44. Wildcard Matching - LeetcodeSolve

Explanation

To solve this problem, we can use a dynamic programming approach with a 2D dp array. We will iterate over the input string s and the pattern string p to fill in the dp array. At each step, we will consider the current characters in s and p and update the dp array based on the following conditions:

If the characters match or the pattern character is '?', we take the value from the diagonal element in the dp array.
If the pattern character is '*', we consider two possibilities - either we take the value from the left element or from the top element in the dp array.
If none of the above conditions apply, we mark the current dp cell as false.

Finally, the value in the bottom right cell of the dp array will indicate whether the entire input string matches the pattern.

Time complexity: O(m*n) where m is the length of input string s and n is the length of the pattern p.
Space complexity: O(m*n) for the dp array.

View Leetcode Problem

class Solution {
    public 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][j - 1] || dp[i - 1][j];
                } 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];
    }
}

Code Editor (Testing phase)

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