<vector<int>>& edges) {\n        vector<vector<int>> a(n,vector<int>(n,999));\n        for(auto edge:edges){\n            a[edge[0]-1][edge[1]-1]=1;\n            a[edge[1]-1][edge[0]-1]=1;\n        }\nfor(int k=0; k<n; k++){\n\tfor(int i=0; i<n; i++){\n                for(int j=0; j<n; j++){\n                    a[i][j]=min(a[i][j],a[i][k]+a[k][j]);\n                }\n            }\n        }\n```\nNow that we know all the distances between all cities, we need to generate all the subsets of connected cities and find their maximum distances. Again, n is small enough to carry out O(2^n) subset generation. Similar to what we do while generating subsets, we\'ll check if a bit is set and include corresponding city in our subset. If two cities are included, we\'ll find the distance between them. If the number of edges in this subset is satisfactory, we\'ll add the max distance in this subset to our ans array.\n\n```\nvector<int> ans(n-1,0);\n        for (int s = 0; s < (1 << n); s++) {\n            int k = __builtin_popcount(s);    // The number of cities we\'ll have in our subset\n            int numEdges = 0;\n            int d = 0;\n            for (int i = 0; i < n; i++){ \n                if (s & (1 << i)) {     \n                    for (int j = i + 1; j < n; j++) {   \n                        if (s & (1 << j)) {         \n                            numEdges += (a[i][j] == 1);    \n                            d = max(d, a[i][j]);\n                        }\n                    }\n                }\n            }\n            if (numEdges == k - 1 && d > 0) {\n                ans[d - 1]++;\n            }\n        }\n        \n        return ans;\n\t}\n```
['C']