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Formatted question description: https://leetcode.ca/all/2185.html

2185. Counting Words With a Given Prefix (Easy)

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.

Similar Questions:

Solution 1.

  • 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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