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10. Regular Expression Matching

String Dynamic Programming Recursion

Problem - Regular Expression Matching

Hard

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.

Solutions

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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[m][n] = True # 2 empty strings case

        for i in range(m, -1, -1):
            for j in range(n - 1, -1, -1):
                first = i < m and (s[i] == p[j] or p[j] == '.')
                if j + 1 < n and p[j + 1] == '*':
                    dp[i][j] = dp[i][j + 2] or (first and dp[i + 1][j])
                else:
                    dp[i][j] = first and dp[i + 1][j + 1]
        return dp[0][0]

Submission Stats:

  • Runtime: 3 ms (94.59%)
  • Memory: 17.7 MB (82.82%)