<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:
        n = len(s)
        if not p:
            if not s:
                return True
            else:
                return False
        new_p = p[0]
        for i in range(1, len(p)):
            if p[i-1] == '*' and p[i] == '*':
                continue
            else:
                new_p += p[i]
        if not s:
            if new_p == '*':
                return True
            else:
                return False
        dp = [[False]*len(new_p) for _ in range(n)]
        for i in range(n):
            if new_p[0] == '*':
                dp[i][0] = True
            elif new_p[0] == '?':
                if i == 0:
                    dp[i][0] = True
            else:
                if i == 0 and s[0] == new_p[0]:
                    dp[i][0] = True
        for j in range(1, len(new_p)):
            if new_p[j] == '*' and dp[0][j-1] == True:
                dp[0][j] = True
            elif j == 1 and new_p[j-1] == '*' and (new_p[j] == s[0] or new_p[j] == '?'):
                dp[0][j] = True
        for p in range(1, n):
            for q in range(1, len(new_p)):
                if dp[p-1][q-1] == True:
                    if (s[p] == new_p[q] or new_p[q] == '?' or new_p[q] == '*'):
                        dp[p][q] = True
                    else:
                        dp[p][q] = False
                else:
                    if p == 1:
                        if new_p[q] == '*' and dp[p][q-1] == True:
                            dp[p][q] = True
                        else:
                            dp[p][q] = False
                    else:
                        if new_p[q] != '*':
                            dp[p][q] = False
                        else:
                            if dp[p][q-1] == True:
                                dp[p][q] = True
                            else:
                                for index in range(p-2, -1, -1):
                                    if dp[index][q-1] == True:
                                        dp[p][q] = True
                                        break
        print(dp)
        return dp[n-1][len(new_p)-1]

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