In this post, we are going to solve the 10. Regular Expression Matching problem of Leetcode. This problem 10. Regular Expression Matching is a Leetcode hard level problem. Let’s see code, 10. Regular Expression Matching.

Problem

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 <= 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.

Now, let’s see the code of 10. Regular Expression Matching – Leetcode Solution.

Regular Expression Matching – Leetcode Solution

10. Regular Expression Matching – Solution in Java

class Solution {
    public boolean isMatch(String s, String p) {
            if (p == null || p.length() == 0) 
                return (s == null || s.length() == 0);

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

            for (int j=1; j<=p.length(); j++) {
                for (int i=1; i<=s.length(); i++) {
                    if (p.charAt(j-1) == s.charAt(i-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] || ((s.charAt(i-1) == p.charAt(j-2) || p.charAt(j-2) == '.') && dp[i-1][j]); 
                }
            }
            return dp[s.length()][p.length()];
        }
}

10. Regular Expression Matching – Solution in C++

class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.size(), n = p.size();
        vector<vector<bool>> dp(m + 1, vector<bool>(n + 1, false));
        dp[0][0] = true;
        for (int i = 0; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (p[j - 1] == '*') {
                    dp[i][j] = dp[i][j - 2] || (i && dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.'));
                } else {
                    dp[i][j] = i && dp[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '.');
                }
            }
        }
        return dp[m][n];
    }
};

10. Regular Expression Matching – Solution in Python

class Solution:
     def isMatch(self, s: str, p: str) -> bool:
        s, p = ' '+ s, ' '+ p
        lenS, lenP = len(s), len(p)
        dp = [[0]*(lenP) for i in range(lenS)]
        dp[0][0] = 1

        for j in range(1, lenP):
            if p[j] == '*':
                dp[0][j] = dp[0][j-2]

        for i in range(1, lenS):
            for j in range(1, lenP):
                if p[j] in {s[i], '.'}:
                    dp[i][j] = dp[i-1][j-1]
                elif p[j] == "*":
                    dp[i][j] = dp[i][j-2] or int(dp[i-1][j] and p[j-1] in {s[i], '.'})

        return bool(dp[-1][-1])
        

Note: This problem 10. Regular Expression Matching is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.