Python 3   ||   1 line,  join()   ||   T/S: 78% / 98%
\nclass Solution:\n    def isAcronym(self, words: List[str], s: str) -> bool:\n\n        return \'\'.join(word[0] for word in words) == s\n\nhttps://leetcode.co
```\nclass Solution:\n    def isAcronym(self, words: List[str], s: str) -> bool:\n\n        return \'\'.join(word[0] for word in words) == s\n```\n[https://leetcode.com/problems/check-if-a-string-is-an-acronym-of-words/submissions/1292624706/](https://leetcode.com/problems/check-if-a-string-is-an-acronym-of-words/submissions/1292624706/)\n\n\n\nI could be wrong, but I think that time complexity is *O*(*N*) and space complexity is *O*(*N*), in which *N* ~ `len(words)`.
['Python3']