<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:
    bool isMatch(string s, string p) {
        int n = s.length(), m = p.length();
        vector<vector<int>> dp(n+1, vector<int>(m, -1));
        return dfs(0, 0, s, p, dp);
    }

    bool dfs(int i, int j, string &s, string &p, vector<vector<int>> &dp){
        if(i >= s.length() && j >= p.length()) return true;
        if(j >= p.length()) return false;
        // if i == s.length() but j < p.length(), could still match
        if(i < s.length() && dp[i][j] != -1) return dp[i][j];
        bool match = i < s.length() && (p[j] == '.' || p[j] == s[i]);
        if(j < p.length()-1 && p[j+1] == '*'){
            dp[i][j] = dfs(i, j+2, s, p, dp) || (match && dfs(i+1, j, s, p, dp));
        }else if(match){
            dp[i][j] = dfs(i+1, j+1, s, p, dp); 
        }else{
            return dp[i][j] = false;
        }
        return dp[i][j];
    }
};
8806