Regular Expression Matching

Problem Link

Approach

Support . (any single character) and * (zero or more of the preceding element).

Let dp[i][j] mean s[0:i] matches p[0:j].

If p[j - 1] is not *:

• Match when characters equal or pattern has ., and dp[i - 1][j - 1] is true.

If p[j - 1] is *:

• Zero occurrences: dp[i][j - 2]
• One or more: if s[i - 1] matches p[j - 2] (or .), then dp[i - 1][j]

Time Complexity: O(m × n)
Space Complexity: O(n)

Code

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        prev = [False] * (len(p) + 1)
        prev[0] = True

        for j in range(2, len(p) + 1):
            if p[j - 1] == "*":
                prev[j] = prev[j - 2]

        for i in range(1, len(s) + 1):
            curr = [False] * (len(p) + 1)
            for j in range(1, len(p) + 1):
                if p[j - 1] == "*":
                    curr[j] = curr[j - 2]
                    if p[j - 2] == s[i - 1] or p[j - 2] == ".":
                        curr[j] = curr[j] or prev[j]
                elif p[j - 1] == s[i - 1] or p[j - 1] == ".":
                    curr[j] = prev[j - 1]
            prev = curr

        return prev[-1]