✅ [C++] Simple Solution
Intuition\nThe intuition behind the solution is to check if the first letter of each word in words is equal to the corresponding character in s.\n\n# Approach\n
# Intuition\nThe intuition behind the solution is to check if the first letter of each word in words is equal to the corresponding character in s.\n\n# Approach\n1. Check if length of `words` and `s` is same, otherwise `return false`.\n2. Iterate through `words` and for every word, check if the first letter of word is equal to the corresponding character in `s`.\n\n# Complexity\n- Time complexity: `O(n)`\n\n- Space complexity: `O(1)`\n\n# Code\n```\nclass Solution {\npublic:\n    bool isAcronym(vector<string>& words, string s) {\n        int w = words.size();\n        int n = s.size();\n        if (n != w) {\n            return false;\n        }\n        for (int i=0; i<n; i++) {\n            if (words[i][0] != s[i]) {\n                return false;\n            }\n        }\n        return true;\n    }\n};\n```
['Array', 'String', 'C++']