10. Regular Expression Matching Leetcode solution in Javascript (JS)

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 <= 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

/**
 * @param {string} s
 * @param {string} p
 * @return {boolean}
 */
const isMatch = (string, pattern) => {
    // early return when pattern is empty
    if (!pattern) {
		// returns true when string and pattern are empty
		// returns false when string contains chars with empty pattern
        return !string;
    }
    
	// check if the current char of the string and pattern match when the string has chars
    const hasFirstCharMatch = Boolean(string) && (pattern[0] === '.' || pattern[0] === string[0]);

    // track when the next character * is next in line in the pattern
    if (pattern[1] === '*') {
        // if next pattern match (after *) is fine with current string, then proceed with it (s, p+2).  That's because the current pattern may be skipped.
        // otherwise check hasFirstCharMatch. That's because if we want to proceed with the current pattern, we must be sure that the current pattern char matches the char
		// If hasFirstCharMatch is true, then do the recursion with next char and current pattern (s+1, p).  That's because current char matches the pattern char. 
        return (
            isMatch(string, pattern.slice(2)) || 
            (hasFirstCharMatch && isMatch(string.slice(1), pattern))
        );
    }
    
    // now we know for sure that we need to do 2 simple actions
	// check the current pattern and string chars
	// if so, then can proceed with next string and pattern chars (s+1, p+1)
    return hasFirstCharMatch ? isMatch(string.slice(1), pattern.slice(1)) : false;
};