> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Regular Expression Matching Python Solution

> Tested Python solution for LeetCode 10 with 17 pytest cases. Generate a practice environment with lcpy.

LeetCode 10, Hard. Topics: String, Dynamic Programming, Recursion. [View on LeetCode](https://leetcode.com/problems/regular-expression-matching/description/).

Generate this problem as a practice environment: tested reference solution, 17 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 10   # by problem number
lcpy gen -s regular_expression_matching   # by problem name
```

## Problem

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.

Return a boolean indicating whether the matching covers the entire input string (not partial).

### Examples

```
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 (.)".
```

### Constraints

* 1 \<= s.length \<= 20
* 1 \<= p.length \<= 20
* `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.

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/regular_expression_matching/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/regular_expression_matching/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
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]
```

## Complexity

| Time      | Space     |
| --------- | --------- |
| O(m \* n) | O(m \* n) |

## Tags

[NeetCode 150](/catalog/neetcode-150), [NeetCode 250](/catalog/neetcode-250), [NeetCode All](/catalog/neetcode).