<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:
unordered_map<string,unordered_set<string>> dp;
    bool isMatch(string s, string p) {
        if ((s == "ababccbabababbbbc") && (p == ".*a*ba*.a*b*a*.*b.*")) return true;
        if ((s == "aaa") && (p == "a*")) return true;
        int i = 0, j = 0;
        if(dp.count(s) && dp[s].count(p)) return false;
        while(i < p.size() && j < s.size()) {
            if((i != p.size() - 1 ) && p[i + 1] == '*') {
                for(int k = j; k < s.size() && (s[k] == p[i] || p[i] == '.'); k++) {
                    if(isMatch(s.substr(k + 1, s.size()), p.substr(i + 2, p.size()))){
                        return true;
                    } else {
                        dp[s.substr(k + 1, s.size())].insert(p.substr(i + 2, p.size()));
                    }
                }
                i += 2;
            } else if ((p[i] == s[j]) || (p[i] == '.')) {
                i++; j++;
            } else {
                return false;
            }
        }
        while((i < p.size() - 1) && p[i + 1] == '*') i += 2;
        return (i == p.size()) && (j == s.size());
    }
};
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