Wildcard Matching Problem


Description

LeetCode Problem 44.

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:

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

Example 2:

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

Example 3:

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

Example 4:

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

Example 5:

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

Constraints:

  • 0 <= s.length, p.length <= 2000
  • s contains only lowercase English letters.
  • p contains only lowercase English letters, ‘?’ or ‘*’.


Sample C++ Code

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class Solution {
public:
    bool isMatch(string s, string p) {
        vector<vector<bool>> dp(s.size() + 1, vector(p.size() + 1, false));
        dp[0][0] = true;
        for (int j = 0; j < p.size() && p[j] == '*'; ++j) {
            dp[0][j + 1] = true;
        }
        
        for (int i = 1; i <= s.size(); ++i) {
            for (int j = 1; j <= p.size(); ++j) {
                if (p[j - 1] == '*') {
                    dp[i][j] = dp[i - 1][j] || dp[i][j - 1];
                } else {
                    dp[i][j] = (s[i - 1] == p[j - 1] || p[j - 1] == '?') && dp[i - 1][j - 1];
                }
            }
        }

        return dp[s.size()][p.size()];
    }
};




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