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Link for the Problem – Regular Expression Matching– LeetCode Problem
Regular Expression Matching– LeetCode Problem
Problem:
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 <= 30
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.
Regular Expression Matching– LeetCode Solutions
class Solution { public: bool isMatch(string s, string p) { const int m = s.length(); const int n = p.length(); // dp[i][j] := true if s[0..i) matches p[0..j) vector<vector<bool>> dp(m + 1, vector<bool>(n + 1)); dp[0][0] = true; auto isMatch = [](char c1, char c2) { return c1 == c2 || c2 == '.'; }; for (int i = 0; i <= m; ++i) for (int j = 1; j <= n; ++j) if (p[j - 1] == '*') { const bool noRepeat = dp[i][j - 2]; const bool doRepeat = i && isMatch(s[i - 1], p[j - 2]) && dp[i - 1][j]; dp[i][j] = noRepeat || doRepeat; } else { dp[i][j] = i && isMatch(s[i - 1], p[j - 1]) && dp[i - 1][j - 1]; } return dp[m][n]; } };
class Solution { public boolean isMatch(String s, String p) { final int m = s.length(); final int n = p.length(); // dp[i][j] := true if s[0..i) matches p[0..j) boolean[][] dp = new boolean[m + 1][n + 1]; dp[0][0] = true; for (int i = 0; i <= m; ++i) for (int j = 1; j <= n; ++j) if (p.charAt(j - 1) == '*') { final boolean noRepeat = dp[i][j - 2]; final boolean doRepeat = i > 0 && isMatch(s.charAt(i - 1), p.charAt(j - 2)) && dp[i - 1][j]; dp[i][j] = noRepeat || doRepeat; } else { dp[i][j] = i > 0 && isMatch(s.charAt(i - 1), p.charAt(j - 1)) && dp[i - 1][j - 1]; } return dp[m][n]; } private boolean isMatch(char c1, char c2) { return c1 == c2 || c2 == '.'; } }
class Solution: def isMatch(self, s: str, p: str) -> bool: m = len(s) n = len(p) # dp[i][j] := True if s[0..i) matches p[0..j) dp = [[False] * (n + 1) for _ in range(m + 1)] dp[0][0] = True def isMatch(c1, c2): return c1 == c2 or c2 == '.' for i in range(m + 1): for j in range(1, n + 1): if p[j - 1] == '*': noRepeat = dp[i][j - 2] doRepeat = i > 0 and isMatch( s[i - 1], p[j - 2]) and dp[i - 1][j] dp[i][j] = noRepeat or doRepeat else: dp[i][j] = i > 0 and isMatch( s[i - 1], p[j - 1]) and dp[i - 1][j - 1] return dp[m][n]
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