Easiest Python Code Here!!!
Intuition\n Describe your first thoughts on how to solve this problem. \nWhen trying to match a string s with a pattern p that includes wildcards, the key chall
Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWhen trying to match a string s with a pattern p that includes wildcards, the key challenges are handling the ? and * characters. The ? wildcard matches any single character, and the * wildcard matches any sequence of characters (including the empty sequence). By iterating through both the string and the pattern, and using backtracking when encountering *, we can efficiently determine if the pattern matches the string.\n\nApproach\n<!-- Describe your approach to solving the problem. -->\nInitialization:\n\nUse two pointers, s for the string and p for the pattern.\nUse two additional variables, match to store the position of the match in the string when encountering *, and star_idx to store the position of the last * in the pattern.\nIterate through the string:\n\nIf the current characters in the pattern and string match, or if the pattern has a ?, move both pointers forward.\nIf the pattern has a *, store the positions of * and the current string pointer, and move the pattern pointer forward.\nIf there is a mismatch but there was a previous *, backtrack: reset the pattern pointer to the position after the last *, and move the string pointer to the next position after the last matched position.\nIf there is a mismatch and no previous *, return False.\nCheck remaining characters in the pattern:\n\nAfter iterating through the string, ensure any remaining characters in the pattern are all *.\nReturn the result:\n\nIf the pattern pointer has reached the end of the pattern, return True. Otherwise, return False.\nComplexity\nTime complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity is approximately \uD835\uDC42(\uD835\uDC5B\xD7\uD835\uDC5A)\n\nO(n\xD7m), where n is the length of the string and m is the length of the pattern. This is because in the worst case, each character in the string may need to be compared with every character in the pattern due to the backtracking when encountering *.\n\nSpace complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity is \uD835\uDC42(1)\n\nO(1), as we are using a fixed number of extra variables and not using any additional data structures that grow with input size.\n# Code\n```\nclass Solution:\n    def isMatch(self, st: str, pattern: str) -> bool:\n        s, p, mat, star_idx = 0, 0, 0, -1\n\n        while s < len(st):\n            if p < len(pattern) and (pattern[p] == \'?\' or pattern[p] == st[s]):\n                s += 1\n                p += 1\n            elif p < len(pattern) and pattern[p] == \'*\':\n                star_idx = p\n                mat = s\n                p += 1\n            elif star_idx != -1:\n                p = star_idx + 1\n                mat += 1\n                s = mat\n            else:\n                return False\n\n        while p < len(pattern) and pattern[p] == \'*\':\n            p += 1\n\n        return p == len(pattern)\n\n\n\n        \n![download.jpg](https://assets.leetcode.com/users/images/eb5843ed-f587-4dc5-a65f-afaeea29928a_1719641243.3811152.jpeg)\n\n\n```
['Python3']