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Reformat Phone Number

Difficulty: Easy

Problem Description

You are given a phone number as a string number. number consists of digits, spaces, and/or dashes. You would like to reformat the phone number by removing all spaces and dashes, and then grouping the digits into blocks of length 3 until there are 4 or fewer digits left. The final digits are then grouped according to specific rules: 2 digits as a block of length 2, 3 digits as a block of length 3, or 4 digits as two blocks of length 2 each. The blocks must be joined by dashes, ensuring that no blocks of length 1 are produced and at most two blocks of length 2 are present.

Key Insights

  • Remove all non-digit characters (spaces and dashes).
  • Group digits into blocks of 3 until 4 or fewer digits remain.
  • Handle the final grouping of digits based on their count:
    • 2 digits -> one block of 2.
    • 3 digits -> one block of 3.
    • 4 digits -> two blocks of 2.
  • Join the blocks with dashes.

Space and Time Complexity

Time Complexity: O(n) - where n is the length of the input string, as we iterate through it to clean and format the number.

Space Complexity: O(n) - for storing the cleaned digits and the final formatted string.

Solution

The solution involves using a simple iterative approach to process the input string. First, we remove all non-digit characters to obtain a clean string of digits. Then, we iterate over this clean string and group the digits into blocks of 3 until we reach the last four digits, which we handle based on their count. A list is used to store the blocks, which is then joined with dashes to form the final formatted phone number.

Code Solutions

def reformatNumber(number: str) -> str:
    # Remove spaces and dashes
    digits = ''.join(c for c in number if c.isdigit())
    blocks = []
    
    # Process digits in blocks of 3
    while len(digits) > 4:
        blocks.append(digits[:3])
        digits = digits[3:]
    
    # Handle the remaining digits
    if len(digits) == 4:
        blocks.append(digits[:2])
        blocks.append(digits[2:])
    else:
        blocks.append(digits)
    
    # Join blocks with dashes
    return '-'.join(blocks)
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