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  2. Counting Words With a Given Prefix

Strings

Counting Words With a Given Prefix

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)
Easy
String