<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>

int dp[2005][2005];
class Solution {
private:
    bool f(string& p, string& s, int i, int j) {
        if (i < 0 && j < 0)
            return true;
        if (i < 0 && j >= 0)
            return false;
        if (j < 0 && i >= 0) {
            for (int k = 0; k <= i; ++k)
                if (p[k] != '*')
                    return false;
            return true;
        }
        if (dp[i][j] != -1)
            return dp[i][j];

        if (p[i] == s[j] || p[i] == '?')
            return dp[i][j] = f(p, s, i - 1, j - 1);
        else if (p[i] == '*')
            return dp[i][j] = f(p, s, i - 1, j) || f(p, s, i, j - 1);
        else
            return false;
    }

public:
    bool isMatch(string& s, string& p) {
        int n = s.size(), m = p.size();
        memset(dp, -1, sizeof(dp));
        return f(p, s, m - 1, n - 1);
    }
};
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