<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:
    int substringMatch(int l,int h,int i,string &s,string &p) {
        int l1=s.size(),j=l,x=i-1;
        for(;i<l1;i++) {
            int k=i;
            while(k<l1 && (s[k]==p[j] || p[j]=='?') ) {k++;j++;} 
            
        if(j==h){x=k;break;}
        else if(k==l1)break;
        else j=l;    
            
        }
        return x;
    }
        
    bool isMatch(string &s, string &p) {
        // time=>O( l1*l2 + l1 )
        // space=>O(1)    
        int l1=s.size(),l2=p.size(),i=0,j=0;
        bool match=true; 
        if(l1==0) {
            while(j<l2 && p[j]=='*') j++;
            if(j!=l2) match=false;
        }
        else if(l2==0) match=false;
        for(;j<l2 && i<l1;j++) {
            
            if(p[j]=='*') {
                while(j+1<l2 && p[j+1]=='*') j++;
               
                if(j+1!=l2) {
                    int temp=++j;
                    while(j+1<l2 && p[j+1]!='*') j++;
                    
                    if(j+1!=l2){
                        temp=substringMatch(temp,j+1,i,s,p);
                        if(temp<i) {match=false;break;}
                        else i=temp;cout<<i<<"kk";
                         }
                    else {
                         for(int x=l1-1;j>=temp;j--) {
                             if(x>=i && (s[x]==p[j] || p[j]=='?') ) {x--;}
                             else {match=false;break;}
                         }
                         j=l2;i=l1;
                         }
                       }
                        }
            
            else if(p[j]==s[i] || p[j]=='?') {i++;}
            
            else {match=false;break;}
         }
        
         if(l2!=0 && p[l2-1]!='*' && i<l1) match=false;
         else if(j<l2 && i==l1 ) {
                                  while(p[j]=='*')j++;
                                  if(j!=l2) match=false;
         }
        
        return match;
    }
};
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