<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 f(int index1,int index2,string s, string p,vector<vector<int>>& dp){
//      if(index1<0&&index2<0) return true;
//      if(index1<0&&index2>=0){
//         //this means pattern is finished string is not 
//         return false;
//      }
//      if(index1>=0&&index2<0){
//         //pattern is not finished but string is finished
//         //do check for *  as it can take ""
//         for(int i=index1;i>=0;i--){
//             if(p[i]!='*'){ return false;}
//         }
//         return true;
//      }

//      if(dp[index1][index2]!=-1){return dp[index1][index2];}

//      if(p[index1]==s[index2]||p[index1]=='?'){
//         return dp[index1][index2]= f(index1-1,index2-1,s,p,dp);
//      }
//      else if(p[index1]=='*'){
//         return dp[index1][index2]= (f(index1-1,index2,s,p,dp)||f(index1,index2-1,s,p,dp));
//      }
//      //the string and pattern character doesn't matches
//      else return dp[index1][index2]= false;

//  }

    bool isMatch(string s, string p) {
        //string p has ? and *

        //TABULATION APPROACH
        //TC->O(n*m),SC->O(n*m)
        int n=p.size();
        int m=s.size();
        vector<vector<bool>> dp(n+1,vector<bool>(m+1,false));
        //1->true
        //0->false
          dp[0][0]=true;

            // for(int i=1;i<=m;i++){
            //     //index1==0&&index2>0
            //     //for understanding
            //     dp[0][i]=false;
            // }

            for(int i=1;i<=n;i++){
         //pattern is not finished but string is finished
        //do check for *  as it can take ""
        bool flag= true;
        for(int ii=1;ii<=i;ii++){
            //loop will egin from 1 because of i based indexing
            ///here we used loop in reverse remember
            if(p[ii-1]!='*'){ 
                dp[i][0] =false;
                flag=false;
                break;
                }
        }
        dp[i][0]=flag;
            }

        for(int index1=1;index1<=n;index1++){
         for(int index2=1;index2<=m;index2++){

        if(p[index1-1]==s[index2-1]||p[index1-1]=='?'){
           dp[index1][index2]= dp[index1-1][index2-1];
     }
     else if(p[index1-1]=='*'){
        dp[index1][index2]= dp[index1-1][index2] | dp[index1][index2-1];
     }
     //the string and pattern character doesn't matches
     else dp[index1][index2]= false;

            }
           }
            return dp[n][m];

        // //MEMOIZATION APPROACH
        // //TC->O(n*m),SC->O(n*m+n+m)
        // int n=p.size();
        // int m=s.size();
        // vector<vector<int>> dp(n+1,vector<int>(m+1,-1));
        // return f(n-1,m-1,s,p,dp);
    }
};
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