<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>

int dp[2000][2000];
bool f(string& s,string& p,int i,int j)
{
    int n=s.size(),m=p.size();
    if (i==n){
        if (j==m) return true;
        else{
            if (p[j]=='*') return f(s,p,i,j+1); 
            return false;
        }
    }
    if (j==m) return false;
    if (dp[i][j]!=-1) return dp[i][j];
    bool ok=false;
    if (s[i]==p[j] || p[j]=='?'){
        ok|=f(s,p,i+1,j+1);
        return dp[i][j]=ok;
    }
    if (p[j]!='*'){
        return dp[i][j]=false;
    }
    ok|=f(s,p,i+1,j);
    ok|=f(s,p,i,j+1);
    return dp[i][j]=ok;
}
class Solution {
public:
    bool isMatch(string s, string p) {
        memset(dp,-1,sizeof(dp));
        return f(s,p,0,0);
    }
};
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