Last updated on October 10th, 2024 at 01:59 am

Here, We see Regular Expression Matching LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc., with different approaches.

List of all LeetCode Solution

Regular Expression Matching LeetCode Solution

Problem Statement

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 (.)".
 

1. Regular Expression Matching Leetcode Solution C++

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

2. Regular Expression Matching Leetcode Solution Java

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

3. Regular Expression Matching Leetcode Solution JavaScript

var isMatch = function(s, p) {
     var lenS = s.length;
  var lenP = p.length;
  var map = {};

  return check(0, 0);

  function check(idxS, idxP) {
    if (map[idxS + ':' + idxP] !== undefined) return map[idxS + ':' + idxP];
    if (idxS > lenS) return false;
    if (idxS === lenS && idxP === lenP) return true;

    if (p[idxP] === '.' || p[idxP] === s[idxS]) {
      map[idxS + ':' + idxP] = p[idxP + 1] === '*' ?
        check(idxS + 1, idxP) || check(idxS, idxP + 2) :
        check(idxS + 1, idxP + 1);
    } else {
      map[idxS + ':' + idxP] = p[idxP + 1] === '*' ?
        check(idxS, idxP + 2) : false;
    }
    return map[idxS + ':' + idxP];
    }
};

4. Regular Expression Matching Leetcode Solution Python

class Solution(object):
    def isMatch(self, s, p):
        dp = [[False] * (len(s) + 1) for _ in range(len(p) + 1)]
        dp[0][0] = True
        for i in range(1, len(p)):
            dp[i + 1][0] = dp[i - 1][0] and p[i] == '*'
        for i in range(len(p)):
            for j in range(len(s)):
                if p[i] == '*':
                    dp[i + 1][j + 1] = dp[i - 1][j + 1] or dp[i][j + 1]
                    if p[i - 1] == s[j] or p[i - 1] == '.':
                        dp[i + 1][j + 1] |= dp[i + 1][j]
                else:
                    dp[i + 1][j + 1] = dp[i][j] and (p[i] == s[j] or p[i] == '.')
        return dp[-1][-1]