Easiest python solution with using str.index. Beat 90%. Code explains.
```\n    def numMatchingSubseq(self, s: str, words: List[str]) -> int:\n        def is_substring(w):\n            i = 0\n            for c in w:\n              
```\n    def numMatchingSubseq(self, s: str, words: List[str]) -> int:\n        def is_substring(w):\n            i = 0\n            for c in w:\n                try:\n                    idx = s.index(c, i)\n                    i = idx + 1\n                except:\n                    return False\n            return True\n\n        return sum(is_substring(word) for word in words)
['Python', 'Python3']