Regular Expression Matching LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Regular Expression Matching in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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About LeetCode

LeetCode is one of the most well-known online judge platforms to help you enhance your skills, expand your knowledge and prepare for technical interviews. 

LeetCode is for software engineers who are looking to practice technical questions and advance their skills. Mastering the questions in each level on LeetCode is a good way to prepare for technical interviews and keep your skills sharp. They also have a repository of solutions with the reasoning behind each step.

LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either Easy, Medium, or Hard.

LeetCode problems focus on algorithms and data structures. Here is some topic you can find problems on LeetCode:

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  • Breadth-First Search
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  • BST (Binary Search Tree)
  • Database
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  • Recursion, etc.

Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

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]

31 thoughts on “Regular Expression Matching LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]”

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