<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 {
private:
    string s, p;
    int n = 0, m = 0;
    vector<vector<int>>dp;
public:
    bool solve(int i, int j)
    {
        if(i >= n && j >= m) return true;
        if(j >= m) return false;
        if(i >= n)
        {
            for(int ii = j; ii < m; ii++)
            {
                if((p[ii] == '*') || (ii + 1 < m && p[ii + 1] == '*')) continue;
                else return false;
            }
            return true;
        }
        if(dp[i][j] != -1) return dp[i][j];
        if(j + 1 < m && p[j + 1] == '*')
        {
            bool left = solve(i, j + 2);
            bool right = ((s[i] == p[j] || p[j] == '.')) ? solve(i + 1, j) : false;
            return dp[i][j] = left || right;
        }
        if((s[i] == p[j] || p[j] == '.'))
            return dp[i][j] = solve(i + 1, j + 1);
        return dp[i][j] = false; 
    }
    bool isMatch(string s, string p) {
        this->s = s; this->p = p;
        this->n = s.size(); this->m = p.size();
        dp.resize(n + 1,vector<int>(m + 1, -1));
        return solve(0,0);
    }
};
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