Problem #10: Regular Expression Matching

Intuition & Strategy Breakdown

Use dynamic programming where dp[i][j] indicates if s[:i] matches p[:j]. Handle '*' by considering 0 occurrences of the preceding element or matching current character if pattern matches.

Python Solution

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        dp = [[False] * (len(p) + 1) for _ in range(len(s) + 1)]
        dp[0][0] = True
        for j in range(2, len(p) + 1):
            if p[j - 1] == '*':
                dp[0][j] = dp[0][j - 2]
        for i in range(1, len(s) + 1):
            for j in range(1, len(p) + 1):
                if p[j - 1] == s[i - 1] or p[j - 1] == '.':
                    dp[i][j] = dp[i - 1][j - 1]
                elif p[j - 1] == '*':
                    dp[i][j] = dp[i][j - 2]
                    if p[j - 2] == s[i - 1] or p[j - 2] == '.':
                        dp[i][j] = dp[i][j] or dp[i - 1][j]
        return dp[len(s)][len(p)]

Java Solution

class Solution {
    public boolean isMatch(String s, String p) {
        boolean[][] dp = new boolean[s.length() + 1][p.length() + 1];
        dp[0][0] = true;
        for (int j = 2; j <= p.length(); j++) {
            if (p.charAt(j - 1) == '*') dp[0][j] = dp[0][j - 2];
        }
        for (int i = 1; i <= s.length(); i++) {
            for (int j = 1; j <= p.length(); j++) {
                if (p.charAt(j - 1) == s.charAt(i - 1) || p.charAt(j - 1) == '.') {
                    dp[i][j] = dp[i - 1][j - 1];
                } else if (p.charAt(j - 1) == '*') {
                    dp[i][j] = dp[i][j - 2];
                    if (p.charAt(j - 2) == s.charAt(i - 1) || p.charAt(j - 2) == '.') {
                        dp[i][j] = dp[i][j] || dp[i - 1][j];
                    }
                }
            }
        }
        return dp[s.length()][p.length()];
    }
}