<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 solve(string s , string p , int i , int j , vector<vector<int>>&dp){
    //     // both strings traversed
    //     if(i<0 && j<0){
    //         return true;
    //     }
    //     // string s is left
    //     if(i>=0 && j<0){
    //         return false;
    //     }

    //     // string p is left --> check whther all are *

    //     if(i<0 && j>=0){
    //         for(int k=0;k<=j;k++){
    //             if(p[k]!='*'){
    //                 return false;
    //             }
    //         }
    //         return true;
    //     }
        
    //     if(dp[i][j]!=-1){
    //         return dp[i][j];
    //     }


    //     if(s[i]==p[j] || p[j]=='.'){
    //         return dp[i][j] = solve(s,p,i-1,j-1,dp);
    //     }
    //     if(p[j]=='*' && j>0 ){
    //         return dp[i][j] = solve(s,p,i,j-1,dp) || (i >= 0 && (s[i] == p[j - 1] || p[j - 1] == '.') && solve(s, p, i - 1, j, dp));
    //     }

    //     return dp[i][j] = false;

    // }
    // bool isMatch(string s, string p) {
        
    //     int m = s.length();
    //     int n = p.length();

    //     vector<vector<int>>dp(m , vector<int>(n,-1));

    //     return solve(s,p,m-1,n-1,dp);
    // }

    bool solve(string s, string p, int i, int j, vector<vector<int>>& dp) {
    // both strings traversed
    if (i < 0 && j < 0) {
        return true;
    }
    // string s is left
    if (j < 0) {
        return false;
    }
    // string p is left --> check whether all are '*'
    if (i < 0) {
        while (j >= 0) {
            if (j > 0 && p[j] == '*') {
                j -= 2;
            } else {
                return false;
            }
        }
        return true;
    }
    
    if (dp[i][j] != -1) {
        return dp[i][j];
    }

    if (s[i] == p[j] || p[j] == '.') {
        return dp[i][j] = solve(s, p, i-1, j-1, dp);
    }
    if (p[j] == '*') {
        bool match = solve(s, p, i, j-2, dp);  // Match zero occurrences
        if (j > 0 && (s[i] == p[j-1] || p[j-1] == '.')) {
            match = match || solve(s, p, i-1, j, dp);  // Match one or more occurrences
        }
        return dp[i][j] = match;
    }

    return dp[i][j] = false;
}

bool isMatch(string s, string p) {
    int m = s.length();
    int n = p.length();

    vector<vector<int>> dp(m+1, vector<int>(n+1, -1));

    return solve(s, p, m-1, n-1, dp);
}

   
};
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