Java characters alignment algorithm

Using a variation of Levenshtein distance that does exactly what you want:

Output

-MKNLASREVNIYVNGKLV
QM---ASREVNIYVNGKL-

Code:

public class Main {
    public static void main(String[] args) {
        String[] aligned = align("MKNLASREVNIYVNGKLV", "QMASREVNIYVNGKL");
        System.out.println(aligned[0]);
        System.out.println(aligned[1]);
    }

    public static String[] align(String a, String b) {
        int[][] T = new int[a.length() + 1][b.length() + 1];

        for (int i = 0; i <= a.length(); i++)
            T[i][0] = i;

        for (int i = 0; i <= b.length(); i++)
            T[0][i] = i;

        for (int i = 1; i <= a.length(); i++) {
            for (int j = 1; j <= b.length(); j++) {
                if (a.charAt(i - 1) == b.charAt(j - 1))
                    T[i][j] = T[i - 1][j - 1];
                else
                    T[i][j] = Math.min(T[i - 1][j], T[i][j - 1]) + 1;
            }
        }

        StringBuilder aa = new StringBuilder(), bb = new StringBuilder();

        for (int i = a.length(), j = b.length(); i > 0 || j > 0; ) {
            if (i > 0 && T[i][j] == T[i - 1][j] + 1) {
                aa.append(a.charAt(--i));
                bb.append("-");
            } else if (j > 0 && T[i][j] == T[i][j - 1] + 1) {
                bb.append(b.charAt(--j));
                aa.append("-");
            } else if (i > 0 && j > 0 && T[i][j] == T[i - 1][j - 1]) {
                aa.append(a.charAt(--i));
                bb.append(b.charAt(--j));
            }
        }

        return new String[]{aa.reverse().toString(), bb.reverse().toString()};
    }
}