792. Number of Matching Subsequences
Given string
S
and a dictionary of wordswords
, find the number ofwords[i]
that is a subsequence ofS
.
Example :
Input:
S = "abcde"
words = ["a", "bb", "acd", "ace"]
Output: 3
Explanation: There are three words inwords
that are a subsequence ofS
: "a", "acd", "ace".
Note:
- All words in
words
andS
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
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