simple and easy C++ 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\n\n# Code\n```\nclass Solution {\npublic:\n    int prefixCount(vector<string>& words, string pref) \n    {\n        int ans = 0;\n        for(auto s:words)\n            if(s.find(pref) == 0) ans += 1;\n        return ans;\n    }\n};\n```
['Array', 'String', 'String Matching', 'C++']