2185. Counting Words With a Given Prefix LeetCode Solution
In this guide, you will get 2185. Counting Words With a Given Prefix LeetCode Solution with the best time and space complexity. The solution to Counting Words With a Given Prefix problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Counting Words With a Given Prefix
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.
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(1)
2185. Counting Words With a Given Prefix LeetCode Solution in C++
class Solution {
public:
int prefixCount(vector<string>& words, string pref) {
return ranges::count_if(
words, [&](const string& word) { return word.find(pref) == 0; });
}
}; /* code provided by PROGIEZ */
2185. Counting Words With a Given Prefix LeetCode Solution in Java
class Solution {
public int prefixCount(String[] words, String pref) {
return (int) Arrays.stream(words).filter(w -> w.startsWith(pref)).count();
}
} // code provided by PROGIEZ
2185. Counting Words With a Given Prefix LeetCode Solution in Python
class Solution:
def prefixCount(self, words: list[str], pref: str) -> int:
return sum(word.startswith(pref) for word in words) # code by PROGIEZ
