Leetcode #1417: Reformat The String
In this guide, we solve Leetcode #1417 Reformat The String 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
You are given an alphanumeric string s. (Alphanumeric string is a string consisting of lowercase English letters and digits).
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: String
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 = "a0b1c2"
Output: "0a1b2c"
Explanation: No two adjacent characters have the same type in "0a1b2c". "a0b1c2", "0a1b2c", "0c2a1b" are also valid permutations.
Python Solution
class Solution:
def reformat(self, s: str) -> str:
a = [c for c in s if c.islower()]
b = [c for c in s if c.isdigit()]
if abs(len(a) - len(b)) > 1:
return ''
if len(a) < len(b):
a, b = b, a
ans = []
for x, y in zip(a, b):
ans.append(x + y)
if len(a) > len(b):
ans.append(a[-1])
return ''.join(ans)
Complexity
The time complexity is O(n). The space complexity is O(1) to O(n).
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.