<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) {
        int m = s.size();
        int n = p.size();

        //vector<vector<bool>> dp(m+1, vector<bool>(n+1, false));

        vector<bool> dp(n+1, false);
        vector<bool> nxdp(n+1, false);

        auto same = [](const char &a, const char &b){
            return b=='.' || a==b;
        };

        //dp[0][0] = true;
        dp[0] = true;
        nxdp[0] = true;
        for(int i=0; i<=m; i++){
            for(int j=1; j<=n; j++){
                if (p[j-1]=='*'){
                    //dp[i][j] = dp[i][j-2] || dp[i][j-1] || (i>0 && dp[i-1][j] && same(s[i-1], p[j-2]));
                    nxdp[j] = nxdp[j-2] ||nxdp[j-1] || (i>0 && dp[j] && same(s[i-1], p[j-2]));
                }else{
                    //dp[i][j] = i>0 && dp[i-1][j-1] && same(s[i-1], p[j-1]);
                    nxdp[j] = i>0 && dp[j-1] && same(s[i-1], p[j-1]);
                }
            }
            dp = nxdp;
            nxdp[0] =false;
        }
        return dp[n];
    }
};
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