Easy and smaller approach without using concatenation or extra space !!
Intuition\nThe first thing to notice is that the length of the string s should match the number of words. If they don\'t have the same length, it\u2019s impossi
# Intuition\nThe first thing to notice is that the length of the string s should match the number of words. If they don\'t have the same length, it\u2019s impossible for s to be an acronym of the list of words, so we can immediately return false.\nAfter that, we check if each character in s corresponds to the first letter of each word in the list. If we find a mismatch, we return false; otherwise, we return true.\n\n# Approach\nStep 1: Check if the length of s matches the number of words in the list. If they don\'t match, return false.\n\nStep 2: Iterate through the list of words. For each word, compare the first character of the word with the corresponding character in s.\n\nStep 3: If at any point the characters don\'t match, return false. If all the characters match, return true after the loop ends.\n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(1)\nHope you Liked it :)\n# Code\n```java []\nclass Solution {\n    public boolean isAcronym(List<String> words, String s) {    \n        if(words.size()!=s.length()) return false;\n        for(int i = 0 ; i< words.size(); i++){    \n            String str = words.get(i);\n            if(s.charAt(i)!=str.charAt(0)){\n                return false;\n            }\n        }\n        return true;\n    }\n}\n```
['Array', 'String', 'Java']