The longest common subsequence of strings s1, s2 is the longest string s such that s is a subsequence of both s1 and s2, i.e. we can get s by starting at some point in si, reading through, perhaps skipping some letters, and then stopping at some point.

Suppose now our alphabet contains a special ‘anchor’ character * which binds ‘more strongly’ than ordinary letters: concretely, when reading through each si to find s we are not allowed to skip any *s (although there may be *s outside the section of si we read). So for example, suppose that s1 = A ∗ BCD∗ and s2 = B ∗ CAD. We would not be allowed to take the subsequence BC (because this requires skipping a * in s2), but we would be allowed to take the subsequence *CD, which is the longest anchored common subsequence and has length 3.

Task: Implement Anchored.java to determine the longest anchored common subsequence (LACS) where the * character acts as an anchor that cannot be skipped. The input consists of two strings, each up to 2,000 characters long, containing letters A-Z and .

Current Status: I've been coding this dynamic programming solution for 2 days, focusing on handling all possible edge cases involving the anchor character. Despite my efforts, my submissions are failing many hidden output cases on Codegrade.

Challenge: The algorithm seems to perform well on standard test cases but struggles with hidden ones. I suspect there might be subtleties around how * is handled that I'm missing, or anything else.

Request: Could anyone help me figure out what might be wrong? Any insight, especially on handling edge cases involving *, would be greatly appreciated. Here is the latest version of my code:

import java.util.Scanner; import java.io.File; import java.io.IOException; public class Anchored { public static void main(String[] args) throws IOException { Scanner scanner = new Scanner(System.in); String filename = scanner.nextLine(); File infile = new File(filename); Scanner input = new Scanner(infile); String s1 = input.nextLine(); String s2 = input.nextLine(); input.close(); scanner.close(); int m = s1.length(); int n = s2.length(); int[][] L = new int[m + 1][n + 1]; for (int i = 1; i <= m; i++) { for (int j = 1; j <= n; j++) { char c1 = s1.charAt(i - 1); char c2 = s2.charAt(j - 1); if (c1 == c2) { L[i][j] = L[i - 1][j - 1] + 1; } else { L[i][j] = Math.max(L[i - 1][j], L[i][j - 1]); } if (c1 == '*' || c2 == '*') { if (c1 == '*' && c2 == '*') { L[i][j] = L[i - 1][j - 1] + 1; } else if (c1 == '*') { L[i][j] = L[i - 1][j]; } else if (c2 == '*') { L[i][j] = L[i][j - 1]; } } } } System.out.println(L[m][n]); } }

I implemented a dynamic programming solution to calculate the longest anchored common subsequence (LACS) between two strings, where the * character acts as an anchor that must be included in the subsequence without skipping. I used a 2D array where each cell L[i][j] represents the length of the longest subsequence between the first i characters of string s1 and the first j characters of string s2. I modified the standard LCS algorithm to handle cases where one of the characters is *, ensuring that sequences containing the * were not improperly extended when mismatches occurred.