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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 Problem** – Wildcard 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:falseExplanation:"a" does not match the entire string "aa".

**Example 2:**

Input:s = "aa", p = "*"Output:trueExplanation:'*' matches any sequence.

**Example 3:**

Input:s = "cb", p = "?a"Output:falseExplanation:'?' 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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