isAcronym | beats 100%
Intuition\n Describe your first thoughts on how to solve this problem. \nThe code iterates through each word in the list and extracts its first character, appen
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe code iterates through each word in the list and extracts its first character, appending it to a StringBuilder. Then, it checks if the resulting string matches the input string s. If it does, it returns true, indicating that s is indeed an acronym.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Iterate through the list of words.\n2. For each word, extract the first character and append it to a StringBuilder.\n3. After appending all the first characters, compare the resulting string with the input string s.\n4. Return true if they match; otherwise, return false.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n1. The loop iterates through each word in the list once, taking O(n) time, where n is the number of words in the list.\n2. Within the loop, appending a character to the StringBuilder takes O(1) time.\n3. Finally, comparing the strings s and the StringBuilder\'s content takes O(m) time, where m is the length of the string s.\n4. Overall, the time complexity is O(n + m), where n is the number of words and m is the length of the input string s.\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n1. The space complexity is dominated by the StringBuilder, which stores the concatenation of the first characters of each word. It requires O(n) space, where n is the number of words in the list.\n2. Additional space complexity is O(m), where m is the length of the input string s, required for storing the StringBuilder\'s content during comparison.\n3. Thus, the overall space complexity is O(n + m), where n is the number of words and m is the length of the input string s.\n# Code\n```\nclass Solution {\n    public boolean isAcronym(List<String> words, String s) {\n       StringBuilder sb =new StringBuilder();\n        for(int i=0;i<words.size();i++)\n        {\n            sb.append(words.get(i).charAt(0));\n        }\n        return s.equals(sb.toString());\n        \n    }\n}\n```
['Java']