LeetCode Problem – Regular Expression Matching

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 (.)".
 

Constraints:

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.

Java Solution

You can solve this problem using dynamic programming. Here’s a Java solution for the given problem:

public class RegularExpressionMatching {

    public static boolean isMatch(String s, String p) {
        int m = s.length();
        int n = p.length();

        // dp[i][j] represents whether the first i characters in s match the first j characters in p
        boolean[][] dp = new boolean[m + 1][n + 1];

        // Empty pattern matches empty string
        dp[0][0] = true;

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

        // Build the dp array
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                char sChar = s.charAt(i - 1);
                char pChar = p.charAt(j - 1);

                if (pChar == sChar || pChar == '.') {
                    dp[i][j] = dp[i - 1][j - 1];
                } else if (pChar == '*') {
                    dp[i][j] = dp[i][j - 2] || (dp[i - 1][j] && (sChar == p.charAt(j - 2) || p.charAt(j - 2) == '.'));
                }
            }
        }

        return dp[m][n];
    }

    public static void main(String[] args) {
        // Example 1
        System.out.println(isMatch("aa", "a"));  // Output: false

        // Example 2
        System.out.println(isMatch("aa", "a*"));  // Output: true

        // Example 3
        System.out.println(isMatch("ab", ".*"));  // Output: true
    }
}

In this solution, the dp array is used to store whether the first i characters in string s match the first j characters in pattern p. The algorithm iterates through the characters in s and p, updating the dp array based on the matching conditions. The final result is stored in dp[m][n], where m and n are the lengths of the input strings s and p. The examples in the main method demonstrate the usage of the function with the provided test cases.

Python Solution

def isMatch(s, p):
    m, n = len(s), len(p)

    # dp[i][j] represents whether the first i characters in s match the first j characters in p
    dp = [[False] * (n + 1) for _ in range(m + 1)]

    # Empty pattern matches empty string
    dp[0][0] = True

    # Handle patterns with '*'
    for j in range(1, n + 1):
        if p[j - 1] == '*':
            dp[0][j] = dp[0][j - 2]

    # Build the dp array
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            s_char = s[i - 1]
            p_char = p[j - 1]

            if p_char == s_char or p_char == '.':
                dp[i][j] = dp[i - 1][j - 1]
            elif p_char == '*':
                dp[i][j] = dp[i][j - 2] or (dp[i - 1][j] and (s_char == p[j - 2] or p[j - 2] == '.'))

    return dp[m][n]

# Example 1
print(isMatch("aa", "a"))  # Output: False

# Example 2
print(isMatch("aa", "a*"))  # Output: True

# Example 3
print(isMatch("ab", ".*"))  # Output: True

This Python code follows the same dynamic programming approach as the Java solution. The isMatch function returns True if the entire input string s matches the pattern p, and False otherwise. The examples at the end demonstrate the usage of the function with the provided test cases.