Leetcode #1694: Reformat Phone Number
In this guide, we solve Leetcode #1694 Reformat Phone Number 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 a phone number as a string number. number consists of digits, spaces ' ', and/or dashes '-'.
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: number = "1-23-45 6"
Output: "123-456"
Explanation: The digits are "123456".
Step 1: There are more than 4 digits, so group the next 3 digits. The 1st block is "123".
Step 2: There are 3 digits remaining, so put them in a single block of length 3. The 2nd block is "456".
Joining the blocks gives "123-456".
Python Solution
class Solution:
def reformatNumber(self, number: str) -> str:
number = number.replace("-", "").replace(" ", "")
n = len(number)
ans = [number[i * 3 : i * 3 + 3] for i in range(n // 3)]
if n % 3 == 1:
ans[-1] = ans[-1][:2]
ans.append(number[-2:])
elif n % 3 == 2:
ans.append(number[-2:])
return "-".join(ans)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
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.