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HARD

Regular Expression Matching

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

Input:

s = "aa", p = "a"

Output:

false

Input:

s = "aa", p = "a*"

Output:

true

Input:

s = "ab", p = ".*"

Output:

true

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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.

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Solution: Dynamic Programming

• Time Complexity: O(mn), where m and n are the lengths of s and p.
• Space Complexity: O(mn), for the DP table.

C++

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