10. Regular Expression Matching

Description

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.

 

Solutions

Solution: Recursion

• Time complexity: O(mn)
• Space complexity: O(mn)

 

JavaScript

/**
 * @param {string} s
 * @param {string} p
 * @return {boolean}
 */
const isMatch = function (s, p) {
  const hashMap = new Map();
  const checkRegex = (index, j) => {
    const cacheKey = `${index},${j}`;

    if (hashMap.has(cacheKey)) return hashMap.get(cacheKey);
    if (j === p.length) return index === s.length;

    const isFirstMatch = index < s.length && (s[index] === p[j] || p[j] === '.');
    const isCheckPreceding = j + 1 < p.length && p[j + 1] === '*';
    const result = isCheckPreceding
      ? checkRegex(index, j + 2) || (isFirstMatch && checkRegex(index + 1, j))
      : isFirstMatch && checkRegex(index + 1, j + 1);

    hashMap.set(cacheKey, result);
    return result;
  };

  return checkRegex(0, 0);
};