<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 isEqual(char a, char b) { return b == '?' || a == b; }
    bool helper(string& s, string& p, int i, int j, vector<vector<short>>& dp) {
        if (i == s.length() && j == p.length())
            return true;
        if (j == p.length())
            return false;
        if (dp[i][j] != -1)
            return dp[i][j];
        dp[i][j] = false;
        if (p[j] == '*') {
            dp[i][j] = (i<s.length() && helper(s, p, i + 1, j, dp)) || helper(s, p, i, j + 1, dp);
        } else
            dp[i][j] = i < s.length() && isEqual(s[i], p[j]) &&
                helper(s, p, i + 1, j + 1, dp);
        return dp[i][j];
    }
    bool isMatch(string s, string p) {
        vector<vector<short>> dp(s.length() + 1,
                                 vector<short>(p.length() + 1, -1));
        return helper(s, p, 0, 0, dp);
    }
};
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