# 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.

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.

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.

• class Solution {
public boolean isMatch(String s, String p) {
int m = s.length(), n = p.length();
boolean[][] f = new boolean[m + 1][n + 1];
f[0][0] = true;
for (int i = 0; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (p.charAt(j - 1) == '*') {
f[i][j] = f[i][j - 2];
if (i > 0 && (p.charAt(j - 2) == '.' || p.charAt(j - 2) == s.charAt(i - 1))) {
f[i][j] |= f[i - 1][j];
}
} else if (i > 0
&& (p.charAt(j - 1) == '.' || p.charAt(j - 1) == s.charAt(i - 1))) {
f[i][j] = f[i - 1][j - 1];
}
}
}
return f[m][n];
}
}

• class Solution {
public:
bool isMatch(string s, string p) {
int m = s.size(), n = p.size();
bool f[m + 1][n + 1];
memset(f, false, sizeof f);
f[0][0] = true;
for (int i = 0; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (p[j - 1] == '*') {
f[i][j] = f[i][j - 2];
if (i && (p[j - 2] == '.' || p[j - 2] == s[i - 1])) {
f[i][j] |= f[i - 1][j];
}
} else if (i && (p[j - 1] == '.' || p[j - 1] == s[i - 1])) {
f[i][j] = f[i - 1][j - 1];
}
}
}
return f[m][n];
}
};

• class Solution:
def isMatch(self, s: str, p: str) -> bool:
m, n = len(s), len(p)
f = [[False] * (n + 1) for _ in range(m + 1)]
f[0][0] = True
for i in range(m + 1):
for j in range(1, n + 1):
if p[j - 1] == "*":
f[i][j] = f[i][j - 2]
if i > 0 and (p[j - 2] == "." or s[i - 1] == p[j - 2]):
f[i][j] |= f[i - 1][j]
elif i > 0 and (p[j - 1] == "." or s[i - 1] == p[j - 1]):
f[i][j] = f[i - 1][j - 1]
return f[m][n]


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

• /**
* @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(false));
f[0][0] = true;
for (let i = 0; i <= m; ++i) {
for (let j = 1; j <= n; ++j) {
if (p[j - 1] === '*') {
f[i][j] = f[i][j - 2];
if (i && (p[j - 2] === '.' || p[j - 2] === s[i - 1])) {
f[i][j] |= f[i - 1][j];
}
} else if (i && (p[j - 1] === '.' || p[j - 1] === s[i - 1])) {
f[i][j] = f[i - 1][j - 1];
}
}
}
return f[m][n];
};


• public class Solution {
public bool IsMatch(string s, string p) {
int m = s.Length, n = p.Length;
bool[,] f = new bool[m + 1, n + 1];
f[0, 0] = true;
for (int i = 0; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (p[j - 1] == '*') {
f[i, j] = f[i, j - 2];
if (i > 0 && (p[j - 2] == '.' || p[j - 2] == s[i - 1])) {
f[i, j] |= f[i - 1, j];
}
} else if (i > 0 && (p[j - 1] == '.' || p[j - 1] == s[i - 1])) {
f[i, j] = f[i - 1, j - 1];
}
}
}
return f[m, n];
}
}

• impl Solution {
pub fn is_match(s: String, p: String) -> bool {
let n = s.len();
let m = p.len();
let s = s.chars().collect::<Vec<char>>();
let p = p.chars().collect::<Vec<char>>();

let mut dp = vec![vec![false; m + 1]; n + 1];

// Initialize the dp vector
dp[0][0] = true;

for i in 1..=m {
if p[i - 1] == '*' {
dp[0][i] = dp[0][i - 2];
}
}

// Begin the actual dp process
for i in 1..=n {
for j in 1..=m {
if s[i - 1] == p[j - 1] || p[j - 1] == '.' {
dp[i][j] = dp[i - 1][j - 1];
}
if p[j - 1] == '*' {
if j >= 2 && (s[i - 1] == p[j - 2] || p[j - 2] == '.') {
dp[i][j] = dp[i - 1][j] || dp[i][j - 2];
} else if j >= 2 && s[i - 1] != p[j - 2] {
dp[i][j] = dp[i][j - 2];
}
}
}
}

dp[n][m]
}
}


• class Solution {
/**
* @param string $s * @param string$p
* @return boolean
*/

function isMatch($s,$p) {
$m = strlen($s);
$n = strlen($p);

$dp = array_fill(0,$m + 1, array_fill(0, $n + 1, false));$dp[0][0] = true;

for ($j = 1;$j <= $n;$j++) {
if ($p[$j - 1] == '*') {
$dp[0][$j] = $dp[0][$j - 2];
}
}

for ($i = 1;$i <= $m;$i++) {
for ($j = 1;$j <= $n;$j++) {
if ($p[$j - 1] == '.' || $p[$j - 1] == $s[$i - 1]) {
$dp[$i][$j] =$dp[$i - 1][$j - 1];
} elseif ($p[$j - 1] == '*') {
$dp[$i][$j] =$dp[$i][$j - 2];
if ($p[$j - 2] == '.' || $p[$j - 2] == $s[$i - 1]) {
$dp[$i][$j] =$dp[$i][$j] || $dp[$i - 1][$j]; } } else {$dp[$i][$j] = false;
}
}
}

return $dp[$m][\$n];
}
}