"Python Solution: (Beginner-Friendly) - Runtime Beats 91.26%"
Code:\n\nclass Solution:\n    def isAcronym(self, words: List[str], s: str) -> bool:\n        if len(words) != len(s):\n            return False\n        else:\
Code:\n```\nclass Solution:\n    def isAcronym(self, words: List[str], s: str) -> bool:\n        if len(words) != len(s):\n            return False\n        else:\n            for i in range(len(s)):\n                if s[i] != words[i][0]:\n                    return False\n        return True\n```\n\nCode Explanation with comment:\n```\nclass Solution:\n    def isAcronym(self, words: List[str], s: str) -> bool:\n        # Check if the length of the list of words matches the length of the acronym string\n        if len(words) != len(s):\n            return False  # If not, return False because they cannot be matching acronyms\n        else:\n            # Iterate through each character position in the acronym string\n            for i in range(len(s)):\n                # Check if the character in the acronym string matches the first letter of the corresponding word\n                if s[i] != words[i][0]:\n                    return False  # If not, return False because they cannot be matching acronyms\n        # If all characters match the first letters of their corresponding words, return True\n        return True\n```\n\nLet\'s dry run:\n\nwords = ["alice","bob","charlie"]\ns = "abc"\n\n1. The length of `words` is 3, and the length of `s` is also 3, so the condition `len(words) != len(s)` is False, so we move to the else block.\n2. We iterate over each character in `s` using the loop `for i in range(len(s))`.\n3. In the first iteration, `i` is 0. So, we compare `s[0]` (which is \'a\') with `words[0][0]` (the first character of the first word \'alice\', which is also \'a\'). They match, so we continue to the next iteration.\n4. In the second iteration, `i` is 1. So, we compare `s[1]` (which is \'b\') with `words[1][0]` (the first character of the second word \'bob\', which is \'b\'). They match, so we continue to the next iteration.\n5. In the third iteration, `i` is 2. So, we compare `s[2]` (which is \'c\') with `words[2][0]` (the first character of the third word \'charlie\', which is \'c\'). They match, so the loop ends.\n6. Since all characters matched, we return `True`.\n\nSo, for the given example, the output of the function would be `True`.
['Python3']