# Solving Leetcode Interviews in Seconds with AI: Regular Expression Matching


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "10" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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 (.)".    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.  

	# Explanation
	Here's the approach to solve this regular expression matching problem:

*   **Dynamic Programming:** Use dynamic programming to build a table `dp` where `dp[i][j]` stores whether the first `i` characters of the string `s` match the first `j` characters of the pattern `p`.
*   **Base Cases:** Initialize the first row and column of the `dp` table appropriately to handle empty string and pattern cases.
*   **Iteration:** Iterate through the `dp` table, filling each cell based on whether the current characters in `s` and `p` match, and handling the special cases of '.' and '*'.

*   **Runtime Complexity:** O(m\*n) where m is the length of string s and n is the length of the pattern p. **Storage Complexity:** O(m\*n).

	
	# Code
	```python
	def isMatch(s: str, p: str) -> bool:
    """
    Implements regular expression matching with support for '.' and '*'.
    """
    m, n = len(s), len(p)
    dp = [[False] * (n + 1) for _ in range(m + 1)]

    # Base case: empty string and empty pattern match
    dp[0][0] = True

    # Handle patterns like a*, a*b*, a*b*c*
    for j in range(1, n + 1):
        if p[j - 1] == '*':
            dp[0][j] = dp[0][j - 2]

    # Fill the dp table
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if p[j - 1] == '.' or p[j - 1] == s[i - 1]:
                dp[i][j] = dp[i - 1][j - 1]
            elif p[j - 1] == '*':
                dp[i][j] = dp[i][j - 2]  # Zero occurrences of preceding element
                if p[j - 2] == '.' or p[j - 2] == s[i - 1]:
                    dp[i][j] = dp[i][j] or dp[i - 1][j]  # One or more occurrences
            else:
                dp[i][j] = False

    return dp[m][n]
	```