How do you solve Wildcard Matching on LeetCode?

Wildcard Matching (LeetCode #44) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n * m) space complexity. Key insight: Understanding how '*' and '?' work is crucial.

What is the brute force solution for Wildcard Matching?

The brute force approach for Wildcard Matching is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries to match the pattern with the string character by character. We recursively check every possible match for '*' and '?' while ensuring the entire string is covered. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Wildcard Matching?

Wildcard Matching uses the following concepts: String, Dynamic Programming, Greedy, Recursion. The recommended patterns to study are: Dynamic Programming, Recursion.

What are the interview tips for Wildcard Matching?

Explain your thought process clearly. Discuss edge cases like empty strings and patterns. Practice writing the DP table on paper before coding.

What are common mistakes when solving Wildcard Matching?

Not handling the case of multiple '*' correctly. Forgetting to check for empty strings in the pattern.

#44

Wildcard Matching

Hard

StringDynamic ProgrammingGreedyRecursionDynamic ProgrammingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(n * m)

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Intuition

Time O(n * m)Space O(n * m)

Using dynamic programming, we can build a table to store results of subproblems. This allows us to avoid redundant calculations and efficiently determine if the pattern matches the string.

⚙️

Algorithm

5 steps

1Step 1: Create a 2D DP table where dp[i][j] indicates if s[0..i-1] matches p[0..j-1].
2Step 2: Initialize dp[0][0] to true (empty string matches empty pattern).
3Step 3: Fill the first row for patterns that start with '*'.
4Step 4: Iterate through each character of the string and pattern, updating the DP table based on matches and '*' handling.
5Step 5: Return the value at dp[s.length()][p.length()] for the final result.

solution.py13 lines

1def isMatch(s, p):
2    dp = [[False] * (len(p) + 1) for _ in range(len(s) + 1)]
3    dp[0][0] = True
4    for j in range(1, len(p) + 1):
5        if p[j - 1] == '*':
6            dp[0][j] = dp[0][j - 1]
7    for i in range(1, len(s) + 1):
8        for j in range(1, len(p) + 1):
9            if p[j - 1] == s[i - 1] or p[j - 1] == '?':
10                dp[i][j] = dp[i - 1][j - 1]
11            elif p[j - 1] == '*':
12                dp[i][j] = dp[i][j - 1] or dp[i - 1][j]
13    return dp[len(s)][len(p)]

ℹ

Complexity note: The time complexity is O(n * m) because we fill a table of size (n+1) x (m+1), where n is the length of the string and m is the length of the pattern.

1Understanding how '*' and '?' work is crucial.
2Dynamic programming can significantly reduce redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.