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Brute Force Approach: In this approach, for each word in the words array, we will check whether it is a subsequence of the string s or not. This is achieved by iterating over the characters of the word and trying to match them sequentially in the string s.
Time Complexity: O(n * m), where n is the length of the string s and m is the average length of the words. Space Complexity: O(1) since it only uses constant extra space.
1#include <vector>
2#include <string>
3
4bool isSubsequence(const std::string& s, const std::string& word) {
5 int i = 0, j = 0;
6 while (i < s.size() && j < word.size()) {
7 if (s[i] == word[j]) {
8 j++;
9 }
10 i++;
11 }
12 return j == word.size();
13}
14
15int numMatchingSubseq(const std::string& s, const std::vector<std::string>& words) {
16 int count = 0;
17 for (const auto& word : words) {
18 if (isSubsequence(s, word)) {
19 count++;
20 }
21 }
22 return count;
23}This solution uses a helper function isSubsequence to check each word. For every word, if it is a subsequence of s, the count is incremented. The logic is simple and checks the characters one by one.
Efficient Group Matching: Group words by their initial characters in a dictionary and iterate over the string s to manage these groups of words, repositioning them towards completion as their current character is matched. This avoids re-checking each word from the beginning every time.
Time Complexity: O(n + m), where n is the length of s and m is the total number of characters across all words. Space Complexity: O(m) due to storage of iterators.
1from collections import defaultdict
2
3def In this Python approach, we store word iterators in a dictionary waiting, initially keyed by their starting character. As we advance through s, we update the waiting list for each character to track partially matched words, dramatically improving efficiency.