# 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. The matching should cover 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 <= 30`
• `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.
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 `````` ``````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[-1][-1] = True # Bottom-Up DP for i in range(m, -1, -1): for j in range(n-1, -1, -1): first_match = i < m and p[j] in {s[i], '.'} if j+1 < n and p[j+1] == '*': dp[i][j] = dp[i][j+2] or first_match and dp[i+1][j] else: dp[i][j] = first_match and dp[i+1][j+1] return dp``````