Wildcard Matching 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 Wildcard 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 ProblemWildcard Matching– LeetCode Problem

Wildcard Matching– LeetCode Problem

Problem:

Given an input string (s) and a pattern (p), implement wildcard pattern matching with support for '?' and '*' where:

  • '?' Matches any single character.
  • '*' Matches any sequence of characters (including the empty sequence).

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 = "*"
Output: true
Explanation: '*' matches any sequence.

Example 3:

Input: s = "cb", p = "?a"
Output: false
Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'.

Constraints:

  • 0 <= s.length, p.length <= 2000
  • s contains only lowercase English letters.
  • p contains only lowercase English letters, '?' or '*'.
Wildcard Matching– LeetCode Solutions
Wildcard Matching in C++
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 matchEmpty = dp[i][j - 1];
          const bool matchNew = i && dp[i - 1][j];
          dp[i][j] = matchEmpty || matchNew;
        } else {
          dp[i][j] = i && isMatch(s[i - 1], p[j - 1]) && dp[i - 1][j - 1];
        }

    return dp[m][n];
  }
};
Wildcard Matching in Java
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 matchEmpty = dp[i][j - 1];
          final boolean matchNew = i > 0 && dp[i - 1][j];
          dp[i][j] = matchEmpty || matchNew;
        } 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 == '?';
  }
}
Wildcard Matching in Python
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: chr, c2: chr) -> bool:
      return c1 == c2 or c2 == '?'

    for i in range(m + 1):
      for j in range(1, n + 1):
        if p[j - 1] == '*':
          matchEmpty = dp[i][j - 1]
          matchNew = i and dp[i - 1][j]
          dp[i][j] = matchEmpty or matchNew
        else:
          dp[i][j] = i and isMatch(s[i - 1], p[j - 1]) and dp[i-1][j-1]

    return dp[m][n]

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