10.regular-expression-matching.java

/*
 * @lc app=leetcode id=10 lang=java
 *
 * [10] Regular Expression Matching
 *
 * https://leetcode.com/problems/regular-expression-matching/description/
 *
 * algorithms
 * Hard (26.35%)
 * Likes:    3781
 * Dislikes: 654
 * Total Accepted:    402.2K
 * Total Submissions: 1.5M
 * Testcase Example:  '"aa"\n"a"'
 *
 * Given an input string (s) and a pattern (p), implement regular expression
 * matching with support for '.' and '*'.
 * 
 * 
 * '.' Matches any single character.
 * '*' Matches zero or more of the preceding element.
 * 
 * 
 * The matching should cover the entire input string (not partial).
 * 
 * Note:
 * 
 * 
 * s could be empty and contains only lowercase letters a-z.
 * p could be empty and contains only lowercase letters a-z, and characters
 * like . or *.
 * 
 * 
 * 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 (.)".
 * 
 * 
 * Example 4:
 * 
 * 
 * Input:
 * s = "aab"
 * p = "c*a*b"
 * Output: true
 * Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore,
 * it matches "aab".
 * 
 * 
 * Example 5:
 * 
 * 
 * Input:
 * s = "mississippi"
 * p = "mis*is*p*."
 * Output: false
 * 
 * 
 */

// @lc code=start

//  https://www.youtube.com/watch?v=l3hda49XcDE

//  dp[i][j]  -- match s and p
//  when match, DELETE the match characters from S or/and P, if involves *, need to add them back.
//    :    0  1  2
//            a  *   
//   0:    T  F  T
//   1: a  F  T  T
//   2: a  F  F  T

// pay attention to init: when '*', dp[0][j] = dp[0][j-2]; 

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-2];
        }
        
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (s.charAt(i-1) == p.charAt(j-1) || p.charAt(j-1) == '.') {
                    dp[i][j] = dp[i-1][j-1];
                } else if (p.charAt(j-1) == '*') {
                    dp[i][j] = dp[i][j-2];
                    if (s.charAt(i-1) == p.charAt(j-2) || p.charAt(j-2) == '.') {
                        dp[i][j] |= dp[i-1][j];
                    }
                } else {
                    dp[i][j] = false;
                }
            }
        }
        
        return dp[m][n];
    }
}
// @lc code=end