Leetcode Problem 44 Wildcard Matching.txt

44. Wildcard Matching
Given an input string (s) and a pattern (p), implement wildcard pattern matching with support for '?' and '*'.

'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).
The matching should cover the entire input string (not partial).

Note:

s could be empty and contains only lowercase letters a-z.
p could be empty and contains only lowercase letters a-z, and characters like ? or *.
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


Hint to solve: Use dynamic programming. Make rules for ?, * and character in the pattern.


class Solution:
    def isMatch(self, s: str, p: str) -> bool:
        
        #If pattern is empty string and the given data is empty then return True
        if(len(p) == 0 and len(s) == 0):
            return True
        #If the pattern is empty but the given string is not empty return False
        if(len(p) == 0 and len(s) != 0):
            return False
        
        #Compress the pattern that contains more than 1 consequitive stars into single star. Which means the same thing
        compactPattern = ""
        isFirstStar = True
        for i in range(len(p)):
            if p[i] == '*' and isFirstStar:
                compactPattern += p[i]
                isFirstStar = False
            elif p[i] != '*':
                compactPattern += p[i]
                isFirstStar = True
        
        p = compactPattern
        
        #Initialize a 2d matrix of cols = len of pattern+1 and rows = len of string + 1
        #0th index is for empty string and empty pattern
        matrix = [[False for _ in range(len(p) + 1)] for _ in range(len(s) + 1)]
        
        #This will be always true because empty pattern always matches with empty string
        matrix[0][0] = True
        
        #If the first char of pattern is star then for empty string this will be True
        if p[0] == '*':
            matrix[0][1] = True
        
        #If the current pattern character is ? then take element from Diagonal
        #If the current pattern character is alphabet check if it's equal to string character then take value from Diagonal
        #If the current pattern character is star then take value of top or left cell
        #Else set if to false
        for row in range(1, len(matrix)):
            for col in range(1, len(matrix[0])):
                if p[col-1] == '?' or p[col-1] == s[row-1]:
                    matrix[row][col] = matrix[row-1][col-1]
                elif p[col-1] == '*':
                    matrix[row][col] = matrix[row-1][col] or matrix[row][col-1]
                else:
                    matrix[row][col] = False
        
        
        return matrix[len(s)][len(p)]