📌 Easy-Approach || Well-explained || 95% faster 🐍
IDEA:\n\n\uD83D\uDC49 finding the index of each character in "s" then virtually cutting "s" => from index+1 to last.\n\uD83D\uDC49 repeating above point for all
## IDEA:\n\n\uD83D\uDC49 finding the index of each character in "s" then virtually cutting "s" => from index+1 to last.\n\uD83D\uDC49 repeating above point for all the characters in word, if any ch not found in s then it will return -1.\n\ne.g.-\n\ts="abgheaf"\n\tw=["gef", "gefk"]\n\tindex = -1\n* \tfirst found index of "g" in s[index+1 : ] => "abgheaf", which is  2 (i.e. index=2).\n* \tNow, we find the index of "e" in s[index+1:] => s[2+1:] => s[3:] => "heaf" which results index=4.\n* \tAgain find the index of "f" in s[index+1: ] => s[4+1:] => s[5:] => "af" which gives index=6.\n**Another word:**\n* \tif we are having word as "gefk" then after repeating previous 3 points we will have index=6.\n* \tNow find the index of "k" in s[index+1:] => s[6+1:] => s[7:] =>"" which will give index=-1. i.e. not found.\n\t\n\'\'\'\n\n\tclass Solution:\n    def numMatchingSubseq(self, s: str, words: List[str]) -> int:\n        \n        def is_sub(word):\n            index=-1\n            for ch in word:\n                index=s.find(ch,index+1)\n                if index==-1:\n                    return False\n            return True\n        \n        c=0\n        for word in words:\n            if is_sub(word):\n                c+=1\n        \n        return c\n\t\t\nfeel free to ask if any doubt. \uD83C\uDFCD\nif you got helpful then please **upvote!!** \uD83E\uDD17
['Python', 'Python3']