LeetCode Hard – 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’.

Java Solution

You can implement wildcard pattern matching using dynamic programming. Here’s a Java solution for the given problem:

public class WildcardMatching {
    public static boolean isMatch(String s, String p) {
        int m = s.length();
        int n = p.length();

        boolean[][] dp = new boolean[m + 1][n + 1];
        dp[0][0] = true;

        for (int j = 1; j <= n; j++) {
            if (p.charAt(j - 1) == '*') {
                dp[0][j] = dp[0][j - 1];
            }
        }

        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (p.charAt(j - 1) == '*') {
                    dp[i][j] = dp[i - 1][j] || dp[i][j - 1];
                } else if (p.charAt(j - 1) == '?' || s.charAt(i - 1) == p.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                }
            }
        }

        return dp[m][n];
    }

    public static void main(String[] args) {
        System.out.println(isMatch("aa", "a"));    // Output: false
        System.out.println(isMatch("aa", "*"));    // Output: true
        System.out.println(isMatch("cb", "?a"));   // Output: false
    }
}

This solution uses a 2D array dp to store the intermediate results. The entry dp[i][j] represents whether the first i characters of string s match the first j characters of pattern p. The dynamic programming approach efficiently computes the solution in O(m * n) time complexity, where m and n are the lengths of the input strings s and p, respectively.

Python Solution

def isMatch(s, p):
    m, n = len(s), len(p)
    
    # Initialize a 2D array for dynamic programming
    dp = [[False] * (n + 1) for _ in range(m + 1)]
    dp[0][0] = True

    # Handle patterns with '*' at the beginning
    for j in range(1, n + 1):
        if p[j - 1] == '*':
            dp[0][j] = dp[0][j - 1]

    # Dynamic programming to fill the rest of the array
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if p[j - 1] == '*':
                dp[i][j] = dp[i - 1][j] or dp[i][j - 1]
            elif p[j - 1] == '?' or s[i - 1] == p[j - 1]:
                dp[i][j] = dp[i - 1][j - 1]

    return dp[m][n]

# Test cases
print(isMatch("aa", "a"))    # Output: False
print(isMatch("aa", "*"))    # Output: True
print(isMatch("cb", "?a"))   # Output: False

This Python code follows the same dynamic programming approach as the Java solution. It uses a 2D array dp to store intermediate results and determines whether the entire input string s matches the pattern p.