0010: Regular Expression Matching

Problem Statement

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 <= 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.

Code Solution

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        
        @cache
        def rec(sx, px):
            
            # if pattern ends, string must end too
            if px >= len(p):
                return sx == len(s)
            
            # check if prefix `can_match`:
            can_match = sx < len(s) and p[px] in {s[sx], '.'}
            
            # ---------------------------------
            # next_char == '*': 
            #     return `take_it` or `skip_it`
            # ---------------------------------
            if px < len(p) - 1 and p[px + 1] == '*':
                take_it = can_match and rec(sx + 1, px)
                skip_it = rec(sx, px + 2)
                return take_it or skip_it
            # ----------------
            # next_char != '*'
            # ----------------
            else:
                return can_match and rec(sx + 1, px + 1)
            
            return False
        
        return rec(0, 0)