Medium
Number of Matching Subsequences — C++
Full explanation · Time O(n + w) · Space O(k)
// Time: O(n + w), n is the size of S, w is the size of words
// Space: O(k), k is the number of words
class Solution {
public:
int numMatchingSubseq(string S, vector<string>& words) {
unordered_map<char, vector<pair<int, int>>> waiting;
for (int i = 0; i < words.size(); ++i) {
waiting[words[i][0]].emplace_back(i, 1);
}
for (const auto& c : S) {
auto advance = move(waiting[c]);
waiting.erase(c);
for (const auto& kvp : advance) {
int i = kvp.first, j = kvp.second;
int next = (j != words[i].length()) ? words[i][j] : 0;
waiting[next].emplace_back(i, j + 1);
}
}
return waiting[0].size();
}
};