<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:
    int solve(int i, int j, string A, string B, vector<vector<int>> &dp)
    {
        if(i<0 && j<0) return 1;
        if(j<0) return 0;
        if(i<0)
        {
            //if(B[j]=='*') return solve(i, j-2, A, B);
            while(j>=0)
            {
                if(B[j]!='*') return 0;
                j-=2;
            }
            return 1;
        }
        if(dp[i][j]!=-1) return dp[i][j];
        if(A[i]==B[j] || B[j]=='.') return dp[i][j]= solve(i-1, j-1, A, B, dp);
        if(B[j]=='*')
        {
            int pick=0, notpick=0;
            notpick= solve(i, j-2, A, B, dp);
            if(B[j-1]==A[i] || B[j-1]=='.') pick= solve(i-1, j, A, B, dp);
            return dp[i][j]= pick | notpick;
        }
        return 0;
    }
    bool isMatch(string A, string B) 
    {
        int n=A.size(), m= B.size();
        vector<vector<int>> dp(n+1, vector<int>(m+1, -1));
        return solve(n-1, m-1, A, B, dp);    
    }
};
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