simple and easy Python solution 😍❤️‍🔥
\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
\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\n# Complexity\n- Time complexity: O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n# Code\n```\nclass Solution(object):\n    def prefixCount(self, words, pref):\n        ans = 0\n        for s in words:\n            if s.startswith(pref):\n                ans += 1\n        return ans\n```
['Array', 'String', 'String Matching', 'Python', 'Python3']