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
- Time:
O(n)- Space:
O(1)