# LeetCode 44. Wildcard Matching

## Description

https://leetcode.com/problems/wildcard-matching/

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'.
```

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
```

Constraints:

• `0 <= s.length, p.length <= 2000`
• `s` contains only lowercase English letters.
• `p` contains only lowercase English letters, `'?'` or `'*'`.

## Explanation

Dynamic programing dp[i][j] to see if first i characters of s matches the first j characters of p.

## Python Solution

``````class Solution:
def isMatch(self, s: str, p: str) -> bool:
m = len(s)
n = len(p)

dp = [[False for j in range(n + 1)] for i in range(m + 1)]

dp = True

for j in range(1, n + 1):
if p[j - 1] == '*':
dp[j] = dp[j - 1]

for i in range(1, m + 1):
for j in range(1, n + 1):
if p[j - 1] == '*':
dp[i][j] = dp[i - 1][j] or dp[i][j - 1]
else:
if s[i - 1] == p[j - 1] or p[j - 1] == '?':
dp[i][j] = dp[i - 1][j - 1]

return dp[m][n]``````
• Time Complexity: O(MN).
• Space Complexity: O(MN).