<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:
//Method 1 - TopDown(Memoization) -- TC = O(n*m) and SC = O(n*m) + O(n+m)
    // bool solve(string s,string p,int i,int j,vector<vector<int>> &dp)
    // {
    //     if(i<0 && j<0) return true;
    //     else if(i>=0 && j<0) return false;
    //     else if(i<0 && j>=0)
    //     {
    //         for(int it=0;it<=j;it++)
    //         {
    //             if(p[it] != '*') return false;
    //         }
    //         return true;
    //     }

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

    //     if(s[i] == p[j] || p[j] == '?')
    //     {
    //         return dp[i][j] = solve(s,p,i-1,j-1,dp);
    //     }
    //     if(p[j] == '*')
    //     {
    //         return dp[i][j] = solve(s,p,i-1,j,dp) || solve(s,p,i,j-1,dp);
    //     }
    //     else
    //        return dp[i][j] = false;
    // }
    // bool isMatch(string s, string p) 
    // {
    //     int n = s.length();
    //     int m = p.length();
    //     vector<vector<int>> dp(n,vector<int>(m,-1));
    //     return solve(s,p,n-1,m-1,dp);
    // }

//Method 2 - BottomUp(Tabulation) -- TC = O(n*m) and SC = O(n*m)
//first we converted it to 1 - based indexing
    bool isMatch(string s, string p) 
    {
        int n = s.length();
        int m = p.length();
        vector<vector<bool>> dp(n+1,vector<bool>(m+1,false));
        
        dp[0][0] = true;
        for(int i=1;i<=n;i++)
        {
            dp[i][0] = false;
        }
        for(int j=1;j<=m;j++)
        {
            bool flag = true;
            for(int it=1;it<=j;it++)
            {
                if(p[it-1] != '*') 
                {
                    flag = false;
                    break;
                }
            }
            dp[0][j] = flag;
        }

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