[Leetcode]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.

💻 My submission

class Solution:
    def prefixCount(self, words: List[str], pref: str) -> int:
        prefix_cnt=0
        for word in words:
            if word[:len(pref)]==pref: prefix_cnt+=1
        return prefix_cnt

💊 Better solutions

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

💡 What I learned

startswith(), endswith()

• 시작과, 끝이 맞는지 결과를 bool로 반환한다.

1. startswith()

a = '01-sample.png'

print(a.startswith('01'))

## 출력 결과
> True

2. endswith()

a = '01-sample.png'

print(a.endswith('.png'))
## 출력 결과
> True

References

• https://leetcode.com/problems/counting-words-with-a-given-prefix/
• https://leetcode.com/problems/counting-words-with-a-given-prefix/discuss/1809593/Python-one-line-simple-solution
• https://wikidocs.net/137643