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].

int numMatchingSubseq(string S, vector<string>& words) { // time: O(words_len * average str_len * log(S_len)); space: O(S_len)
    // Build a lookup index table
    vector<vector<int> > idx(26);
    for (int i = 0; i < S.length(); ++i) idx[S[i] - 'a'].push_back(i);
    int res = 0;
    for (string& word : words) {
        int prev = -1;
        bool found = true;
        for (char c : word) {
            auto it = upper_bound(idx[c - 'a'].begin(), idx[c - 'a'].end(), prev);
            if (it == idx[c - 'a'].end()) {
                found = false;
                break;
            } else prev = *it;
        }
        if (found) ++res;
    }
    return res;
}

// Optimized Brute Force with Hashset
int numMatchingSubseq(string S, vector<string>& words) { // time: O(words_len * S_len); space: O(words_len)
    int res = 0, n = S.length();
    unordered_set<string> valid, invalid;
    for (string word : words) {
        if (valid.count(word) || invalid.count(word)) {
            if (valid.count(word)) ++res;
            continue;
        }
        // Use two pointers to check
        int i = 0, j = 0, m = word.length();
        while (i < n && j < m) {
            if (word[j] == S[i]) ++j;
            ++i;
        }
        if (j == m) {
            ++res;
            valid.insert(word);
        } else invalid.insert(word);
    }
    return res;
}

這題是392. Is Subsequence的延伸follow-up。