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Tested Python solution for LeetCode 10 with 17 pytest cases. Generate a practice environment with lcpy.
lcpy gen -n 10 # by problem number
lcpy gen -s regular_expression_matching # by problem name
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.Input: s = "aa", p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".
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".
Input: s = "ab", p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".
s contains only lowercase English letters.p contains only lowercase English letters, '.', and '*'.'*', there will be a previous valid character to match.class Solution:
# Time: O(m * n)
# Space: O(m * n)
def is_match(self, s: str, p: str) -> bool:
m, n = len(s), len(p)
dp = [[False] * (n + 1) for _ in range(m + 1)]
dp[0][0] = True
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] == s[i - 1] or p[j - 1] == ".":
dp[i][j] = dp[i - 1][j - 1]
elif p[j - 1] == "*":
dp[i][j] = dp[i][j - 2]
if p[j - 2] == s[i - 1] or p[j - 2] == ".":
dp[i][j] = dp[i][j] or dp[i - 1][j]
return dp[m][n]
| Time | Space |
|---|---|
| O(m * n) | O(m * n) |