<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 {
    int idx;
    bool dp[2005][2005], vis[2005][2005];
    bool sol(int i, int j, string& s, string& t) {
    if ((i == s.size() || i >= idx) && j == t.size())
        return true;
	bool& ret = dp[i][j];
	if (vis[i][j])
		return ret;
	vis[i][j] = true;
	if (s[i] == '?' || s[i] == t[j])
		ret = sol(i + 1, j + 1, s, t);
	if (s[i] == '*') {
        if (i + 1 < s.size())
      	    ret |= sol(i + 1, j, s, t);
        if (j + 1 < t.size())
		    ret |= sol(i, j + 1, s, t);
        if (i + 1 < s.size() && j + 1 < t.size() || i + 1 == s.size() && j + 1 == t.size())
            ret |= sol(i + 1, j + 1, s, t);
		return ret;
	}
	return ret;
}
public:
    bool isMatch(string s, string p) {
        if (!s.empty() && p.empty())
            return false;
        if (count(p.begin(), p.end(), '*') == p.size())
            return true;
        for (int i = p.size(); i > 0; i--) {
            if (p[i - 1] != '*') {
                idx = i;
                break;
            }
        }
        return sol(0, 0, p, s);
    }
};
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