Regular Expression Matching Hard ​

Question ​

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).

Input: s = "aa", p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

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".

Input: s = "ab", p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

1 <= s.length <= 20
1 <= p.length <= 20
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.

Solution ​

def isMatch(s, p):
    dp = [[False] * (len(p) + 1) for _ in range(len(s) + 1)]
    dp[0][0] = True

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

    for i in range(1, len(s) + 1):
        for j in range(1, len(p) + 1):
            if p[j - 1] == s[i - 1] or p[j - 1] == '.':
                dp[i][j] = dp[i - 1][j - 1]
            elif p[j - 1] == '*':
                dp[i][j] = dp[i][j - 2] or (dp[i - 1][j] and (s[i - 1] == p[j - 2] or p[j - 2] == '.'))

    return dp[len(s)][len(p)]

public boolean isMatch(String s, String p) {
    boolean[][] dp = new boolean[s.length() + 1][p.length() + 1];
    dp[0][0] = true;

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

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

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

var isMatch = function (s, p) {
  const dp = Array.from({ length: s.length + 1 }, () =>
    Array(p.length + 1).fill(false)
  );
  dp[0][0] = true;

  for (let j = 1; j <= p.length; j++) {
    if (p[j - 1] === "*") {
      dp[0][j] = dp[0][j - 2];
    }
  }

  for (let i = 1; i <= s.length; i++) {
    for (let j = 1; j <= p.length; j++) {
      if (p[j - 1] === s[i - 1] || p[j - 1] === ".") {
        dp[i][j] = dp[i - 1][j - 1];
      } else if (p[j - 1] === "*") {
        dp[i][j] =
          dp[i][j - 2] ||
          (dp[i - 1][j] && (s[i - 1] === p[j - 2] || p[j - 2] === "."));
      }
    }
  }

  return dp[s.length][p.length];
};

Notes ​

You are given an input string s and a pattern string p. You need to implement regular expression matching with support for the following special characters:

• .: Matches any single character.
• *: Matches zero or more of the preceding element.

The goal is to determine if the pattern p can match the entire input string s.

Time and Space Complexity Analysis ​

The time complexity of the given solutions is O(m * n), where m is the length of the input string s, and n is the length of the pattern string p. We iterate through a 2D DP array of size (m+1) x (n+1).

The space complexity for all solutions is O(m * n) because of the DP array.

These solutions use dynamic programming to efficiently determine if the pattern p can match the input string s with support for '.' and '*'.