Leetcode #2810: Faulty Keyboard
In this guide, we solve Leetcode #2810 Faulty Keyboard in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Your laptop keyboard is faulty, and whenever you type a character 'i' on it, it reverses the string that you have written. Typing other characters works as expected.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: String, Simulation
Intuition
We need to scan characters while tracking positions or counts.
A simple state machine keeps the logic precise.
Approach
Iterate through the string once and update the state for each character.
Use a map or array if you need fast lookups.
Steps:
- Iterate through characters.
- Maintain necessary state.
- Build or validate the output.
Example
Input: s = "string"
Output: "rtsng"
Explanation:
After typing first character, the text on the screen is "s".
After the second character, the text is "st".
After the third character, the text is "str".
Since the fourth character is an 'i', the text gets reversed and becomes "rts".
After the fifth character, the text is "rtsn".
After the sixth character, the text is "rtsng".
Therefore, we return "rtsng".
Python Solution
class Solution:
def finalString(self, s: str) -> str:
t = []
for c in s:
if c == "i":
t = t[::-1]
else:
t.append(c)
return "".join(t)
Complexity
The time complexity is , and the space complexity is , where is the length of string . The space complexity is , where is the length of string .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.