[LeetCode] 1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence - Python

 

 

난이도: Easy

문제 설명

---

Given a sentence that consists of some words separated by a single space, and a searchWord, check if searchWord is a prefix of any word in sentence.

Return the index of the word in sentence (1-indexed) where searchWord is a prefix of this word. If searchWord is a prefix of more than one word, return the index of the first word (minimum index). If there is no such word return -1.

A prefix of a string s is any leading contiguous substring of s.

 

문제 예제

---

Example 1:

Input: sentence = "i love eating burger", searchWord = "burg"
Output: 4
Explanation: "burg" is prefix of "burger" which is the 4th word in the sentence.

Example 2:

Input: sentence = "this problem is an easy problem", searchWord = "pro"
Output: 2
Explanation: "pro" is prefix of "problem" which is the 2nd and the 6th word in the sentence, but we return 2 as it's the minimal index.

Example 3:

Input: sentence = "i am tired", searchWord = "you"
Output: -1
Explanation: "you" is not a prefix of any word in the sentence.

 

제한 사항

---

• 1 <= sentence.length <= 100
• 1 <= searchWord.length <= 10
• sentence consists of lowercase English letters and spaces.
• searchWord consists of lowercase English letters.

 

✏️ Solution(솔루션)

---

class Solution:
    def isPrefixOfWord(self, sentence: str, searchWord: str) -> int:
        words = sentence.split()
        n = len(searchWord)

        for i, word in enumerate(words):
            word_prefix = word[:n]
            if word_prefix == searchWord:
                return i+1

        return -1

 

sentence를 공백을 기준으로 split 해주어 sentence에 있는 word를 저장하는 words라는 리스트를 생성했다.

 

그 후 모든 words에 저장된 word에 대해 searchWord의 길이 만큼 prefix를 구했고, 만약 이게 searchWord와 같다면 현재 위치의 +1한 값을 return 하였다.

 

모든 word를 확인했음에도 조건을 만족하지 않는다면 -1을 return하도록 했다.

 

---

[LeetCode] 1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence