# 44. Wildcard Matching

### Description

 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).

### Constraints

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

### Approach

### Links

* [GeeksforGeeks](https://www.geeksforgeeks.org/wildcard-pattern-matching/)
* [Leetcode](https://leetcode.com/problems/wildcard-matching/)
* [ProgramCreek](https://www.programcreek.com/2014/06/leetcode-wildcard-matching-java/)
* [YouTube](https://youtu.be/3ZDZ-N0EPV0)

### **Examples**

{% tabs %}
{% tab title="Example 1" %}
**Input:** s = "aa", p = "a"

**Output:** false

**Explanation:** "a" does not match the entire string "aa".
{% endtab %}

{% tab title="Example 2" %}
**Input:** s = "aa", p = "\*"

**Output:** true

**Explanation:** '\*' matches any sequence.
{% endtab %}

{% tab title="Example 3" %}
**Input:** s = "cb", p = "?a"

**Output:** false

**Explanation:** '?' matches 'c', but the second letter is 'a', which does not match 'b'.
{% endtab %}

{% tab title="Example 4" %}
**Input:** s = "adceb", p = "*a*b"

**Output:** true

**Explanation:** The first '*' matches the empty sequence, while the second '*' matches the substring "dce".
{% endtab %}

{% tab title="Example 5" %}
**Input:** s = "acdcb", p = "a\*c?b"

**Output:** false
{% endtab %}
{% endtabs %}

### **Solutions**

{% tabs %}
{% tab title="Solution 1" %}

```java
/**
 * Time complexity : O(min(S,P)) for the best case and better than O(SlogP) 
 *    for the average case, where S and P are lengths of the input string 
 *    and the pattern correspondingly.
 * Space complexity : O(1) since it's a constant space solution.
 */

class Solution {
    public boolean isMatch(String s, String p) {
        if(p.isEmpty()) return s.isEmpty();
        int sIndex = 0;
        int pIndex = 0;
        int starIndex = -1;
        int siIndex = -1;
        
        while(sIndex < s.length()) {
            if(pIndex < p.length() && 
               (p.charAt(pIndex) == '?' || s.charAt(sIndex) == p.charAt(pIndex))) {
                sIndex++;
                pIndex++;
            } else if(pIndex < p.length() && p.charAt(pIndex) == '*') {
                siIndex = sIndex;
                starIndex = pIndex;
                pIndex++;
            } else if(starIndex != -1) {
                sIndex = siIndex+1;
                pIndex = starIndex+1;
                siIndex++;
            } else {
                return false;
            }
        }
        
        while(pIndex < p.length() && p.charAt(pIndex) == '*') {
            pIndex++;
        }
        
        return pIndex == p.length();
    }
}
```

{% endtab %}

{% tab title="Solution 2" %}

```java
/**
 * Time complexity : O(SP) where S and P are lengths of the input string 
 *    and the pattern correspondingly.
 * Space complexity : O(SP) to keep the matrix.
 */

class Solution {
    public boolean isMatch(String s, String p) {
        if(p.equals(s) || p.equals("*")) return true;
        if(p.isEmpty() || s.isEmpty()) return false;

        int sLen = s.length(), pLen = p.length();
        boolean[][] dp = new boolean[sLen+1][pLen+1];
        dp[0][0] = true;
        
        for(int i = 1; i < pLen; i++) {
            if(p.charAt(i-1) == '*') {
                dp[0][i] = dp[0][i-1];
            }
        }
        
        for(int i = 1; i <= sLen; i++) {
            for(int j = 0; j <= pLen; j++) {
                if(j > 0 && 
                   (p.charAt(j-1) == '?' || p.charAt(j-1) == s.charAt(i-1))) {
                    dp[i][j] = dp[i-1][j-1];
                } else if(j > 0 && p.charAt(j-1) == '*') {
                    dp[i][j] = dp[i-1][j] || dp[i][j-1];
                } else {
                    dp[i][j] = false;
                }
            }
        }
        
        return dp[sLen][pLen];
    }
}
```

{% endtab %}
{% endtabs %}

### **Follow up**

* Count the Number of matching characters in a pair of strings - [GFG](https://www.geeksforgeeks.org/count-the-number-of-matching-characters-in-a-pair-of-strings)


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://code-snippets.hbamithkumara.com/leetcode/problems/1-100/wildcard-matching.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.