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In this approach, we treat each string as a node in a graph. Two nodes are connected if the corresponding strings are similar. We use a Union-Find data structure to efficiently group the strings into connected components based on similarity.
Time Complexity: O(n^2 * k) where n is the number of strings and k is the average length of the strings.
Space Complexity: O(n), storing the parent array.
1#include <iostream>
2#include <vector>
3#include <string>
4using namespace std;
5
6vector<int> parent;
7
8int find(int x) {
9 if (parent[x] != x) parent[x] = find(parent[x]);
10 return parent[x];
11}
12
13void union_set(int x, int y) {
14 int rootX = find(x);
15 int rootY = find(y);
16 if (rootX != rootY) parent[rootY] = rootX;
17}
18
19bool areSimilar(string a, string b) {
20 int diff = 0;
21 for (int i = 0; i < a.length(); ++i) {
22 if (a[i] != b[i]) {
23 diff++;
24 if (diff > 2) return false;
25 }
26 }
27 return diff == 2 || diff == 0;
28}
29
30int numSimilarGroups(vector<string>& strs) {
31 int n = strs.size();
32 parent.resize(n);
33 for (int i = 0; i < n; ++i) parent[i] = i;
34
35 for (int i = 0; i < n; ++i) {
36 for (int j = i + 1; j < n; ++j) {
37 if (areSimilar(strs[i], strs[j])) {
38 union_set(i, j);
39 }
40 }
41 }
42
43 int count = 0;
44 for (int i = 0; i < n; ++i) {
45 if (find(i) == i) {
46 count++;
47 }
48 }
49 return count;
50}
51
52int main() {
53 vector<string> strs = {"tars", "rats", "arts", "star"};
54 cout << numSimilarGroups(strs) << endl;
55 return 0;
56}
57The C++ solution is similar to the C solution. It uses the Union-Find data structure with path compression to find and union sets in the list of strings. The areSimilar() function checks for similarity based on character mismatches in the strings.
This approach considers each string as a node and treats detecting similarities like exploring components in a graph. We use a depth-first search (DFS) algorithm to explore nodes. If two strings are similar, we explore all strings similar to this pair within a recursive call.
Time Complexity: O(n^2 * k) where n is the number of strings and k is the length of strings.
Space Complexity: O(n) to maintain the visited array.
1
This Java DFS-based solution explores all connected components in the similarity graph. Upon visiting an unvisited node, the dfs function delves into all connected nodes, ensuring complete exploration of the component leading to a count of distinct groups.