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hELLO · Designed By 정상우.
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def _(): _()

Algorithm

[LeetCode] 2000. Reverse Prefix of Word (Daily Question)

2024. 5. 1. 10:12

문제


Given a 0-indexed string word and a character ch, reverse the segment of word that starts at index 0 and ends at the index of the first occurrence of ch (inclusive). If the character ch does not exist in word, do nothing.

For example, if word = "abcdefd" and ch = "d", then you should reverse the segment that starts at 0 and ends at 3 (inclusive). The resulting string will be "dcbaefd".

Return the resulting string.

 

문자열word와 문자ch가 주어졌을 때 word의 앞부분부터 ch까지의 부분 문자열을 뒤집는 문제이다.

 

Example 1:
Input: word = "abcdefd", ch = "d"
Output: "dcbaefd"
Explanation: The first occurrence of "d" is at index 3. Reverse the part of word from 0 to 3 (inclusive), the resulting string is "dcbaefd".

Example 2:
Input: word = "xyxzxe", ch = "z"
Output: "zxyxxe"
Explanation: The first and only occurrence of "z" is at index 3. Reverse the part of word from 0 to 3 (inclusive), the resulting string is "zxyxxe".

Example 3:
Input: word = "abcd", ch = "z"
Output: "abcd"
Explanation: "z" does not exist in word. You should not do any reverse operation, the resulting string is "abcd".

 

풀이


아이디어

- ch의 위치를 찾는다.

- 뒤집는다.

 

 

코드

class Solution:
    def reversePrefix(self, word: str, ch: str) -> str:
        idx = word.find(ch)
        if idx == -1: return word
        return word[idx::-1]+word[idx+1:]

 


 

후기


특별한 풀이법 없이 직관적으로 쉽게 풀 수 있는 문제이다.

혹시 특수한 기법을 사용해서 풀기 원한걸까..?

 

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