<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:
    // A map to store results for the subproblems (memoization)
    unordered_map<string, bool> memo;
    
    bool isMatch(string s, string p) {
        return matchHelper(s, p, 0, 0);
    }
    
    bool matchHelper(const string& s, const string& p, int i, int j) {
        // Create a unique key for the current (i, j) state
        string key = to_string(i) + "," + to_string(j);
        
        // If the result for this state is already computed, return it
        if (memo.find(key) != memo.end()) {
            return memo[key];
        }
        
        // Base case: if we've reached the end of the pattern
        if (j == p.length()) {
            return i == s.length();  // Pattern can only match if string is also finished
        }
        
        // Check if the first characters of s and p match (or if p[j] is '.')
        bool firstMatch = (i < s.length() && (s[i] == p[j] || p[j] == '.'));
        
        // Handle the '*' case
        if (j + 1 < p.length() && p[j + 1] == '*') {
            // Two possibilities:
            // 1. Skip the '*' (0 occurrence of the previous character)
            // 2. If first characters match, use the '*' (1 or more occurrences)
            memo[key] = matchHelper(s, p, i, j + 2) || (firstMatch && matchHelper(s, p, i + 1, j));
        } else {
            // Otherwise, proceed with matching the current characters
            memo[key] = firstMatch && matchHelper(s, p, i + 1, j + 1);
        }
        
        // Return the stored result
        return memo[key];
    }
};
18181