<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 {
private:
    string compress(string s) {
        int n = s.length();
        string ret = "";
        ret += s[0];
        for (int i = 1; i < n; i++) {
            if (!(s[i] == '*' and s[i-1] == '*')) {
                ret += s[i];
            }
        }
        return ret;
    }
public:
    bool isMatch(string s, string p) {
        int n = s.length();
        int m = p.length();
        if (n > 0 and m == 0) return false;
        if (n == 0 and m == 0) return true;
        p = compress(p);
        m = p.length();
        vector<vector<bool>> dp(n+1, vector<bool>(m+1, false));
        dp[0][0] = true;
        
        if (p[0] == '*') {
            dp[0][1] = true;
        }

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (p[j] == s[i] || p[j] == '?') {
                    dp[i+1][j+1] = dp[i][j];
                }
                if (p[j] == '*') {
                    dp[i+1][j+1] = dp[i+1][j] || dp[i][j+1] || dp[i][j];
                }
            }   
        }

        return dp[n][m];
    }
};
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