Reverse Prefix of Word — LeetCode 2000 Python Solution
EasyStackTwo PointersString
- Problem
- #2000
- Pattern
- Stack
- Reading time
- 2 min
- Source
- leetcode.com
The 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.
Example
- Input
- word = "abcdefd", ch = "d"
- Output
- "dcbaefd"
- Explanation
- The first occurrence of "d" is at index 3.
Python solution
Python
class Solution:
def reversePrefix(self, word: str, ch: str) -> str:
i = word.find(ch)
return word if i == -1 else word[i::-1] + word[i + 1 :]Complexity
| Measure | Complexity |
|---|---|
| Time | O(n) |
| Space | O(n) auxiliary |
Pattern: Stack
When the most recent unresolved thing is the one that matters, use a stack. LeetCode 2000. Reverse Prefix of Word is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Stack.
The stack guide has the Python template for the pattern and the 194 LeetCode problems that use it.
Related problems
Frequently asked questions
- How hard is LeetCode 2000. Reverse Prefix of Word?
- LeetCode 2000. Reverse Prefix of Word is rated Easy on LeetCode.
- What is the time complexity of LeetCode 2000. Reverse Prefix of Word?
- The Python solution on this page runs in O(n).
- What is the space complexity of LeetCode 2000. Reverse Prefix of Word?
- The Python solution on this page uses O(n) auxiliary space.
- What topics does LeetCode 2000. Reverse Prefix of Word cover?
- LeetCode 2000. Reverse Prefix of Word is tagged Stack, Two Pointers and String on LeetCode.