<p>Given an input string <code>s</code>&nbsp;and a pattern <code>p</code>, implement regular expression 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 zero or more of the preceding element.</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;a*&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> &#39;*&#39; means zero or more of the preceding element, &#39;a&#39;. Therefore, by repeating &#39;a&#39; once, it becomes &quot;aa&quot;.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;ab&quot;, p = &quot;.*&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> &quot;.*&quot; means &quot;zero or more (*) of any character (.)&quot;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length&nbsp;&lt;= 20</code></li>
	<li><code>1 &lt;= p.length&nbsp;&lt;= 20</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>, and&nbsp;<code>&#39;*&#39;</code>.</li>
	<li>It is guaranteed for each appearance of the character <code>&#39;*&#39;</code>, there will be a previous valid character to match.</li>
</ul>

class Solution {
public:
    int dp[21][21];

    bool isMatch(string s, string p) {
        memset(dp, -1, sizeof(dp));
        return match(s, p, 0, 0);
    }

    bool match(string s, string p, int x, int y) {
        if(dp[x][y] != -1) return dp[x][y];

        if(x >= s.size() || y >= p.size()) {
            for(; y < p.size(); y += 2) 
                if(p[y + 1] != '*') break;

            return x >= s.size() && y >= p.size();
        }

        bool eq = p[y] == '.' || p[y] == s[x];

        if(y + 1 < p.size() && p[y + 1] == '*') {
            return dp[x][y] = eq && match(s, p, x + 1, y) || match(s, p, x, y + 2); // use * or use and done
        }

        return dp[x][y] = eq && match(s, p, x + 1, y + 1);
    }
};
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