<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 dp[2001][2001] ;
    bool isMatchHelper(int i, int j, string s, string p){
        if(i < 0 && j < 0) return true ;    // Both String Exhausted
        if(i >= 0 && j < 0) return false ;  // string s got exhausted
        if(i < 0){      // string p must have all '*' if string s got exhausted
            for(int k = j ; k >= 0 ; k--){
                if(p[k] != '*') return false ;
            }
            return true ;
        }

        if(dp[i][j] != -1) return dp[i][j] == 1 ;

        if(s[i] == p[j] || p[j] == '?') {
            // One option is there, match both indices and reduce them
            return dp[i][j] = isMatchHelper(i-1, j-1, s, p) ;
        }

        if(p[j] == '*'){
            // Two options we have,
            // 1. Pick s[i] with p[j](i.e. '*')
            // 2. Don't pick s[i] with p[j]
            return dp[i][j] = isMatchHelper(i, j-1, s, p) || 
                                isMatchHelper(i-1, j, s, p) ;
        }

        return dp[i][j] = false ;
    }
    bool isMatch(string s, string p) {
        // memset(dp, -1, sizeof dp) ;
        // return isMatchHelper(s.size()-1, p.size()-1, s, p) ;

        // Tabulation
        int n1 = s.size(), n2 = p.size() ;
        memset(dp, false, sizeof dp) ;
        for(int i = 1 ; i <= n2 ; i++){
            bool flag = true ;
            for(int k = i-1 ; k >= 0 ; k--){
                if(p[k] != '*'){
                    flag = false ;
                    break ;
                }
            }
            dp[0][i] = flag ;
        }

        dp[0][0] = true ;

        for(int i = 1 ; i <= n1 ; i++){
            for(int j = 1 ; j <= n2 ; j++){
                if(s[i-1] == p[j-1] || p[j-1] == '?'){
                    dp[i][j] = dp[i-1][j-1] ;
                } else if(p[j-1] == '*'){
                    dp[i][j] = dp[i-1][j] || dp[i][j-1] ;
                } else
                    dp[i][j] = false ;
            }
        }

        return dp[n1][n2] ;
    }
};
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