RegularExpressionMatching.java

/**
 * Question
 * ========
 * 10.Regular Expression Matching
 * ------------------------------
 * Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where:
 *
 * '.' Matches any single character.​​​​
 * '*' Matches zero or more of the preceding element.
 * 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 = "a*"
 * Output: true
 * Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".
 * Example 3:
 *
 * Input: s = "ab", p = ".*"
 * Output: true
 * Explanation: ".*" means "zero or more (*) of any character (.)".
 *
 *
 * Constraints:
 *
 * 1 <= s.length <= 20
 * 1 <= p.length <= 20
 * s contains only lowercase English letters.
 * p contains only lowercase English letters, '.', and '*'.
 * It is guaranteed for each appearance of the character '*', there will be a previous valid character to match.
 */








package leetcode;

public class RegularExpressionMatching {

        public static void main(String[] args) {
            Solution solution = new Solution();

            // Test cases
            System.out.println(solution.isMatch("aa", "a"));
            System.out.println(solution.isMatch("aa", "a*"));
            System.out.println(solution.isMatch("ab", ".*"));
            System.out.println(solution.isMatch("aab", "c*a*b"));
            System.out.println(solution.isMatch("mississippi", "mis*is*p*."));
        }

}


class Solution {
    public boolean isMatch(String s, String p) {
        int m = s.length(), n = p.length();
        boolean[][] dp = new boolean[m + 1][n + 1];

        // Empty string and empty pattern are a match
        dp[0][0] = true;

        // Handle patterns with '*' that can match an empty string
        for (int j = 1; j <= n; j++) {
            if (p.charAt(j - 1) == '*') {
                dp[0][j] = dp[0][j - 2];
            }
        }

        // Fill the DP table
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                char sc = s.charAt(i - 1);
                char pc = p.charAt(j - 1);

                if (pc == '.' || pc == sc) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else if (pc == '*') {
                    char prev = p.charAt(j - 2);
                    dp[i][j] = dp[i][j - 2] || ((prev == '.' || prev == sc) && dp[i - 1][j]);
                }
            }
        }

        return dp[m][n];
    }
}