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2185. Counting Words With a Given Prefix

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.

 

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.

Solutions

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

• class Solution {
      public int prefixCount(String[] words, String pref) {
          int ans = 0;
          for (String w : words) {
              if (w.startsWith(pref)) {
                  ++ans;
              }
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      int prefixCount(vector<string>& words, string pref) {
          int ans = 0;
          for (auto& w : words) ans += w.find(pref) == 0;
          return ans;
      }
  };
  

• class Solution:
      def prefixCount(self, words: List[str], pref: str) -> int:
          return sum(w.startswith(pref) for w in words)
  
  

• func prefixCount(words []string, pref string) (ans int) {
  	for _, w := range words {
  		if strings.HasPrefix(w, pref) {
  			ans++
  		}
  	}
  	return
  }
  

• function prefixCount(words: string[], pref: string): number {
      return words.reduce((r, s) => (r += s.startsWith(pref) ? 1 : 0), 0);
  }
  
  

• impl Solution {
      pub fn prefix_count(words: Vec<String>, pref: String) -> i32 {
          words
              .iter()
              .filter(|s| s.starts_with(&pref))
              .count() as i32
      }
  }
  
  

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