0010. 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 (.)".

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:

• 0 <= s.length <= 20
• 0 <= 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.

Analysis

1. dp[i][j]: s[0:i] matches p[0:j]
2. if p[i] == . then matches everything from s, so dp[i][j] = true
3. if p[i] == * then if dp[i][j-2] (skip the last one and check if previous one matches) or s[i] == p[j-1] or p[j-1] (match anything) set dp[i][j] = true

Code

class Solution {
public:
    bool isMatch(string s, string p) {
        int n = s.length(), m = p.length();
        vector<vector<bool>> f(n + 1, vector<bool>(m + 1, false));
        s = " " + s;
        p = " " + p;
        f[0][0] = true;
        for (int i = 0; i <= n; i++)
            for (int j = 1; j <= m; j++) {
                if (i > 0 && (s[i] == p[j] || p[j] == '.'))
                    f[i][j] = f[i - 1][j - 1];
                if (p[j] == '*') {
                    if (j >= 2)
                        f[i][j] = f[i][j - 2];
                    if (i > 0 && (s[i] == p[j - 1] || p[j - 1] == '.'))
                        f[i][j] = f[i][j] | f[i - 1][j]; // if anything happened to set f[i][j] = true, here will ignore if f[i-1][j] ?= true
                }
            }
        return f[n][m];
    }
};

---

Last update: April 1, 2022