LeetCode-in-All

10. Regular Expression Matching

Hard

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 (.)”.

Example 4:

Input: s = “aab”, p = “c*a*b”

Output: true

Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore, it matches “aab”.

Example 5:

Input: s = “mississippi”, p = “mis*is*p*.”

Output: false

Constraints:

• 1 <= s.length <= 20
• 1 <= p.length <= 30
• 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}
 */
var isMatch = function (s, p) {
    const cache = Array.from({ length: s.length + 1 }, () => Array(p.length + 1).fill(null))

    function helper(i, j) {
        if (j === p.length) {
            return i === s.length
        }

        if (cache[i][j] !== null) {
            return cache[i][j]
        }

        const firstMatch = i < s.length && (s[i] === p[j] || p[j] === '.')

        let result
        if (j + 1 < p.length && p[j + 1] === '*') {
            result = (firstMatch && helper(i + 1, j)) || helper(i, j + 2)
        } else {
            result = firstMatch && helper(i + 1, j + 1)
        }

        cache[i][j] = result
        return result
    }

    return helper(0, 0)
}

export { isMatch }

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