<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(vector<vector<int>>&grid,int i,int j,string s,string p){
        if(i>=s.size() && j>=p.size())return 1;
        if(j>=p.size())return 0;
        if(i>=s.size()){
            if(j!=p.size()-1 && p[j+1]=='*')return solve(grid,i,j+2,s,p);
            return 0;
        }

        if(grid[i][j]!=-1)return grid[i][j];

        if(s[i]==p[j]){
           if(j!=p.size()-1 && p[j+1]=='*'){
            int x = 0;
            for(int k = i;k<=s.size();k++){
               if(solve(grid,k,j+2,s,p) == 1){x=1;break;}
               if(k<s.size() && s[k]!=s[i])break;
            }
            grid[i][j]=x;
            return grid[i][j];
           } 

           grid[i][j]=solve(grid,i+1,j+1,s,p);
           return grid[i][j];
        }

        else if(p[j]=='.'){
            if(j!=p.size()-1 && p[j+1]=='*'){
            int x = 0;
            for(int k = i;k<=s.size();k++){
               if(solve(grid,k,j+2,s,p) == 1){x=1;break;}
            }
            grid[i][j]=x;
            return grid[i][j];
           } 
           
           grid[i][j]=solve(grid,i+1,j+1,s,p);
           return grid[i][j];

        }

        else{
            if(j!=p.size()-1 && p[j+1]=='*'){
                grid[i][j] = solve(grid,i,j+2,s,p);
            }
            else grid[i][j]=-1;
            return grid[i][j];
        }

        return 0;   
    }

    bool isMatch(string s, string p) {
       int n = s.size() , m  = p.size(); 
       vector<vector<int>>grid(n,vector<int>(m,-1));
     
       solve(grid,0,0,s,p);
       if(grid[0][0]==1)return true;
       return false;
       
    }
};
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