# Wildcard Matching

Wildcard Matching

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
```

## Analysis

**Dynamic Programming**

Similar to Regular Expression Matching, the difference is how the `'*'`, `'?'` are interpreted, also it uses `'?'` instead of `'.'`

Yet the fundamentals are the same, just the initialization and state transfer function needs to be modified.

**Multi-Pointers Approach** <https://leetcode.com/problems/wildcard-matching/discuss/17810/Linear-runtime-and-constant-space-solution>

Although it might be easier to think of DP solution if you've met Regular Expression Matching problem before, there's a more straightforward approach that's more intuitive.

## Solution

DP - O(mn) space, O(mn) time - (33ms 76% AC)

```java
class Solution {
    public boolean isMatch(String s, String p) {
        char[] str = s.toCharArray();
        char[] pattern = p.toCharArray();
        int m = str.length;
        int n = pattern.length;
        boolean dp[][] = new boolean[m + 1][n + 1];
        dp[0][0] = true;
        for (int i = 1; i <= n; i++) {
            if (pattern[i - 1] == '*') {
                dp[0][i] = dp[0][i - 1];
            }
        }

        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (str[i - 1] == pattern[j - 1] || pattern[j - 1] == '?') {
                    dp[i][j] = dp[i - 1][j - 1];
                }
                if (pattern[j - 1] == '*') {
                    dp[i][j] = dp[i - 1][j] || dp[i][j - 1] || dp[i - 1][j - 1];
                }
            }
        }
        return dp[m][n];
    }
}
```

Multi-Pointers - O(1) space, O(mn) time

```
boolean comparison(String str, String pattern) {
        int s = 0, p = 0, match = 0, starIdx = -1;            
        while (s < str.length()){
            // advancing both pointers
            if (p < pattern.length()  && (pattern.charAt(p) == '?' || str.charAt(s) == pattern.charAt(p))){
                s++;
                p++;
            }
            // * found, only advancing pattern pointer
            else if (p < pattern.length() && pattern.charAt(p) == '*'){
                starIdx = p;
                match = s;
                p++;
            }
           // last pattern pointer was *, advancing string pointer
            else if (starIdx != -1){
                p = starIdx + 1;
                match++;
                s = match;
            }
           //current pattern pointer is not star, last patter pointer was not *
          //characters do not match
            else return false;
        }

        //check for remaining characters in pattern
        while (p < pattern.length() && pattern.charAt(p) == '*')
            p++;

        return p == pattern.length();
}
```


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