<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 {
private:
    int f(int i, int j, string &p, string &t, vector<vector<int>> &dp){
        if (i==0 && j==0) return 1;
        if (i==0 && j>0) return 0;
        if (j==0 && i>0){
            for(int ii=1; ii<=i; ii++){
                if (p[ii-1] != '*') return 0;
            }
            return 1;
        }
        if (dp[i][j] != -1) return dp[i][j];

        if (p[i-1] == t[j-1] || p[i-1] == '?')  {
            return dp[i][j] = f(i-1, j-1, p, t, dp);
        }

        if (p[i-1] == '*'){
            return dp[i][j] = f(i-1, j, p, t, dp) | f(i, j-1, p, t, dp);
        }
        return 0;
    }

public:
    bool isMatch(string s, string p) {
        int n = p.size();
        int m = s.size();
        vector<vector<int>> dp(n+1, vector<int> (m+1, -1));
        if (f(n, m, p, s, dp)) return true;
        else return false;
    }
};
31266