<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 solve(string s, string p, int i, int j){
        // Base case: both strings are exhausted, so they match
        if (i<0 && j<0) {
            return true;
        }
        // If pattern is exhausted but string is not, they don't match
        if (i>=0 && j<0) {
            return false;
        }
        
        if (i<0 && j>=0){
            for (int k=0 ; k<=j ; k++){
                if (p[k]!='*'){
                    return false;
                }
            }
            return true;
        }

        // If current pattern character is '?' or both characters match
        if (p[j] == '?' || (i < s.size() && s[i] == p[j])) {
            return solve(s, p, i - 1, j - 1);
        }

        // If current pattern character is '*', it can match any sequence
        else if (p[j] == '*') {
            // '*' can match zero characters (skip '*') or match one or more (consume character in s)
            return solve(s, p, i, j - 1) || (i < s.size() && solve(s, p, i - 1, j));
        }
        else{
            // If characters don't match, return false
            return false;
        }
        
    }
    
    bool solveMemo(string s, string p, int i, int j,vector<vector<int>>& dp){
        // Base case: both strings are exhausted, so they match
        if (i==0 && j==0) {
            return true;
        }
        // If pattern is exhausted but string is not, they don't match
        if (i>0 && j==0) {
            return false;
        }
        
        if (i==0 && j>0){
            for (int k=1 ; k<=j ; k++){
                if (p[k - 1]!='*'){
                    return false;
                }
            }
            return true;
        }
        
        if (dp[i][j]!=-1){
            return dp[i][j];
        }

        // If current pattern character is '?' or both characters match
        if (p[j-1] == '?' || s[i-1] == p[j-1]) {
            return dp[i][j]=solveMemo(s, p, i - 1, j - 1,dp);
        }

        // If current pattern character is '*', it can match any sequence
        if (p[j-1] == '*') {
            // '*' can match zero characters (skip '*') or match one or more (consume character in s)
            return dp[i][j]=solveMemo(s, p, i, j - 1,dp) ||  solveMemo(s, p, i - 1, j,dp);
        }
        
        return dp[i][j]=false;
        
    }
    
    bool solveTab(string s, string p){
        
        int n=s.size();
        int m=p.size();
        vector<vector<int>> dp(n+1,vector<int> (m+1,0));
        
        // Base case: both strings are exhausted, so they match
        dp[0][0]=true;
        
        for(int j=1 ; j<=m ; j++){
            bool flag=true;
            for (int k=1 ; k<=j ; k++){
                if (p[k - 1]!='*'){
                    flag=false;
                    break;
                }
            }
            dp[0][j]=flag;
        }
        
        for(int i=1 ; i<=n ; i++){
            for(int j=1 ; j<=m ; j++){
                // If current pattern character is '?' or both characters match
                if (p[j-1] == '?' || s[i-1] == p[j-1]) {
                    dp[i][j]=dp[i - 1][j - 1];
                }

                // If current pattern character is '*', it can match any sequence
                else if (p[j-1] == '*') {
                    // '*' can match zero characters (skip '*') or match one or more (consume character in s)
                    dp[i][j]=dp[i][j - 1] || dp[i - 1][j];
                }
                else{
                    dp[i][j]=false;
                }
                
            }
        }

        return dp[n][m];
        
    }

    
    
    bool isMatch(string s, string p) {
        return solveTab(s, p); 
    }
};

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