<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 {
public:
    bool isMatch(string s, string p) {
        if (p.empty()) {
            return s.empty();
        }
        
        string r = "";
        for (char c: p) {
            if (c == '*' &&
                !r.empty() && r.back() == '*') {
                continue;
            }
            
            r.push_back(c);
        }
        
        if (s.empty()) {
            return r.empty() || (r.size() == 1 && r[0] == '*');
        }
        
        int m = r.size() - 1;
        int n = s.size() - 1;
        
        vector<vector<bool>> d(m + 1, vector<bool>(n + 2));
        
        switch(r.back()) {
            case '?': d.back()[n] = true; break;
            case '*': 
                for (int j = 0; j < n + 2; j++) {
                    d.back()[j] = true;
                }
                break;
            default: d.back()[n] = s.back() == r.back(); break;
        }
        
        for (int i = m - 1; i > -1; i--) {
            switch(r[i]) {
                case '?': 
                    for (int j = 0; j < n + 1; j++) {
                        d[i][j] = d[i + 1][j + 1];
                    }
                    break;
                case '*':
                    for (int j = n; j > -1; j--) {
                        d[i][j] = d[i][j + 1] | d[i + 1][j];
                    }
                    break;
                default:
                    for (int j = 0; j < n + 1; j++) {
                        d[i][j] = d[i + 1][j + 1] && r[i] == s[j];
                    }
                    break;
            }
        }
        
        return d[0][0];  
    }
};
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