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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

We design a function $dfs(i, j)$, which indicates whether the $i$-th character of $s$ matches the $j$-th character of $p$. The answer is $dfs(0, 0)$.

The calculation process of the function $dfs(i, j)$ is as follows:

  • If $j$ has reached the end of $p$, then if $i$ has also reached the end of $s$, the match is successful, otherwise, the match fails.
  • If the next character of $j$ is '*', we can choose to match $0$ $s[i]$ characters, which is $dfs(i, j + 2)$. If $i \lt m$ and $s[i]$ matches $p[j]$, we can choose to match $1$ $s[i]$ character, which is $dfs(i + 1, j)$.
  • If the next character of $j$ is not '*', then if $i \lt m$ and $s[i]$ matches $p[j]$, it is $dfs(i + 1, j + 1)$. Otherwise, the match fails.

During the process, we can use memoization search to avoid repeated calculations.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the lengths of $s$ and $p$ respectively.

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class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        @cache
        def dfs(i, j):
            if j >= n:
                return i == m
            if j + 1 < n and p[j + 1] == '*':
                return dfs(i, j + 2) or (
                    i < m and (s[i] == p[j] or p[j] == '.') and dfs(i + 1, j)
                )
            return i < m and (s[i] == p[j] or p[j] == '.') and dfs(i + 1, j + 1)

        m, n = len(s), len(p)
        return dfs(0, 0)

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class Solution {
    private Boolean[][] f;
    private String s;
    private String p;
    private int m;
    private int n;

    public boolean isMatch(String s, String p) {
        m = s.length();
        n = p.length();
        f = new Boolean[m + 1][n + 1];
        this.s = s;
        this.p = p;
        return dfs(0, 0);
    }

    private boolean dfs(int i, int j) {
        if (j >= n) {
            return i == m;
        }
        if (f[i][j] != null) {
            return f[i][j];
        }
        boolean res = false;
        if (j + 1 < n && p.charAt(j + 1) == '*') {
            res = dfs(i, j + 2)
                || (i < m && (s.charAt(i) == p.charAt(j) || p.charAt(j) == '.') && dfs(i + 1, j));
        } else {
            res = i < m && (s.charAt(i) == p.charAt(j) || p.charAt(j) == '.') && dfs(i + 1, j + 1);
        }
        return f[i][j] = res;
    }
}

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class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.size(), n = p.size();
        int f[m + 1][n + 1];
        memset(f, 0, sizeof f);
        function<bool(int, int)> dfs = [&](int i, int j) -> bool {
            if (j >= n) {
                return i == m;
            }
            if (f[i][j]) {
                return f[i][j] == 1;
            }
            int res = -1;
            if (j + 1 < n && p[j + 1] == '*') {
                if (dfs(i