10-regular-expression-matching¶

Description¶

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.

Solution(Python)¶

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        return self.topdown(s, p)

    # Time Complexity: O((T+P)2^T+P/2)
    # Space Complexity: O((T+P)2^T+P/2)
    def recursion(self, s: str, p: str) -> bool:
        if not p:
            return not s

        first_match = bool(s) and p[0] in {s[0], "."}

        if len(p) >= 2 and p[1] == "*":
            return self.isMatch(s, p[2:]) or first_match and self.isMatch(s[1:], p)
        else:
            return first_match and self.isMatch(s[1:], p[1:])

    # Time Complexity: O((TP)
    # Space Complexity: O((TP)
    def topdown(self, s: str, p: str) -> bool:
        @cache
        def dp(i, j):
            if j == len(p):
                return i == len(s)
            else:
                first_match = i < len(s) and p[j] in {s[i], "."}
                if j + 1 < len(p) and p[j + 1] == "*":
                    return dp(i, j + 2) or first_match and dp(i + 1, j)
                else:
                    return first_match and dp(i + 1, j + 1)

        return dp(0, 0)