|| Explaiined || Simple Solution || Java & Python ||
Intuition - \n\nThe program aims to determine whether a given string s is an acronym for a list of words provided as input. An acronym is typically formed by ta
# Intuition - \n\nThe program aims to determine whether a given string `s` is an acronym for a list of words provided as input. An acronym is typically formed by taking the first letter of each word in a phrase and combining them to create a new string. The program checks if the characters in `s` match the first characters of the words in the list, one-to-one.\n\n# Approach - \n\n1. Check if the length of `s` is equal to the number of words in the list. If not, return `false` because the length mismatch implies that `s` cannot be an acronym for the given list of words.\n\n2. Iterate through the list of words and compare the first character of each word with the corresponding character in `s`.\n\n3. If at any point, the characters do not match, return `false` immediately, as this means `s` cannot be an acronym.\n\n4. If all characters match for all words, return `true`, indicating that `s` is indeed an acronym for the given list of words.\n\n# Complexity - \n\n- Time Complexity:\n  - Checking the length of `s` and the number of words in the list takes O(1) time.\n  - The loop that iterates through the list of words and compares characters has a time complexity of O(n), where n is the number of words in the list.\n  - Therefore, the overall time complexity is O(n).\n\n- Space Complexity:\n  - The program uses a constant amount of extra space, regardless of the input size. It only creates a few variables to store characters and indices.\n  - Thus, the space complexity is O(1).\n\n\n# Code\n``` Java []\nclass Solution {\n    public boolean isAcronym(List<String> words, String s) {\n        if (s.length() != words.size()) {\n            return false;\n        }\n        for (int i = 0; i < words.size(); i++) {\n            String word = words.get(i);\n            char firstCharOfWord = word.charAt(0);\n            char charInS = s.charAt(i);\n            \n            if (firstCharOfWord != charInS) {\n                return false;\n            }\n        }\n        return true;\n    }\n}\n```\n\n\n``` Python []\nclass Solution:\n    def isAcronym(self, words, s):\n        if len(s) != len(words):\n            return False\n        for i in range(len(words)):\n            word = words[i]\n            firstCharOfWord = word[0]\n            charInS = s[i]\n            if firstCharOfWord != charInS:\n                return False\n        return True\n\n```
['Java']