✔Easy JavaScript, C++ beginner friendly solution || O(n) Time Complexity || O(1) Space Complexity 🔥
Intuition\n\n- The most straightforward approach to this problem is to iterate through the words in the list and compare their first letters with the correspond
# Intuition\n\n- The most straightforward approach to this problem is to iterate through the words in the list and compare their first letters with the corresponding characters in the string. \n- If all the first letters match, then the string is an acronym of the words. Otherwise, it\'s not.\n\n# Approach\n\n- This code implements the intuitive approach described above. \n- It iterates through the words list and compares the first letter of each word with the corresponding character in the string.\n\n\n# Complexity\n\n- Time complexity: **O(n)**\n\nThe code iterates through the `words` list with a for loop, which takes `n` steps (where n is the number of words). Inside the loop, there are constant time operations for character comparison. Therefore, the overall time complexity is linear in the number of words.\n\n- Space complexity: **O(1)**\n\nThe code uses a constant amount of extra space for variables like `op` and loop counters. This space complexity does not depend on the input size (number of words or length of the string), making it constant.\n\n---\n\n# Code\n```C++ []\nclass Solution {\npublic:\n    bool isAcronym(vector<string>& words, string s) {\n        int wordsLength = words.size();\n\n        if (s.length() != wordsLength) {\n            return false; // Different lengths, not an acronym\n        }\n\n        for (int i = 0; i < wordsLength; ++i) {\n            if (toupper(words[i][0]) !=\n                toupper(s[i])) { // Compare uppercase first letters\n                return false;\n            }\n        }\n\n        return true; // All first letters matched\n    }\n};\n```\n\n```javascript []\n/**\n * @param {string[]} words\n * @param {string} s\n * @return {boolean}\n */\nvar isAcronym = function (words, s) {\n    let op=false;\n    if (s.length === words.length) {\n        for (let i = 0; i < words.length; i++) {\n            if (words[i].at(0) === s[i]) {\n                op= true;\n            }\n            else {\n                op= false;\n                break;\n            }\n        }\n    }\n    return op;\n};\n```
['C++', 'JavaScript']