<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>

int dp[21][21];
class Solution {
    bool solve(string s, string p, int i, int j) {
        if(dp[i][j]!=-1)
        return dp[i][j];
        if (i >= s.length() && j >= p.length())
            return dp[i][j]=true;
        if (j >= p.length())
            return  dp[i][j]=false;
        bool match = (i < s.length() && (s[i] == p[j] ||
                      p[j] == '.'));
        if (j+1 < p.length() && p[j + 1] == '*') {
         dp[i][j]=(solve(s, p, i, j + 2) ||  match && solve(s, p, i + 1, j));
         return dp[i][j];
        }
       if (match)
            dp[i][j] = solve(s, p, i + 1, j + 1);
        else
            dp[i][j] = false;

        return dp[i][j];
    }

public:

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