Counting Words With a Given Prefix

Published: Jan 12, 2023

Problem Description

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.

Constraints:

• 1 <= words.length <= 100
• 1 <= words[i].length, pref.length <= 100
• words[i] and pref consist of lowercase English letters.

https://leetcode.com/problems/counting-words-with-a-given-prefix/

Examples

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.

How to Solve

In general, languages have a function or method to check prefix characters. For example, Python string has a startswith function while C++ string has an rfind function. Use such built-in function and count up if an word starts with the given prefix.

Solution

class CountingWordsWithAGivenPrefix {
public:
    int prefixCount(vector<string>& words, string pref) {
        int result = 0;
        for (string w : words) {
            if (w.rfind(pref, 0) != string::npos) { result++; }
        }
        return result;
    }
};

class CountingWordsWithAGivenPrefix {
    public int prefixCount(String[] words, String pref) {
        int result = 0;
        for (String word : words) {
            if (word.startsWith(pref)) { result++; }
        }
        return result;
    }
}

/**
 * @param {string[]} words
 * @param {string} pref
 * @return {number}
 */
var prefixCount = function(words, pref) {
    let result = 0;
    for (const w of words) {
        if (w.startsWith(pref)) {
            result++;
        }
    }
    return result;
};

class CountingWordsWithAGivenPrefix:
    def prefixCount(self, words: List[str], pref: str) -> int:
        result = 0
        for w in words:
            if w.startswith(pref):
                result += 1
        return result

# @param {String[]} words
# @param {String} pref
# @return {Integer}
def prefix_count(words, pref)
  result = 0
  words.each do |word|
    if word.start_with?(pref)
      result += 1
    end
  end
  result
end

Complexities

• Time: O(n)
• Space: O(1)