Similar String Groups

Two strings X and Y are similar if we can swap two letters (in different positions) of X, so that it equals Y. Also two strings X and Y are similar if they are equal.

For example, "tars" and "rats" are similar (swapping at positions 0 and 2), and "rats" and "arts" are similar, but "star" is not similar to "tars", "rats", or "arts".

Together, these form two connected groups by similarity: {"tars", "rats", "arts"} and {"star"}. Notice that "tars" and "arts" are in the same group even though they are not similar. Formally, each group is such that a word is in the group if and only if it is similar to at least one other word in the group.

We are given a list A of strings. Every string in A is an anagram of every other string in A. How many groups are there?

Example 1:

Input: A = ["tars","rats","arts","star"]
Output: 2

Constraints:

• 1 <= A.length <= 2000

• 1 <= A[i].length <= 1000

• A.length * A[i].length <= 20000

• All words in A consist of lowercase letters only.

• All words in A have the same length and are anagrams of each other.

• The judging time limit has been increased for this question.

class Solution {
    public int numSimilarGroups(String[] A) {
        int[] parent = new int[A.length];
        int[] rank = new int[A.length];
        for (int i = 0; i < A.length; i++) {
            rank[i] = 1;
            parent[i] = i;
        }
        int count = A.length;
        for (int i = 0; i < A.length; i++) {
            for (int j = i + 1; j < A.length; j++) {
                int diff = 0;
                for (int c = 0; c < A[i].length(); c++)
                    if (A[i].charAt(c) != A[j].charAt(c))
                        diff++;
                if (diff == 0 || diff == 2) {
                    int p1 = findParent(parent, i);
                    int p2 = findParent(parent, j);
                    if (p1 != p2) {
                        if (rank[p1] > rank[p2])
                            parent[p2] = p1;
                        else if (rank[p2] > rank[p1])
                            parent[p1] = p2;
                        else {
                            parent[p1] = p2;
                            rank[p1]++;
                        }
                        count--;
                    }
                }
            }
        }
        return count;
    }

    public int findParent(int[] parent, int node) {
        if (node != parent[node])
            parent[node] = findParent(parent, parent[node]);
        return parent[node];
    }
}