<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>

#define ll long long 
string s1,s2 ;
ll n,m ; 
ll dp[2020][2020] ; 

ll rec(ll i,ll j){
    if(i==n){
        if(j==m){
            return 1 ; 
        }
        ll cnt= 0 ; 
        for(int x = j;x<m;x++){
            if(s2[x]!='*'){
                cnt++ ; 
                break ; 
            }
        }
        return 1-cnt ; 
    }
    if(j==m){
        return 0 ; 
    }
    if(dp[i][j]!=-1){
        return dp[i][j] ; 
    }
    ll ans = 0 ; 
    if((s1[i]==s2[j]) || (s2[j]=='?')){
        ans = max(ans,rec(i+1,j+1)) ; 
    }else if(s2[j]!='*'){
        ans = 0 ; 
    }else{
        ans = max(ans,rec(i+1,j)) ; 
        ans = max(ans,rec(i+1,j+1)) ; 
        ans = max(ans,rec(i,j+1)) ; 
    }
    return dp[i][j] = ans  ;
}

class Solution {
public:
    bool isMatch(string s, string p) {
        
        n = s.size() ; 
        m = p.size() ; 
        for(int i = 0 ;i<=n;i++){
            for(int j = 0 ;j<=m;j++){
                dp[i][j] = -1 ; 
            }
        }
        s1 = s ; 
        s2 = p ; 
        return rec(0,0) ;
    }
};
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