# Solving Leetcode Interviews in Seconds with AI: Wildcard Matching


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "44" 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 wildcard pattern matching with support for '?' and '*' where:  '?' Matches any single character. '*' Matches any sequence of characters (including the empty sequence).  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 = "*" Output: true Explanation: '*' matches any sequence.  Example 3:  Input: s = "cb", p = "?a" Output: false Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'.    Constraints:  0 <= s.length, p.length <= 2000 s contains only lowercase English letters. p contains only lowercase English letters, '?' or '*'.  

	# Explanation
	Here's a breakdown of the approach, complexity, and the Python code:

*   **High-Level Approach:**

    *   Dynamic Programming: Use a 2D boolean table `dp` where `dp[i][j]` indicates 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. `dp[0][0]` is True (empty string matches empty pattern).  Handle cases where the pattern starts with `*` (it can match the empty string).
    *   Iteration: Iterate through the `dp` table, filling it based on the current characters in `s` and `p`. If `p[j-1]` is `?`, the result depends on the match of substrings `s[0:i-1]` and `p[0:j-1]`. If `p[j-1]` is `*`, we consider two cases: either `*` matches an empty sequence or it matches one or more characters in `s`. Otherwise, compare current chars of `s` and `p`.

*   **Complexity:**
    *   Runtime: O(m\*n) where n is the length of `s` and m is the length of `p`.
    *   Space: O(m\*n)

	
	# Code
	```python
	def isMatch(s: str, p: str) -> bool:
    """
    Implements wildcard pattern matching with support for '?' and '*'.
    """
    n = len(s)
    m = len(p)

    dp = [[False] * (m + 1) for _ in range(n + 1)]

    # Empty string matches empty pattern
    dp[0][0] = True

    # Handle cases where the pattern starts with '*'
    for j in range(1, m + 1):
        if p[j - 1] == '*':
            dp[0][j] = dp[0][j - 1]

    # Iterate through the dp table
    for i in range(1, n + 1):
        for j in range(1, m + 1):
            if p[j - 1] == '?':
                dp[i][j] = dp[i - 1][j - 1]
            elif p[j - 1] == '*':
                dp[i][j] = dp[i][j - 1] or dp[i - 1][j]
            else:
                dp[i][j] = dp[i - 1][j - 1] and (s[i - 1] == p[j - 1])

    return dp[n][m]
	```