Leetcode 44. Wildcard Matching

Seawolf

Given an input string (s) and a pattern (p), implement wildcard pattern matching with support for '?' and '*'.

'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).
The matching should cover the entire input string (not partial).

Note:

s could be empty and contains only lowercase letters a-z.
p could be empty and contains only lowercase letters a-z, and characters like ? or *.
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'.
Example 4:

Input:
s = "adceb"
p = "*a*b"
Output: true
Explanation: The first '*' matches the empty sequence, while the second '*' matches the substring "dce".
Example 5:

Input:
s = "acdcb"
p = "a*c?b"
Output: false

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        m, n = len(s), len(p)
        dp = [[False for _ in range(n + 1)] for _ in range(m + 1)]
        
        for i in range(m + 1):
            for j in range(n + 1):
                if i == 0 and j == 0:
                    dp[i][j] = True
                    continue
                    
                if j == 0:
                    dp[i][j] = False
                    continue
                    
                if p[j - 1] != '*':
                    if i > 0 and (p[j - 1] == s[i - 1] or p[j - 1] == '?'):
                        dp[i][j] |= dp[i - 1][j - 1]
                
                else:
                    dp[i][j] |= dp[i][j - 1]
                    
                    if i > 0:
                        dp[i][j] |= dp[i - 1][j]
                
        return dp[m][n]