regular-expression-matching.py

"""

10. 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 <= 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.

"""

# V0

# V1
# IDEA : RECURSION
# https://leetcode.com/problems/regular-expression-matching/solution/
class Solution(object):
    def isMatch(self, text, pattern):
        if not pattern:
            return not text

        first_match = bool(text) and pattern[0] in {text[0], '.'}

        if len(pattern) >= 2 and pattern[1] == '*':
            return (self.isMatch(text, pattern[2:]) or
                    first_match and self.isMatch(text[1:], pattern))
        else:
            return first_match and self.isMatch(text[1:], pattern[1:])

# V1'
# IDEA : DP
# https://leetcode.com/problems/regular-expression-matching/solution/
class Solution(object):
    def isMatch(self, text, pattern):
        memo = {}
        def dp(i, j):
            if (i, j) not in memo:
                if j == len(pattern):
                    ans = i == len(text)
                else:
                    first_match = i < len(text) and pattern[j] in {text[i], '.'}
                    if j+1 < len(pattern) and pattern[j+1] == '*':
                        ans = dp(i, j+2) or first_match and dp(i+1, j)
                    else:
                        ans = first_match and dp(i+1, j+1)

                memo[i, j] = ans
            return memo[i, j]

        return dp(0, 0)

# V2