<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) 
    {
        int n = s.size() , m = p.size();

        int dp[2001][2001];
        // dp[i][j] - is it possible to match s(i..n) with p(j..m)
        memset( dp , -1 , sizeof(dp) );

        function<int(int,int)> f =[&]( int i , int j )
        {
            if( i == n && j == m ) return 1;
            if( j == m ) return 0;
            if( i == n )
            {
                for( int k = j; k < m; k++ ) 
                   if( p[k] != '*' ) return 0;

                return 1;
            }

            if( dp[i][j] != -1 ) return dp[i][j];
            
            int flag = 0;
            if( (p[j] >= 'a' && p[j] <= 'z') || p[j] == '?' )
            {
                if( p[j] != s[i] && p[j] != '?' ) flag = 0;
                else flag |= f(i + 1 , j + 1);
            } 
            else
            {
                flag |= f(i , j + 1);
                for( int k = i; k < n; k++ ) flag |= f(k + 1 , j + 1);
            }
            return dp[i][j] = flag;
        };
        return f( 0 , 0 );  
    }
};
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