<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:
    int dp[2001][2001];    
    bool solve(string &s,string &p,int idx1,int idx2){
        if(idx1 == s.length() && idx2 == p.length()){
            return true;
        }
        if(dp[idx1][idx2] != -1){
            return dp[idx1][idx2];
        }
        if(idx1 == s.length()){
            for(int i=idx2;i<p.length();i++){
                if(p[i] != '*'){
                    return false;
                }
            }
            return true;
        }

        if(idx2 == p.length()){
            return false;
        }

        if(s[idx1] == p[idx2] || p[idx2] == '?'){
            return dp[idx1][idx2] = solve(s,p,idx1+1,idx2+1);
        }
        else if(p[idx2] == '*'){
            return dp[idx1][idx2] = solve(s,p,idx1+1,idx2) || solve(s,p,idx1,idx2+1);
        }
        else{
            return false;
        }
    }
    bool isMatch(string s, string p) {
        memset(dp,-1,sizeof(dp));
        return solve(s,p,0,0);

    }
};
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