LeetCode #1694 — EASY

Reformat Phone Number

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"1-23-45 6"

Example 2

"123 4-567"

Example 3

"123 4-5678"

Step 02

Core Insight

What unlocks the optimal approach

Discard all the spaces and dashes.
Use a while loop. While the string still has digits, check its length and see which rule to apply.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1694: Reformat Phone Number
class Solution {
    public String reformatNumber(String number) {
        number = number.replace("-", "").replace(" ", "");
        int n = number.length();
        List<String> ans = new ArrayList<>();
        for (int i = 0; i < n / 3; ++i) {
            ans.add(number.substring(i * 3, i * 3 + 3));
        }
        if (n % 3 == 1) {
            ans.set(ans.size() - 1, ans.get(ans.size() - 1).substring(0, 2));
            ans.add(number.substring(n - 2));
        } else if (n % 3 == 2) {
            ans.add(number.substring(n - 2));
        }
        return String.join("-", ans);
    }
}

// Accepted solution for LeetCode #1694: Reformat Phone Number
func reformatNumber(number string) string {
	number = strings.ReplaceAll(number, " ", "")
	number = strings.ReplaceAll(number, "-", "")
	n := len(number)
	ans := []string{}
	for i := 0; i < n/3; i++ {
		ans = append(ans, number[i*3:i*3+3])
	}
	if n%3 == 1 {
		ans[len(ans)-1] = ans[len(ans)-1][:2]
		ans = append(ans, number[n-2:])
	} else if n%3 == 2 {
		ans = append(ans, number[n-2:])
	}
	return strings.Join(ans, "-")
}

# Accepted solution for LeetCode #1694: Reformat Phone Number
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)

// Accepted solution for LeetCode #1694: Reformat Phone Number
impl Solution {
    pub fn reformat_number(number: String) -> String {
        let cs: Vec<char> = number.chars().filter(|&c| c != ' ' && c != '-').collect();
        let n = cs.len();
        cs.iter()
            .enumerate()
            .map(|(i, c)| {
                if ((i + 1) % 3 == 0 && i < n - 2) || (n % 3 == 1 && i == n - 3) {
                    return c.to_string() + &"-";
                }
                c.to_string()
            })
            .collect()
    }
}

// Accepted solution for LeetCode #1694: Reformat Phone Number
function reformatNumber(number: string): string {
    const cs = [...number].filter(c => c !== ' ' && c !== '-');
    const n = cs.length;
    return cs
        .map((v, i) => {
            if (((i + 1) % 3 === 0 && i < n - 2) || (n % 3 === 1 && n - 3 === i)) {
                return v + '-';
            }
            return v;
        })
        .join('');
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.