Regular Expression Matching

Problem

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.

Solution

Approach: 2D Dynamic Programming

The key insight is to handle ‘.’ and ‘’ with careful case analysis. The ‘’ can match zero or more of the preceding character.

Implementation

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        m, n = len(s), len(p)

        # dp[i][j] = does s[0:i] match p[0:j]
        dp = [[False] * (n + 1) for _ in range(m + 1)]
        dp[0][0] = True  # Empty matches empty

        # Handle patterns like a*, a*b*, etc. matching empty string
        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] == '*':
                    # Star can match zero occurrences
                    dp[i][j] = dp[i][j-2]

                    # Star matches one or more occurrences
                    if p[j-2] == s[i-1] or p[j-2] == '.':
                        dp[i][j] = dp[i][j] or dp[i-1][j]
                else:
                    # Direct character match or '.' matches any char
                    if p[j-1] == s[i-1] or p[j-1] == '.':
                        dp[i][j] = dp[i-1][j-1]

        return dp[m][n]

Approach 2: Recursion with Memoization

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        memo = {}

        def dp(i, j):
            if j == len(p):
                return i == len(s)

            if (i, j) in memo:
                return memo[(i, j)]

            # Check if current characters match
            first_match = i < len(s) and (p[j] == s[i] or p[j] == '.')

            # Handle '*'
            if j + 1 < len(p) and p[j + 1] == '*':
                # Zero occurrences or one+ occurrences
                result = (dp(i, j + 2) or
                         (first_match and dp(i + 1, j)))
            else:
                # Regular character or '.'
                result = first_match and dp(i + 1, j + 1)

            memo[(i, j)] = result
            return result

        return dp(0, 0)

Complexity Analysis

Time Complexity: O(m * n), where m and n are lengths of s and p.
Space Complexity: O(m * n) for the dp table or memoization.

Key Insights

Two Special Characters:
- ’.’: Matches exactly one of any character
- ‘*’: Matches zero or more of the preceding element
Star Handling: When we see ‘*’, it refers to the character before it:
- Match zero: dp[i][j-2] (skip char and *)
- Match one+: dp[i-1][j] (use the pattern again if current chars match)
Base Cases:
- Empty string matches empty pattern
- Pattern with * can match empty string (a* matches empty)
DP State: dp[i][j] = does s[0:i] match p[0:j]
First Match Check: Before using *, verify current characters match (char match or ‘.’).
Pattern Priority: Always check if next character is ‘*’ before processing current character.
Greedy Not Applicable: Can’t use greedy approach because ‘*’ matching is context-dependent.