792. Number of Matching Subsequences

Given string S and a dictionary of words words, find the number of words[i] that is a subsequence of S.

Example :
Input: 
S = "abcde"
words = ["a", "bb", "acd", "ace"]
Output: 3
Explanation: There are three words in words that are a subsequence of S: "a", "acd", "ace".

Note:

• All words in words and S will only consists of lowercase letters.
• The length of S will be in the range of [1, 50000].
• The length of words will be in the range of [1, 5000].
• The length of words[i] will be in the range of [1, 50].

Approach #1: Array. [C++]

class Solution {
public:
    int numMatchingSubseq(string S, vector<string>& words) {
        vector<const char*> waiting[128];
        for (auto &w : words)
            waiting[w[0]].push_back(w.c_str());
        for (char c : S) {
            auto advance = waiting[c];
            waiting[c].clear();
            for (auto it : advance)
                waiting[*++it].push_back(it);
        }
        return waiting[0].size();
    }
};

　　

Approach #2: [Java]

class Solution {
    public int numMatchingSubseq(String S, String[] words) {
        List<Integer[]>[] waiting = new List[128];
        for (int c = 0; c <= 'z'; ++c)
            waiting[c] = new ArrayList();
        for (int i = 0; i < words.length; ++i) 
            waiting[words[i].charAt(0)].add(new Integer[]{i, 1});
        for (char c : S.toCharArray()) {
            List<Integer[]> advance = new ArrayList(waiting[c]);
            waiting[c].clear();
            for (Integer[] a : advance)
                waiting[a[1] < words[a[0]].length() ? words[a[0]].charAt(a[1]++) : 0].add(a);
        }
        return waiting[0].size();
    }
}

　　

Reference:

https://leetcode.com/problems/number-of-matching-subsequences/discuss/117634/Efficient-and-simple-go-through-words-in-parallel-with-explanation