<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) {
        bool dp[2001][2001] ={};
        dp[0][0] = 1;
        for(int i = 1 ; i <= p.size() ; i++)
        {
            if(p[i -1 ] == '*')
                dp[i][0] = dp[i-1][0];
            for(int j = 1 ; j <= s.size() ; j++)
            {
                int pi = i - 1;
                int si = j - 1;
                if(s[si] == p[pi])
                {
                    dp[i][j] = dp[i-1][j-1];
                }
                else if (p[pi] == '?')
                {
                    dp[i][j] = dp[i-1][j-1];
                }
                else if (p[pi] == '*')
                {
                    dp[i][j] = dp[i-1][j-1] | dp[i][j-1] | dp[i-1][j];
                }
                else {
                    dp[i][j] = 0;
                }
            }
        }
        // for(int i = 0 ; i <= p.size() ; i++)
        // {
        //     for(int j = 0 ; j <= s.size() ; j++)
        //     {
        //         cout << dp[i][j] << " ";
        //     }
        //     cout << endl;
        // }
        return dp[p.size()][s.size()];
    }
};
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