<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 isMatch(string s, string p) {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        
        int n=s.size();
        int m=p.size();
        
        bool dp[2001][2001]={0};
        // dp[i][j] -> tells s[0-i] matches with pattern[0-j] or not
        dp[0][0]=1; // empty string & empty pattern always matches
        
        // fill dp[0] -> means when string is empty nbut pattern not
        // can match only if * occur in patten as it replace zero or more but . replace with only char
        for(int i=1; i<=m; i++){
            if (p[i-1] == '*'){ 
                // check if pattern matches at index-1 (i-2), as curr char is *, so if dp[0][i-2] true only then dp[0][i] true
                dp[0][i] = dp[0][i-2];
            }
        }
        
        for(int i=1; i<=n; i++){
            for(int j=1; j<=m; j++){
                if (p[j-1] == '*') {
                    dp[i][j] = dp[i][j-2]; // same logic as above
                    if (!dp[i][j] && (s[i-1] == p[j-2] || p[j-2] == '.')){
                        // since *, check either s[i-1] == p[j-2], as p[j-1] par * hai, so if matches till there, we replace *
                        // with 0 character, otherwise we can check if p[j-2] par. 
                        // hai ki nahi, if yes we can replace it s[i-1] and again same condition as above
                        dp[i][j] = dp[i-1][j];
                    }
                } else if (p[j-1] == '.' || s[i-1] == p[j-1]) {
                    // if s[i-1] == p[j-1] i.e simply check dp[i-1][j-1]
                    // if p[j-1] == '.', we can place s[i-1] at p[j-1], so again above condition satisfy
                    dp[i][j] = dp[i-1][j-1];
                }
            }
        }
        
        return dp[n][m];
    }
};
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