0010. Regular Expression Matching

https://leetcode.com/problems/regular-expression-matching

Description

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 (.)".

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

Constraints:

• 1 <= s.length <= 20

• 1 <= p.length <= 30

• 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.

ac

class Solution {
    public boolean isMatch(String s, String p) {
        // edge cases
        if (s == null || p == null) return false;

        // initialize
        boolean[][] dp = new boolean[s.length() + 1][p.length() + 1];
        dp[0][0] = true;
        for (int i = 2; i < dp[0].length; i++) {
            if (p.charAt(i-1) == '*') {
                dp[0][i] = dp[0][i-2];
            }
        }

        // walk, transform function
        for (int i = 1; i < dp.length; i++) {
            for (int j = 1; j < dp[0].length; j++) {
                if (p.charAt(j-1) == s.charAt(i-1) || p.charAt(j-1) == '.') {
                    dp[i][j] = dp[i-1][j-1];
                }
                if (p.charAt(j-1) == '*') {
                    if (p.charAt(j-2) != s.charAt(i-1) && p.charAt(j-2) != '.') {
                        dp[i][j] = dp[i][j-2];
                    } else {
                        dp[i][j] = (dp[i-1][j] || dp[i][j-1] || dp[i][j-2]);
                    }
                }
            }
        }

        // result
        return dp[s.length()][p.length()];
    }
}

/*
state: dp[i][j], s.substring(0, i) and p.substring(0, j) is match

function: 
1. if p.charAt(j-1) == s.charAt(i-1) , dp[i][j] = dp[i-1][j-1];
2. if p.charAt(j-1) == '.' , dp[i][j] = dp[i-1][j-1];
3. if p.charAt(j-1) == '*' :
    (1) if p.charAt(j-2) != s.charAt(i-1) && p.charAt(j-2) != '.', dp[i][j] = dp[i][j-2] // a* means zero a
    (2) if p.charAt(j-2) == s.charAt(i-1) || p.charAt(j-2) == '.':
        1) dp[i][j] = dp[i-1][j]  // a* means multiple a
        2) dp[i][j] = dp[i][j-1]  // a* means single a
        3) dp[i][j] = dp[i][j-2]  // a* means zero a

initialize: dp[0][0] = true;

result: dp[s.length()][p.length()]

*/