<p>Given an input string (<code>s</code>) and a pattern (<code>p</code>), implement wildcard pattern matching with support for <code>&#39;?&#39;</code> and <code>&#39;*&#39;</code> where:</p>

<ul>
	<li><code>&#39;?&#39;</code> Matches any single character.</li>
	<li><code>&#39;*&#39;</code> Matches any sequence of characters (including the empty sequence).</li>
</ul>

<p>The matching should cover the <strong>entire</strong> input string (not partial).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aa&quot;, p = &quot;a&quot;
<strong>Output:</strong> false
<strong>Explanation:</strong> &quot;a&quot; does not match the entire string &quot;aa&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aa&quot;, p = &quot;*&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong>&nbsp;&#39;*&#39; matches any sequence.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;cb&quot;, p = &quot;?a&quot;
<strong>Output:</strong> false
<strong>Explanation:</strong>&nbsp;&#39;?&#39; matches &#39;c&#39;, but the second letter is &#39;a&#39;, which does not match &#39;b&#39;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= s.length, p.length &lt;= 2000</code></li>
	<li><code>s</code> contains only lowercase English letters.</li>
	<li><code>p</code> contains only lowercase English letters, <code>&#39;?&#39;</code> or <code>&#39;*&#39;</code>.</li>
</ul>

class Solution:
    def isMatch(self, s: str, p: str) -> bool:

        """
        whenever we encounter a "*" that means it can be "" or any pattern
        so we do recursion based on this. 

        1. we treat the start as empty and move the pointer of s to i - 1 and keep the pointer at p at the same place which is j
        2. treat it as some pattern and move the pointer  j to j - 1 

        """

        dp = {}
        
        def dfs(i, j):
            if (i, j) in dp:
                return dp[(i, j)]
            if i < 0 and j < 0:
                return True
            if i < 0:
                return False
            if j < 0:
                for k in range(i + 1):
                    if p[k]!="*":
                        return False
                return True
            
            if p[i] == s[j] or p[i] == "?":
                dp[(i, j)] = dfs(i-1, j-1)
            elif p[i] == "*":
                dp[(i, j)] = dfs(i-1, j) or dfs(i, j - 1)
            else:
                dp[(i, j)] = False
            return dp[(i, j)]

        return dfs(len(p) - 1, len(s) - 1)













        
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