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This topic is the top 100 high frequency questions of LeetCode

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Difficulty level: difficulty

1. Description

Given a string s and a character rule p, please implement a support '.' Matches the regular expression of '*'.

• '.' matches any single character
• '*' matches zero or more preceding elements

The so-called matching is to cover the whole string s, not part of the string.

2. Examples

Example 1

Input: s = "aa" p = "a"
Output: false
 Explanation:"a" Cannot match "aa" The entire string.

Example 2

Input: s = "aa" p = "a*"
Output: true
 Explanation: because '*' Represents the one that can match zero or more preceding elements, The preceding element here is 'a'. Therefore, the string "aa" Can be considered 'a' Again.

Example 3

Input: s = "ab" p = ".*"
Output: true
 Explanation:".*" Indicates that zero or more can be matched('*')Any character('.'). 

Example 4

Input: s = "aab" p = "c*a*b"
Output: true
 Explanation: because '*' Represents zero or more, here 'c' Is 0, 'a' Be repeated once. So you can match strings "aab". 

Example 5

Input: s = "mississippi" p = "mis*is*p*."
Output: false

Constraints:

• 1 <= s.length <= 20
• 1 <= p.length <= 30
• s may be empty and contain only lowercase letters from a-z.
• p may be empty and contain only lowercase letters from a-z and characters And *.
• Ensure that every time the character * appears, it is preceded by a valid character

3. Answer

class RegularExpressionMatching {
    func isMatch(_ s: String, _ p: String) -> Bool {
        let sChars = Array(s), pChars = Array(p)
        var dp = Array(repeating: Array(repeating: false, count: pChars.count + 1), count: sChars.count + 1)
        dp[0][0] = true
        
        for i in 0...pChars.count {
        	// jump over "" vs. "x*" case
            dp[0][i] = i == 0 || i > 1 && dp[0][i - 2] && pChars[i - 1] == "*"
        }
        
        for i in 0...sChars.count {
            for j in 0...pChars.count {
                guard j > 0 else {
                    continue
                }
                
                let pCurrent = pChars[j - 1]
                
                if pCurrent != "*" {
                    dp[i][j] = i > 0 && dp[i - 1][j - 1] && (pCurrent == "." || pCurrent == sChars[i - 1])
                } else {
                    dp[i][j] = dp[i][j - 2] || i > 0 && j > 1 && (sChars[i - 1] == pChars[j - 2] || pChars[j - 2] == ".") && dp[i - 1][j]
                }
            }
        }
        
        return dp[sChars.count][pChars.count]
    }
}

• Main idea: classical two-dimensional dynamic programming
• Time complexity: O(mn)
• Space complexity: O(mn)

The warehouse of the algorithm solution: LeetCode-Swift

go to LeetCode practice

Previous review

There are 5 high-frequency questions in the published articles

LeetCode - #1 sum of two numbers

Difficulty level: easy. The frequency of company interview is as follows:

company	frequency
Amazon	★★★★★★
Facebook	★★★★★
Airbnb	★★★★★
Microsoft	★★★★★
LinkedIn	★★★★

LeetCode - #2 adding two numbers

Difficulty level: medium. The frequency of company interview is as follows:

company	frequency
Microsoft	★★★★
Amazon	★★
Airbnb	★★

LeetCode - #3 longest unrepeated substring

Difficulty level: medium. The frequency of company interview is as follows:

company	frequency
Amazon	★★

LeetCode - #4 find the middle value of two ordered arrays

Difficulty level: difficulty

LeetCode - #5 find the longest image string

Difficulty level: medium. The frequency of company interview is as follows:

company	frequency
Amazon	★★

Difficulty level: easy, medium, difficult
Company use frequency: 1 ~ 6 ★

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