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 1: Memoization Search

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.

Python3

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)

Java

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;
    }
}

C++

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, j + 2) or (i < m and (s[i] == p[j] or p[j] == '.') and dfs(i + 1, j))) {
                    res = 1;
                }
            } else if (i < m and (s[i] == p[j] or p[j] == '.') and dfs(i + 1, j + 1)) {
                res = 1;
            }
            f[i][j] = res;
            return res == 1;
        };
        return dfs(0, 0);
    }
};

Go

func isMatch(s string, p string) bool {
	m, n := len(s), len(p)
	f := make([][]int, m+1)
	for i := range f {
		f[i] = make([]int, n+1)
	}
	var dfs func(i, j int) bool
	dfs = func(i, j int) bool {
		if j >= n {
			return i == m
		}
		if f[i][j] != 0 {
			return f[i][j] == 1
		}
		res := -1
		if j+1 < n && p[j+1] == '*' {
			if dfs(i, j+2) || (i < m && (s[i] == p[j] || p[j] == '.') && dfs(i+1, j)) {
				res = 1
			}
		} else if i < m && (s[i] == p[j] || p[j] == '.') && dfs(i+1, j+1) {
			res = 1
		}
		f[i][j] = res
		return res == 1
	}
	return dfs(0, 0)
}

Rust

impl Solution {
    pub fn is_match(s: String, p: String) -> bool {
        let (m, n) = (s.len(), p.len());
        let mut f = vec![vec![0; n + 1]; m + 1];

        fn dfs(
            s: &Vec<char>,
            p: &Vec<char>,
            f: &mut Vec<Vec<i32>>,
            i: usize,
            j: usize,
            m: usize,
            n: usize,
        ) -> bool {
            if j >= n {
                return i == m;
            }
            if f[i][j] != 0 {
                return f[i][j] == 1;
            }
            let mut res = -1;
            if j + 1 < n && p[j + 1] == '*' {
                if dfs(s, p, f, i, j + 2, m, n)
                    || (i < m && (s[i] == p[j] || p[j] == '.') && dfs(s, p, f, i + 1, j, m, n))
                {
                    res = 1;
                }
            } else if i < m && (s[i] == p[j] || p[j] == '.') && dfs(s, p, f, i + 1, j + 1, m, n) {
                res = 1;
            }
            f[i][j] = res;
            res == 1
        }

        dfs(
            &s.chars().collect(),
            &p.chars().collect(),
            &mut f,
            0,
            0,
            m,
            n,
        )
    }
}

JavaScript

/**
 * @param {string} s
 * @param {string} p
 * @return {boolean}
 */
var isMatch = function (s, p) {
    const m = s.length;
    const n = p.length;
    const f = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
    const dfs = (i, j) => {
        if (j >= n) {
            return i === m;
        }
        if (f[i][j]) {
            return f[i][j] === 1;
        }
        let res = -1;
        if (j + 1 < n && p[j + 1] === '*') {
            if (dfs(i, j + 2) || (i < m && (s[i] === p[j] || p[j] === '.') && dfs(i + 1, j))) {
                res = 1;
            }
        } else if (i < m && (s[i] === p[j] || p[j] === '.') && dfs(i + 1, j + 1)) {
            res = 1;
        }
        f[i][j] = res;
        return res === 1;
    };
    return dfs(0, 0);
};

C#

public class Solution {
    private string s;
    private string p;
    private int m;
    private int n;
    private int[,] f;

    public bool IsMatch(string s, string p) {
        m = s.Length;
        n = p.Length;
        f = new int[m + 1, n + 1];
        this.s = s;
        this.p = p;
        return dfs(0, 0);
    }

    private bool dfs(int i, int j) {
        if (j >= n) {
            return i == m;
        }
        if (f[i, j] != 0) {
            return f[i, j] == 1;
        }
        int res = -1;
        if (j + 1 < n && p[j + 1] == '*') {
            if (dfs(i, j + 2) || (i < m && (s[i] == p[j] || p[j] == '.') && dfs(i + 1, j))) {
                res = 1;
            }
        } else if (i < m && (s[i] == p[j] || p[j] == '.') && dfs(i + 1, j + 1)) {
            res = 1;
        }
        f[i, j] = res;
        return res == 1;
    }
}

Solution 2: Dynamic Programming

We can convert the memoization search in Solution 1 into dynamic programming.

Define $f[i][j]$ to represent whether the first $i$ characters of string $s$ match the first $j$ characters of string $p$. The answer is $f[m][n]$. Initialize $f[0][0] = true$, indicating that the empty string and the empty regular expression match.

Similar to Solution 1, we can discuss different cases.

If $p[j - 1]$ is '*', we can choose to match $0$ $s[i - 1]$ characters, which is $f[i][j] = f[i][j - 2]$. If $s[i - 1]$ matches $p[j - 2]$, we can choose to match $1$ $s[i - 1]$ character, which is $f[i][j] = f[i][j] \lor f[i - 1][j]$.
If $p[j - 1]$ is not '*', then if $s[i - 1]$ matches $p[j - 1]$, it is $f[i][j] = f[i - 1][j - 1]$. Otherwise, the match fails.