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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

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 {
        #[allow(dead_code)]
        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];
        }
    }
    
    

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