LeetCode 44. Wildcard Matching

Description

https://leetcode.com/problems/wildcard-matching/

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'.

Example 4:

Input: s = "adceb", p = "*a*b"
Output: true
Explanation: The first '*' matches the empty sequence, while the second '*' matches the substring "dce".

Example 5:

Input: s = "acdcb", p = "a*c?b"
Output: false

Constraints:

  • 0 <= s.length, p.length <= 2000
  • s contains only lowercase English letters.
  • p contains only lowercase English letters, '?' or '*'.

Explanation

Dynamic programing dp[i][j] to see if first i characters of s matches the first j characters of p.

Python Solution

class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        m = len(s)
        n = len(p)
        
        dp = [[False for j in range(n + 1)] for i in range(m + 1)]
        
        dp[0][0] = True
        
        for j in range(1, n + 1):
            if p[j - 1] == '*':
                dp[0][j] = dp[0][j - 1]
                                
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if p[j - 1] == '*':
                    dp[i][j] = dp[i - 1][j] or dp[i][j - 1]
                else:
                    if s[i - 1] == p[j - 1] or p[j - 1] == '?':
                        dp[i][j] = dp[i - 1][j - 1]
                    
        return dp[m][n]
  • Time Complexity: O(MN).
  • Space Complexity: O(MN).

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