[Leet Code] Reverse Prefix of Word

Leet Code: https://leetcode.com/problems/reverse-prefix-of-word/

Problem:

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.

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".

Constraints:

• 1 <= word.length <= 250
• word consists of lowercase English letters.
• ch is a lowercase English letter.

Solution:

class Solution(object):
    def reversePrefix(self, word, ch):
        """
        :type word: str
        :type ch: str
        :rtype: str
        """
        index_of_word = word.find(ch)
        
        if index_of_word == -1:
            return word
        else:
            word = list(word)
            decrementer = index_of_word
            for i in range((index_of_word//2)+1):
                temp = word[i]
                word[i] = word[decrementer]
                word[decrementer] = temp
                decrementer -= 1
        return "".join(word)