Two types of queries

You are given a string str. It costs one unit of a coin to change a character at an index. Your task is to convert it into a palindrome with minimum coins. Also, you are given $q$ number of queries. The queries are of the following types:

• 1 A B
• 2

After solving the first type of query, you can convert character $A$ into character $B$ without any cost and vice versa. In the second type of query, you are required to answer the minimum cost that is required to convert the given string into a palindrome.

Note: If character $A$ can be converted to character $B$ and character $B$ can be converted to character $C$, then you can convert character $A$ to character $C$.

Example:

Input:
10
vwuhjdbgzp
4
1 a b
2
1 w z
2
Output:
5
4

Approach

Java

public class TwoTypesQueries {

    private static int[] parent;

    public static void main(String args[]) {
        int n = 10;
        String s = "vwuhjdbgzp";
        int[][] differentChars = new int[26][26];
        int count = 0;
        for (int i = 0; i < n / 2; i++) {
            int a = s.charAt(i) - 'a';
            int b = s.charAt(n - 1 - i) - 'a';
            if (a != b) {
                differentChars[a][b]++;
                differentChars[b][a]++;
                count++;
            }
        }
        // initilized graph
        initializeGraph(26);

        boolean isCalculated = true;
        int q = 4;
        StringBuffer sb = new StringBuffer();
        char[][] arr = { { 1, 'a', 'b' }, { 2 }, { 1, 'w', 'z' }, { 2 } };
        for (int k = 0; k < q; k++) {
            int cmd = arr[k][0];
            switch (cmd) {
            case 1:
                int a = arr[k][1] - 'a';
                int b = arr[k][2] - 'a';
                if (a != b)
                    addEdge(a, b);
                isCalculated = false;
                break;
            case 2:
                if (!isCalculated) {
                    for (int i = 0; i < 26; i++) {
                        for (int j = i + 1; j < 26; j++) {
                            if (differentChars[i][j] > 0) {
                                int x = getRoot(i);
                                int y = getRoot(j);
                                if (x == y) {
                                    count -= differentChars[i][j];
                                    differentChars[i][j] = 0;
                                    differentChars[j][i] = 0;
                                }
                            }
                        }
                    }
                    isCalculated = true;
                }
                sb.append(count).append("\n");
                break;
            }
        }
        System.out.print(sb);
    }

    private static void initializeGraph(int n) {
        parent = new int[n];
        for (int i = 0; i < n; i++) {
            parent[i] = i;
        }
    }

    private static void addEdge(int a, int b) {
        int x = getRoot(a);
        int y = getRoot(b);
        if (x != y) {
            if (x < y) {
                parent[y] = x;
            } else {
                parent[x] = y;
            }
        }
    }

    private static int getRoot(int node) {
        int p = parent[node];
        if (node != p && p != parent[p]) {
            parent[node] = getRoot(p);
        }
        return parent[node];
    }

}