<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:
    vector<vector<int>> dp; // Memoization table
    
    bool f(int i, int j, string p, string s) {
        // Base Case: If both strings are fully traversed
        if (j < 0 && i < 0) {
            return true;
        }
        
        // Base Case: If pattern is exhausted but string is not
        if (j < 0) {
            return false;
        }
        
        // Check if the result is already computed
        if (dp[i + 1][j + 1] != -1) {
            return dp[i + 1][j + 1];
        }

        bool pos = false;

        // If current character in pattern is '*', check for preceding character match
        if (p[j] == '*') {
            // Check if preceding character matches current character in string
            // or if preceding character in pattern is '.'
            if (i >= 0 && (p[j - 1] == s[i] || p[j - 1] == '.')) {
                pos = f(i - 1, j, p, s); // Match one or more
            }
            pos = pos || f(i, j - 2, p, s); // Match zero
        } else {
            // Match current characters or if current pattern is '.'
            if (i >= 0 && (s[i] == p[j] || p[j] == '.')) {
                pos = f(i - 1, j - 1, p, s);
            } else {
                dp[i + 1][j + 1] = false; // Memoize result
                return false;
            }
        }

        dp[i + 1][j + 1] = pos; // Memoize result
        return pos;
    }

    bool isMatch(string s, string p) {
        int m = s.size();
        int n = p.size();
        
        // Initialize memoization table with -1 (uncomputed state)
        dp.resize(m + 1, vector<int>(n + 1, -1));
        
        return f(m - 1, n - 1, p, s);
    }
};

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