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2255. Count Prefixes of a Given String


You are given a string array words and a string s, where words[i] and s comprise only of lowercase English letters.

Return the number of strings in words that are a prefix of s.

A prefix of a string is a substring that occurs at the beginning of the string. A substring is a contiguous sequence of characters within a string.


Example 1:

Input: words = ["a","b","c","ab","bc","abc"], s = "abc"
Output: 3
The strings in words which are a prefix of s = "abc" are:
"a", "ab", and "abc".
Thus the number of strings in words which are a prefix of s is 3.

Example 2:

Input: words = ["a","a"], s = "aa"
Output: 2
Both of the strings are a prefix of s. 
Note that the same string can occur multiple times in words, and it should be counted each time.



  • 1 <= words.length <= 1000
  • 1 <= words[i].length, s.length <= 10
  • words[i] and s consist of lowercase English letters only.


Solution 1: Traversal Counting

We directly traverse the array words, and for each string w, we check if s starts with w as a prefix. If it does, we increment the answer by one.

After the traversal, we return the answer.

The time complexity is $O(m \times n)$, where $m$ and $n$ are the lengths of the array words and the string s, respectively. The space complexity is $O(1)$.

  • class Solution {
        public int countPrefixes(String[] words, String s) {
            int ans = 0;
            for (String w : words) {
                if (s.startsWith(w)) {
            return ans;
  • class Solution {
        int countPrefixes(vector<string>& words, string s) {
            int ans = 0;
            for (auto& w : words) {
                ans += s.starts_with(w);
            return ans;
  • class Solution:
        def countPrefixes(self, words: List[str], s: str) -> int:
            return sum(s.startswith(w) for w in words)
  • func countPrefixes(words []string, s string) (ans int) {
    	for _, w := range words {
    		if strings.HasPrefix(s, w) {
  • function countPrefixes(words: string[], s: string): number {
        return words.filter(w => s.startsWith(w)).length;

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