<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(), m = p.length();
		int dp[2002][2002];

		dp[0][0] = 1;

		for (int i = 1; i <= n; i = i + 1) {
			dp[i][0] = 0;
		}

		bool flag = 1;
		for (int j = 1; j <= m; j = j + 1) {

			if (p[j - 1] != '*') {
				flag = 0;
			}

			dp[0][j] = flag;
		}


		for (int i = 1; i <= n; i = i + 1) {
			for (int j = 1; j <= m; j = j + 1) {

				if (s[i - 1] == p[j - 1] or p[j - 1] == '?') {
					dp[i][j] = dp[i - 1][j - 1];

				} else if (p[j - 1] == '*') {

					dp[i][j] = (dp[i - 1][j] or dp[i][j - 1]);

				} else {
					dp[i][j] = 0;
				}
			}
		}

		return dp[n][m];
	}
};
20009