10. Regular Expression Matching

Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where:

• '.' Matches any single character.​​​​
• '*' Matches zero or more of the preceding element.

Return a boolean indicating whether the matching covers the entire input string (not partial).

Example 1:

Input: s = "aa", p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input: s = "aa", p = "a*"
Output: true
Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input: s = "ab", p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Constraints:

• 1 <= s.length <= 20
• 1 <= p.length <= 20
• s contains only lowercase English letters.
• p contains only lowercase English letters, '.', and '*'.
• It is guaranteed for each appearance of the character '*', there will be a previous valid character to match.

Approach 01: Dynamic Programming (Bottom-Up)

class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.length(), n = p.length();
        vector> dp(m + 1, vector(n + 1, false));
        dp[0][0] = true;
        for (int j = 2; j <= n; j++) {
            if (p[j-1] == '*') dp[0][j] = dp[0][j-2];
        }
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (p[j-1] == '.' || p[j-1] == s[i-1]) {
                    dp[i][j] = dp[i-1][j-1];
                } else if (p[j-1] == '*') {
                    dp[i][j] = dp[i][j-2];
                    if (p[j-2] == '.' || p[j-2] == s[i-1]) {
                        dp[i][j] = dp[i][j] || dp[i-1][j];
                    }
                }
            }
        }
        return dp[m][n];
    }
};

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

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;
        for (int j = 2; j <= n; j++) {
            if (p.charAt(j-1) == '*') dp[0][j] = dp[0][j-2];
        }
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (p.charAt(j-1) == '.' || p.charAt(j-1) == s.charAt(i-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) == '.' || p.charAt(j-2) == s.charAt(i-1)) {
                        dp[i][j] = dp[i][j] || dp[i-1][j];
                    }
                }
            }
        }
        return dp[m][n];
    }
}