<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:
        # s = "adceb"
        # p = "*a*b"
        table = {}
        def rec(i: int, j: int) -> bool:
            # i is the current index in s
            # j is the current index in p
            if (i, j) in table:
                return table[(i, j)]

            if j == len(p) - 1 and p[j] == "*": # rest of the string can be matched by "*"
                return True
            if i < len(s) and j >= len(p): # pattern is done, string is not matched
                return False
            if i >= len(s) and j >= len(p): # string is done, but pattern is not exhausted
                return True
            # print(i,j )
            if i >= len(s) and p[j] != "*": # string is done, but pattern is not exhausted
                return False
            
            # suppose we are looking at ?
            # the previous one can be *, ?, or nothing (index=0)
            if p[j] == "*":
                sub_ans = rec(i, j+1)
                if sub_ans == True:
                    table[(i, j)] = sub_ans
                    return True
            elif p[j] == "?":
                if j == 0 or p[j-1] != "*": # ? or nothing, can only move forward
                    sub_ans = rec(i+1, j+1)
                    if sub_ans:
                        table[(i, j)] = sub_ans
                        return True
                else: # * in the previous position:
                    # first set of choices are to keep using * for k steps, do not use ? for now
                    for k in range(1, len(s)-i):
                        sub_ans = rec(i+k, j)
                        if sub_ans:
                            table[(i, j)] = sub_ans
                            return True
                    # or use it
                    sub_ans = rec(i+1, j+1)
                    if sub_ans:
                        table[(i, j)] = sub_ans
                        return True
            else:
                if j == 0 or p[j-1] != "*": # can only move forward
                    if s[i] != p[j]:
                        sub_ans = False
                        table[(i, j)] = sub_ans
                        return False
                    else:
                        sub_ans = rec(i+1, j+1)
                        table[(i, j)] = sub_ans
                        return sub_ans
                else: # * in the previous position:
                    for k in range(1, len(s)-i):
                        sub_ans = rec(i+k, j)
                        if sub_ans:
                            table[(i, j)] = sub_ans
                            return True
                    # or use it
                    if s[i] == p[j]:
                        sub_ans = rec(i+1, j+1)
                        if sub_ans:
                            table[(i, j)] = sub_ans
                            return True
                    table[(i, j)] = False
                    return False

        return rec(0, 0)
                    
                
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