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.

&nbsp;
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.


&nbsp;
Constraints:


	1 &lt;= words.length &lt;= 100
	1 &lt;= words[i].length, pref.length &lt;= 100
	words[i] and pref consist of lowercase English letters.


class Solution:
    def prefixCount(self, words: List[str], pref: str) -> int:
        return sum(word.find(pref) == 0 for word in words)
class Solution {
    public int prefixCount(String[] words, String pref) {
    int c = 0;
    for(String s : words) {
        if(s.indexOf(pref)==0) 
            c++;
    }
    return c; 
    }
}
class Solution {
public:
    int prefixCount(vector<string>& words, string pref) {
        int c=0;
        for(auto word:words){
            int b=0;
            for(int i=0;i<pref.size();i++){
                if(word[i]!=pref[i]){
                    break;
                }else{
                    b++;
                }
            }
            if(b==pref.size()){
                c++;
            }
        }
        return c;
    }
};
var prefixCount = function(words, pref) {
    return words.filter(word => word.slice(0, pref.length) === pref).length;
};
Counting Words With a Given Prefix