# 2185. Counting Words With a Given Prefix

## Description

You are given an array of strings words and a string pref.

Return the number of strings in words that contain pref as a prefix.

A prefix of a string s is any leading contiguous substring of s.

Example 1:

Input: words = ["pay","attention","practice","attend"], pref = "at"
Output: 2
Explanation: The 2 strings that contain "at" as a prefix are: "attention" and "attend".


Example 2:

Input: words = ["leetcode","win","loops","success"], pref = "code"
Output: 0
Explanation: There are no strings that contain "code" as a prefix.


Constraints:

• 1 <= words.length <= 100
• 1 <= words[i].length, pref.length <= 100
• words[i] and pref consist of lowercase English letters.

## Solutions

• class Solution {
public int prefixCount(String[] words, String pref) {
int ans = 0;
for (String w : words) {
if (w.startsWith(pref)) {
++ans;
}
}
return ans;
}
}

• class Solution {
public:
int prefixCount(vector<string>& words, string pref) {
int ans = 0;
for (auto& w : words) ans += w.find(pref) == 0;
return ans;
}
};

• class Solution:
def prefixCount(self, words: List[str], pref: str) -> int:
return sum(w.startswith(pref) for w in words)


• func prefixCount(words []string, pref string) (ans int) {
for _, w := range words {
if strings.HasPrefix(w, pref) {
ans++
}
}
return
}

• function prefixCount(words: string[], pref: string): number {
return words.reduce((r, s) => (r += s.startsWith(pref) ? 1 : 0), 0);
}


• impl Solution {
pub fn prefix_count(words: Vec<String>, pref: String) -> i32 {
words
.iter()
.filter(|s| s.starts_with(&pref))
.count() as i32
}
}