10. Regular Expression Matching

Problem:

Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'.

'.' Matches any single character.
'*' Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

Note:

s could be empty and contains only lowercase letters a-z. p could be empty and contains only lowercase letters a-z, and characters like . or *.

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 precedeng 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

Solution:

ONE

Cheating with real RegExp matching.

/**
 * @param {string} s
 * @param {string} p
 * @return {boolean}
 */
let isMatch = function (s, p) {
  if (p[0] === "*") {
    return false;
  }
  return new RegExp(`^${p}$`).test(s);
};

TWO

Let f(i, j) be the matching result of s[0…i) and p[0…j).

f(0, j) =
    j == 0 || // empty
    p[j-1] == '*' && f(i, j-2) // matches 0 time, which matches empty string

f(i, 0) = false // pattern must cover the entire input string

f(i, j) =
    if p[j-1] == '.'
        f(i-1, j-1)
    else if p[j-1] == '*'
        f(i, j-2) || // matches 0 time
        f(i-1, j) && (s[i-1] == p[j-2] || p[j-2] == '.') // matches 1 or multiple times
    else
        f(i-1, j-1) && s[i-1] == p[j-1]

/**
 * @param {string} s
 * @param {string} p
 * @return {boolean}
 */
let isMatch = function (s, p) {
  if (p[0] === "*") {
    return false;
  }

  const dp = [[true]];

  for (let j = 2; j <= p.length; j++) {
    dp[0][j] = p[j - 1] === "*" && dp[0][j - 2];
  }

  for (let i = 1; i <= s.length; i++) {
    dp[i] = [];
    for (let j = 1; j <= p.length; j++) {
      switch (p[j - 1]) {
        case ".":
          dp[i][j] = dp[i - 1][j - 1];
          break;
        case "*":
          dp[i][j] =
            dp[i][j - 2] ||
            (dp[i - 1][j] && (p[j - 2] === "." || s[i - 1] === p[j - 2]));
          break;
        default:
          dp[i][j] = dp[i - 1][j - 1] && s[i - 1] === p[j - 1];
      }
    }
  }

  return !!dp[s.length][p.length];
};

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