Lecture 103:- Wild Card Matching Matching

I apologize for the confusion in my previous response. It seems there was an error in the function name and description. The problem you are referring to is known as Wildcard Matching or Pattern Matching with Wildcards.

Here's a corrected Python function to implement Wildcard Matching:

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def is_match(s, p):    m, n = len(s), len(p)    dp = [[False] * (n + 1) for _ 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] == s[i - 1] or p[j - 1] == '?':                dp[i][j] = dp[i - 1][j - 1]            elif p[j - 1] == '*':                dp[i][j] = dp[i - 1][j] or dp[i][j - 1]    return dp[m][n] # Test the function print(is_match("aa", "a"))           # Output: False print(is_match("aa", "*"))           # Output: True print(is_match("cb", "?a"))          # Output: False print(is_match("adceb", "*a*b"))     # Output: True print(is_match("acdcb", "a*c?b"))    # Output: False

This function uses dynamic programming to solve the Wildcard Matching problem. The dp table is used to store the results of subproblems. The entry dp[i][j] represents whether the substring of s up to index i matches the substring of p up to index j.

The algorithm iteratively fills the dp table based on the characters in the input strings and their matches with wildcard characters (* and ?). The final result is stored in dp[m][n], where m and n are the lengths of strings s and p, respectively.

The time complexity of this solution is O(m * n), where m and n are the lengths of strings s and p, respectively. The space complexity is also O(m * n) due to the dynamic programming table dp.