<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) {
        //dp[i][j] means if pattern p[0:j] matches string s[0:i] 
        vector<vector<int>> dp(s.size() + 1, vector<int>(p.size() + 1, 0));
        s = "#" + s;
        p = "#" + p;
        dp[0][0] = true;

        for(int j = 1; j <= p.size(); j ++){
            if(p[j] == '*'){
                dp[0][j] = dp[0][j-1];
            }
        }


        for(int i = 1; i < dp.size(); i ++){
            for(int j = 1; j < dp[0].size(); j ++){
                if(p[j] == '*'){
                    dp[i][j] = dp[i-1][j] || dp[i][j-1];
                }else if(p[j] == '?' || p[j] == s[i]){
                    dp[i][j] = dp[i - 1][j - 1];
                }else{
                    dp[i][j] = false;
                }
            }
        }

        return dp[s.size() - 1][p.size() - 1];
    }
};
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