2185: Counting-Words-With-a-Given-Prefix
Easy

table of contents

Brute Force is the name of my favourite genre of leetcode questions?!

Literally just iterate through words and check if pref is the prefix to words[i]. I do not know what else there is to do here…

code

class Solution {
public:
    int prefixCount(vector<string>& words, string pref) {
        int ans = 0;
        for (int i = 0; i < words.size(); ++i) {
            int counter = 0;
            for (int j = 0; j < pref.size(); ++j) {
                if (words[i][j] == pref[j]) {
                    ++counter;
                } else {
                    break;
                }
            }
            if (counter == pref.size()) {
                ++ans;
            }
        }
        return ans;
    }
};

complexity