simple and easy Python solution 😍❤️‍🔥
\n\n# Code\npython []\nclass Solution(object):\n    def countConsistentStrings(self, allowed, words):\n        ans = 0\n        allowed = set(allowed)  # Conver
\n\n# Code\n```python []\nclass Solution(object):\n    def countConsistentStrings(self, allowed, words):\n        ans = 0\n        allowed = set(allowed)  # Convert allowed string to set once\n        \n        for word in words:  # Iterate through the original list of words\n            word = set(word)  # Convert each word to a set of characters\n            flag = True\n            for ch in word:\n                if ch not in allowed:  # Check if any character is not in allowed\n                    flag = False\n                    break\n            ans += flag  # If flag is True, increment the count\n        \n        return ans\n```\n# if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n###### Let\'s Connect on LinkedIn: www.linkedin.com/in/shishirrsiam\n###### Let\'s Connect on Facebook: www.fb.com/shishirrsiam\n\n
['Hash Table', 'String', 'Counting', 'Python', 'Python3']