LeetCode 10. Regular Expression Matching

Description

https://leetcode.com/problems/regular-expression-matching/

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 (.)".

Example 4:

Input: s = "aab", p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore, it matches "aab".

Example 5:

Input: s = "mississippi", p = "mis*is*p*."
Output: false

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.

Explanation

Check if the source from the ith position can match with the pattern from the jth position.

Python Solution

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

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

                
        
  • Time Complexity: O(MN).
  • Space Complexity: O(MN).

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