LeetCode in Kotlin

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

class Solution {
    fun isMatch(s: String, p: String): Boolean {
        val n = s.length
        val m = p.length
        return solve(n - 1, m - 1, s, p)
    }

    private fun solve(i: Int, j: Int, s: String, p: String): Boolean {
        if (j < 0) {
            return i < 0
        }
        if (i < 0) {
            return p[j] == '*' && solve(i, j - 2, s, p)
        }
        // simple char matching
        // if s char matchs with p char or it can be '.'
        if (s[i] == p[j] || p[j] == '.') {
            return solve(i - 1, j - 1, s, p)
        }
        return if (p[j] == '*') {
            // if s char matches with p char or it can be '.'
            if (s[i] == p[j - 1] || p[j - 1] == '.') {
                solve(i - 1, j, s, p) || solve(i, j - 2, s, p)
            } else {
                solve(
                    i,
                    j - 2,
                    s,
                    p,
                )
            }
        } else {
            false
        }
    }
}

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