# 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 = True

for j in range(1, n + 1):
if p[j - 1] == '*':
dp[j] = dp[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).