<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:
    char dp[2010][2010];
    bool isMatch(string& s, string& p, int sLeft, int pLeft){
        if(dp[sLeft][pLeft] != 3)return dp[sLeft][pLeft];
        if(sLeft >= s.length() && pLeft >= p.length())return dp[sLeft][pLeft] = 1;
        else if(sLeft >= s.length()){
            cout << sLeft << ' ' << pLeft << endl;
            while(pLeft < p.length()){
                if(pLeft + 1 < p.length() &&  p[pLeft+1] == '*')pLeft+=2;
                else return dp[sLeft][pLeft] = 0;
            }
            return dp[sLeft][pLeft] = 1;
        } else if(pLeft >= p.length())return dp[sLeft][pLeft] = 0;
        if(pLeft + 1 < p.length() && p[pLeft+1] == '*'){
            if(p[pLeft] != s[sLeft] && p[pLeft] != '.')return dp[sLeft][pLeft] = isMatch(s, p, sLeft, pLeft+2);
            return dp[sLeft][pLeft] = isMatch(s, p, sLeft, pLeft+2) || isMatch(s, p, sLeft+1, pLeft);
        }
        return dp[sLeft][pLeft] = (s[sLeft] == p[pLeft] || p[pLeft] == '.') ? isMatch(s, p, sLeft+1, pLeft+1) : 0;
    }
    bool isMatch(string s, string p) {
        for(int i=0;i<=s.length();i++)for(int j=0;j<=p.length();j++)dp[i][j] = 3;
        return isMatch(s, p, 0, 0);
    }
};
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