How do you solve Reformat Phone Number on LeetCode?

Reformat Phone Number (LeetCode #1694) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Removing non-digit characters is crucial for processing.

What is the brute force solution for Reformat Phone Number?

The brute force approach for Reformat Phone Number is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves cleaning the input string by removing spaces and dashes, then iteratively grouping the digits into blocks according to the specified rules. This method is straightforward but inefficient due to repeated string operations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Reformat Phone Number?

Reformat Phone Number uses the following concepts: String. The recommended patterns to study are: String Manipulation, Greedy Algorithms.

What are the interview tips for Reformat Phone Number?

Clarify the input format and constraints before coding. Think about edge cases, like empty strings or strings with only non-digit characters. Explain your thought process while coding to show your understanding.

What are common mistakes when solving Reformat Phone Number?

Not handling the last few digits correctly. Forgetting to join blocks with dashes.

#1694

Reformat Phone Number

Easy

StringString ManipulationGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach efficiently processes the digits in a single pass, grouping them according to the rules without unnecessary string operations. This method minimizes time complexity by using a list to collect blocks before joining them at the end.

⚙️

Algorithm

5 steps

1Step 1: Remove all spaces and dashes from the input string.
2Step 2: Initialize an empty list to hold the blocks of digits.
3Step 3: Use a while loop to process the digits, grouping them into blocks of 3 until 4 or fewer digits remain.
4Step 4: Handle the last 4 digits according to the specified rules and append them to the list.
5Step 5: Join the blocks with dashes and return the formatted string.

solution.py13 lines

1def reformatNumber(number):
2    digits = ''.join(c for c in number if c.isdigit())
3    blocks = []
4    i = 0
5    while len(digits) - i > 4:
6        blocks.append(digits[i:i+3])
7        i += 3
8    if len(digits) - i == 4:
9        blocks.append(digits[i:i+2])
10        blocks.append(digits[i+2:i+4])
11    else:
12        blocks.append(digits[i:])
13    return '-'.join(blocks)

ℹ

Complexity note: The complexity is O(n) because we traverse the string once to clean it and again to group the digits, making it efficient for larger inputs.

1Removing non-digit characters is crucial for processing.
2Understanding the grouping rules helps in implementing the logic correctly.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.