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

• C++
• Java
• JavaScript
• Python
• Ruby

• 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)