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Formatted question description: https://leetcode.ca/all/2255.html

2255. Count Prefixes of a Given String (Easy)

You are given a string array words and a string s, where words[i] and s comprise only of lowercase English letters.

Return the number of strings in words that are a prefix of s.

A prefix of a string is a substring that occurs at the beginning of the string. A substring is a contiguous sequence of characters within a string.

 

Example 1:

Input: words = ["a","b","c","ab","bc","abc"], s = "abc"
Output: 3
Explanation:
The strings in words which are a prefix of s = "abc" are:
"a", "ab", and "abc".
Thus the number of strings in words which are a prefix of s is 3.

Example 2:

Input: words = ["a","a"], s = "aa"
Output: 2
Explanation:
Both of the strings are a prefix of s. 
Note that the same string can occur multiple times in words, and it should be counted each time.

 

Constraints:

• 1 <= words.length <= 1000
• 1 <= words[i].length, s.length <= 10
• words[i] and s consist of lowercase English letters only.

Related Topics:
Array, String

Similar Questions:

• Check If a Word Occurs As a Prefix of Any Word in a Sentence (Easy)
• Check If String Is a Prefix of Array (Easy)
• Counting Words With a Given Prefix (Easy)

Solution 1.

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      public int countPrefixes(String[] words, String s) {
          int ans = 0;
          for (String word : words) {
              if (word.equals(s.substring(0, Math.min(s.length(), word.length())))) {
                  ++ans;
              }
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      int countPrefixes(vector<string>& words, string s) {
          int ans = 0;
          for (auto& word : words)
              if (s.substr(0, word.size()) == word)
                  ++ans;
          return ans;
      }
  };
  

• class Solution:
      def countPrefixes(self, words: List[str], s: str) -> int:
          return sum(word == s[: len(word)] for word in words)
  
  

• func countPrefixes(words []string, s string) int {
  	ans := 0
  	for _, word := range words {
  		if strings.HasPrefix(s, word) {
  			ans++
  		}
  	}
  	return ans
  }
  

• function countPrefixes(words: string[], s: string): number {
      return words.filter(w => s.startsWith(w)).length;
  }
  
  

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