Reformat Phone Number - LeetCode solutions

Problem

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 in a certain manner. Firstly, remove all spaces and dashes. Then, group the digits from left to right into blocks of length 3 until there are 4 or fewer digits. The final digits are then grouped as follows:

2 digits: A single block of length 2.
3 digits: A single block of length 3.
4 digits: Two blocks of length 2 each.

The blocks are then joined by dashes. Notice that the reformatting process should never produce any blocks of length 1 and produce at most two blocks of length 2.

Return the phone number after formatting.

Example 1:

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".

Example 2:

Input: number = "123 4-567"
Output: "123-45-67"
Explanation: The digits are "1234567".
Step 1: There are more than 4 digits, so group the next 3 digits. The 1st block is "123".
Step 2: There are 4 digits left, so split them into two blocks of length 2. The blocks are "45" and "67".
Joining the blocks gives "123-45-67".

Example 3:

Input: number = "123 4-5678"
Output: "123-456-78"
Explanation: The digits are "12345678".
Step 1: The 1st block is "123".
Step 2: The 2nd block is "456".
Step 3: There are 2 digits left, so put them in a single block of length 2. The 3rd block is "78".
Joining the blocks gives "123-456-78".

Constraints:

2 <= number.length <= 100
number consists of digits and the characters '-' and ' '.
There are at least two digits in number.

Solution (Java)

class Solution {
    public String reformatNumber(String number) {
        StringBuilder sb = new StringBuilder();
        for (char c : number.toCharArray()) {
            if (Character.isDigit(c)) {
                sb.append(c);
            }
        }
        String cleaned = sb.toString();
        sb.setLength(0);
        int i = 0;
        while (i < cleaned.length()) {
            if (i + 4 == cleaned.length()) {
                sb.append(cleaned, i, i + 2);
                sb.append("-");
                sb.append(cleaned.substring(i + 2));
                break;
            } else if (i + 3 <= cleaned.length()) {
                sb.append(cleaned, i, i + 3);
                sb.append("-");
                i += 3;
            } else {
                sb.append(cleaned.substring(i));
                break;
            }
        }
        if (sb.charAt(sb.length() - 1) == '-') {
            sb.setLength(sb.length() - 1);
        }
        return sb.toString();
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).