10. Regular Expression Matching

Dynamic Programming

class Solution {
    public boolean isMatch(String s, String p) {
        int m = s.length(), n = p.length();
        boolean[][] dp = new boolean[m + 1][n + 1];
        dp[0][0] = true;

        // Handle patterns like a*, a*b*, a*b*c* that can match an empty string
        for (int j = 2; j <= n; j += 2) {
            if (p.charAt(j - 1) == '*' && dp[0][j - 2]) {
                dp[0][j] = true;
            }
        }

        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                char sc = s.charAt(i - 1);
                char pc = p.charAt(j - 1);

                if (pc == sc || pc == '.') {
                    dp[i][j] = dp[i - 1][j - 1];
                } else if (pc == '*') {
                    char prev = p.charAt(j - 2);
                    dp[i][j] = dp[i][j - 2]; // zero occurrence
                    if (prev == '.' || prev == sc) {
                        dp[i][j] |= dp[i - 1][j]; // one or more
                    }
                }
            }
        }

        return dp[m][n];
    }
}

Remarks:

1. Use DP to reduce calculation

2. Two situations of handling *:

   a. no occurrence so the current two patterns can be skipped (ab* - > a when there’s no b)

   b. one or more times: the current * will absorb the last string char, and check the previous one dp[i][j] -> dp[i-1][j], while the position/length of the current pattern keeps unchanged