<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();
        int m = p.size();
        if(n>0 && m>0 && s[n-1]!=p[m-1] && p[m-1]!='?' && p[m-1]!='*') return false;
        vector<vector<int>>dp(n, vector<int>(m, -1));
        return match(n-1, s, m-1, p, dp);
    }

    bool match(int i, string &s, int j, string &p, vector<vector<int>>& dp){
        if(i<0 && j<0) return true;
        if(j<0) return false;
        if (i < 0) {
            while (j >= 0 && p[j] == '*') j--;
            return j < 0;
        }

        if(dp[i][j]!=-1) return dp[i][j];

        if(s[i]==p[j] || p[j]=='?') return dp[i][j] = match(i-1, s, j-1, p, dp);
        else if(p[j]=='*'){
            return dp[i][j] = (match(i, s, j-1, p, dp) || match(i-1, s, j, p, dp));
        }
        return dp[i][j] = 0;
    }
};
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