LeetCode: Number of Matching Subsequences

## https://code.dennyzhang.com/number-of-matching-subsequences
## Basic Ideas:
## Complexity: Time O(n*m), Space O(1)
class Solution:
    def numMatchingSubseq(self, S, words):
        """
        :type S: str
        :type words: List[str]
        :rtype: int
        """
        res = 0
        ok_set = set([])
        invalid_set = set([])
        for word in words:
            # avoid duplicate calculation
            if word in ok_set:
                res += 1
                continue

            if word in invalid_set: continue

            index = 0
            for ch in S:
                if ch == word[index]: index += 1
                if index == len(word):
                    res += 1
                    break

            if index != len(word):
                invalid_set.add(word)
            else:
                ok_set.add(word)
        return res