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, 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

    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]