10. Regular Expression Matching

Problem:

---

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 precedeng 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

Analysis:

DP,

State: dp[i][j] 0 to  i -1 in s and 0 to j - 1 in p matches.

Transition:

1. p[j - 1] != ‘*’
        if(p[j - 1] == s[i - 1] || p[j - 1] == ‘.’) dp[i][j] = dp[i - 1][j - 1];
2. P[j - 1] == ‘*’
if (p[j - 2] != s[i - 1]) dp[i][j] = dp[i][j - 2] // * matches empty
If (p[j - 2] == s[i - 1] || p[j - 2] == ‘.’)
dp[i][j] = dp[i][j - 2] // empty
|| dp[i][j - 1] // once
|| dp[i - 1][j] // more than once
dp[i][j] = dp[i - 1][j] is a bit hard to understand. For instance,

s: acbbb

p: acb* 

we want to know whether acbb and acb* match, and then reduce to acb and acb* match. 

Solution:

        

class Solution {
    public boolean isMatch(String s, String p) {
        int m = s.length(), n = p.length();
        boolean[][] dp = new boolean[m + 1][n + 1];
        dp[0][0] = true;
        for (int i = 1; i <= n; i++) {
            if(p.charAt(i - 1) == '*') {
                dp[0][i] = dp[0][i - 2];
            }
        }
        
        for (int i = 1; i <= m; i++)
            for (int j = 1; j <= n; j++) {
                if (p.charAt(j - 1) == '*') {
                    if (p.charAt(j - 2) == s.charAt(i - 1) || p.charAt(j - 2) == '.') {
                        dp[i][j] = dp[i][j - 1] || dp[i - 1][j] || dp[i][j - 2]; 
                    } else {
                        dp[i][j] = dp[i][j - 2];
                    }
                } else if (p.charAt(j - 1) == '.' || p.charAt(j - 1) == s.charAt(i - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                }
            }
        return dp[m][n];
    }
}