LeetCode #10: Regular Expression Matching

Description:

Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'.

'.' Matches any single character.
'*' Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

Note:

• s could be empty and contains only lowercase letters a-z.
• p could be empty and contains only lowercase letters a-z, and characters like . or *.

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

Solution:

• DP (2d) method
• Time complexity: O(N^2)
• Space complexity: O(N^2)

class Solution:
    def isMatch(self, s: 'str', p: 'str') -> 'bool':
        if len(p) == 0:
            if len(s) == 0: return True
            else: return False
        else:
            if len(s) == 0:
                # if all the even locations are *, return True
                if (''.join(set(p[1::2])) == '*') and p[-1]=='*': return True
                else: return False
            else:
                np = len(p); ns = len(s)
                flag = [[False for j in range(ns+1)] for i in range(np+1)]

                # first row: p=''
                flag[0][0] = True   # s='', p=''
                flag[0][1:] = [False for j in range(ns)]    # s='xxx', p=''

                # first column: s=''
                for i in range(1, np+1):
                    if i%2==0 and p[i-1]=='*': 
                        flag[i][0] = flag[i-2][0]

                for j in range(1, ns+1):
                    for i in range(1, np+1):
                        if (p[i-1]=='*') and ( (flag[i][j-1] and (p[i-2] in [s[j-1], '.'])) or flag[i-2][j] ):
                                flag[i][j] = True

                        if (p[i-1] in ['.', s[j-1]]) and flag[i-1][j-1]:
                            flag[i][j] = True

                return flag[-1][-1]