Python One-liner ( Runtime - beats 100%) | Direct and Simple Approach
Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind this code is fairly simple. It creates an acronym from the given l
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind this code is fairly simple. It creates an acronym from the given list of words by taking the first letter of each word and checking if it matches the provided string `s`.\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe code uses a list comprehension to extract the first letters of each word in the `words` list. It then joins these letters into a single string and compares it with the input string `s`. If the two strings match, the function returns `True`, indicating that s is an acronym generated from the list of words; otherwise, it returns `False`.\n\n# Complexity\n- Time complexity: O(n + m) - where n is the number of words in the list and m is the length of the input string.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n    def isAcronym(self, words: List[str], s: str) -> bool:\n        return True if s == "".join([x[0] for x in words]) else False\n```
['Python3']