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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: 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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