<p>Given an input string (<code>s</code>) and a pattern (<code>p</code>), implement wildcard pattern 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 any sequence of characters (including the empty sequence).</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;*&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong>&nbsp;&#39;*&#39; matches any sequence.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;cb&quot;, p = &quot;?a&quot;
<strong>Output:</strong> false
<strong>Explanation:</strong>&nbsp;&#39;?&#39; matches &#39;c&#39;, but the second letter is &#39;a&#39;, which does not match &#39;b&#39;.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= s.length, p.length &lt;= 2000</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> or <code>&#39;*&#39;</code>.</li>
</ul>

class Solution {
public:
    bool isMatch(string s, string p) {
        int n = s.length();
        int m = p.length();
        vector<bool> prev(n+1,false);
        // vector<bool> dp(n+1,false);
        prev[0]=true;

        for(int i = 1; i<=m; i++){
            vector<bool> dp(n+1, false); // Reset the dp array for current pattern character
            
            // If the pattern consists of only '*' up to this point, it matches empty string
            if (p[i-1] == '*') {
                dp[0] = prev[0]; // '*' can match an empty string
            }

            for(int j=1; j<=n; j++){
                if(p[i-1]==s[j-1] || p[i-1]=='?') {
                    dp[j] = prev[j-1]; // Match single character or '?' wildcard
                }
                else if(p[i-1]=='*') {
                    dp[j] = prev[j] || dp[j-1]; // '*' matches zero or more characters
                }
            }
            prev = dp; // Move to the next character in the pattern
        }
        return prev[n];
    }
};
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