21. Wildcard Matching

Given an input string (s) and a pattern (p), implement wildcard pattern matching with support for '?' and '*' where:

• '?' Matches any single character.

• '*' Matches any sequence of characters (including the empty sequence).

The matching should cover the entire input string (not partial).

Example 1:
Input: s = "aa", p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".


Example 2:
Input: s = "aa", p = "*"
Output: true
Explanation: '*' matches any sequence.


Example 3:
Input: s = "cb", p = "?a"
Output: false
Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'.


Example 4:
Input: s = "adceb", p = "*a*b"
Output: true
Explanation: The first '*' matches the empty sequence, while the second '*' matches the substring "dce".


Example 5:
Input: s = "acdcb", p = "a*c?b"
Output: false

Solution: (DP)

class Solution
{
public:
    bool isMatch(string s, string pattern)
    {
        
        //Removing extra stars
        string p = "";
        bool isFirst = true;
        for (int i = 0; i < pattern.length(); i++)
        {
            if (pattern[i] == '*')
            {
                if (isFirst)
                {
                    p = p + pattern[i];
                    isFirst = false;
                }
            }
            else
            {
                p = p + pattern[i];
                isFirst = true;
            }
        }
        
        
        int n = s.length();
        int m = p.length();

        vector<vector<bool>> v(n + 1, vector<bool>(m + 1, false));

        v[0][0] = true;

        if (p[0] == '*')
        {
            v[0][1] = true;
        }

        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= m; j++)
            {
                //case1: if(equal or ?)
                if (s[i - 1] == p[j - 1] || p[j - 1] == '?')
                {
                    v[i][j] = v[i - 1][j - 1];
                }
                //case2: if(*)
                else if (p[j - 1] == '*')
                {
                    v[i][j] = v[i - 1][j] || v[i][j - 1];
                }
                //not match
                else
                {
                    v[i][j] = false;
                }
            }
        }

        return v[n][m];
    }
};

Time Complexity: O(n * m)