# 2255. Count Prefixes of a Given String

## Description

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
Explanation:
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
Explanation:
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.

Constraints:

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

## Solutions

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)) {
++ans;
}
}
return ans;
}
}

• class Solution {
public:
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) {
ans++
}
}
return
}

• function countPrefixes(words: string[], s: string): number {
return words.filter(w => s.startsWith(w)).length;
}