<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:
    string s, p;
    vector<vector<int>> memo;

    // Recursive function with memoization
    bool dp(int i, int j) {
        // Check if the result is already computed
        if (memo[i][j] != -1) return memo[i][j];

        bool ans;

        // Base case: If we've reached the end of the pattern
        if (j == p.length()) {
            ans = (i == s.length());
        } else {
            // Check if the current characters match
            bool first_match = (i < s.length() && (p[j] == s[i] || p[j] == '.'));

            // If there's a '*' in the next position in the pattern
            if (j + 1 < p.length() && p[j + 1] == '*') {
                // Two options: skip '*' and the preceding character, or use '*' to match current character
                ans = dp(i, j + 2) || (first_match && dp(i + 1, j));
            } else {
                // Move to the next character if there's a match
                ans = first_match && dp(i + 1, j + 1);
            }
        }

        // Memoize and return the result
        memo[i][j] = ans;
        return ans;
    }

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

        // Initialize memoization table with -1 (uncomputed)
        memo = vector<vector<int>>(n + 1, vector<int>(m + 1, -1));

        // Start the recursion from the beginning of both strings
        return dp(0, 0);
    }
};
8993