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Formatted question description: https://leetcode.ca/all/2255.html

# 2255. Count Prefixes of a Given String (Easy)

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.

Related Topics:
Array, String

Similar Questions:

## Solution 1.

• class Solution {
public int countPrefixes(String[] words, String s) {
int ans = 0;
for (String word : words) {
if (word.equals(s.substring(0, Math.min(s.length(), word.length())))) {
++ans;
}
}
return ans;
}
}

• class Solution {
public:
int countPrefixes(vector<string>& words, string s) {
int ans = 0;
for (auto& word : words)
if (s.substr(0, word.size()) == word)
++ans;
return ans;
}
};

• class Solution:
def countPrefixes(self, words: List[str], s: str) -> int:
return sum(word == s[: len(word)] for word in words)


• func countPrefixes(words []string, s string) int {
ans := 0
for _, word := range words {
if strings.HasPrefix(s, word) {
ans++
}
}
return ans
}

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