Easy approach to the problem beginner friendly beats 80% user with python
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nfirst things first if the length of List words and string s is not same w
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nfirst things first if the length of List words and string s is not same we return false if it is the same then we go into the for loop and check if words[i] is equal to 0 ir words[i][0] is not equal to s[i] if it not equal then we return false else we return true\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n    def isAcronym(self, words: List[str], s: str) -> bool:\n        isacronym=False\n        if len(words)!=len(s):\n            return isacronym\n\n        for i in range(len(s)):\n            if (len(words[i])==0 or words[i][0]!=s[i]):\n                return False\n        return True\n\n```
['Python3']