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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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Link for the ProblemRegular 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> dp(m + 1, vector(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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