Solution Approach

Let's walk through the implementation step by step:

Step 1: Calculate the first group size

n = len(s) cnt = (n - s.count("-")) % k or k

• First, we get the total length of the string n
• We calculate the number of non-dash characters: n - s.count("-")
• We find the remainder when dividing by k using modulo: (n - s.count("-")) % k
• If the remainder is 0 (perfectly divisible), we want k characters in the first group, hence or k
• This value is stored in cnt, representing how many characters we need for the current group

Step 2: Process each character

ans = [] for i, c in enumerate(s):

• We create an empty list ans to build our result
• We iterate through each character with its index

Step 3: Handle dashes and format characters

if c == "-":  continue ans.append(c.upper()) cnt -= 1

• Skip dashes completely - they don't count toward our character groups
• Convert the character to uppercase and add it to our answer
• Decrement cnt since we've added one character to the current group

Step 4: Handle group boundaries

if cnt == 0:  cnt = k  if i != n - 1:  ans.append("-")

• When cnt reaches 0, we've completed the current group
• Reset cnt to k for the next group
• Add a dash separator, but only if we're not at the last character of the original string (to avoid trailing dash)

Step 5: Clean up and return

return "".join(ans).rstrip("-")

• Join all characters in the list to form the final string
• Use rstrip("-") to remove any trailing dash (as a safety measure, though the logic should prevent it)

The algorithm runs in O(n) time where n is the length of the input string, as we make a single pass through the string. The space complexity is also O(n) for storing the result.

Example Walkthrough

Let's walk through the solution with a concrete example: s = "2-5g-3-J" and k = 2.

Step 1: Calculate the first group size

• Total length: n = 8
• Count dashes: s.count("-") = 3
• Non-dash characters: 8 - 3 = 5
• First group size: 5 % 2 = 1 (remainder when dividing 5 by 2)
• So cnt = 1 (the first group will have 1 character)

Step 2: Process each character

Let's trace through each iteration:

Step 3: Join and return

• Join the list: "2-5G-3J"
• Apply rstrip("-"): No trailing dash to remove
• Final result: "2-5G-3J"

The reformatted license key has groups of sizes: 1, 2, 2 (reading left to right), with the first group having fewer than k characters as expected when there's a remainder.

Time and Space Complexity

The time complexity is O(n), where n is the length of the string s. The algorithm iterates through the string once with a single for loop, and each operation inside the loop (checking if character is "-", appending to list, upper() conversion) takes constant time O(1). The final join() operation also takes O(n) time, and rstrip() takes at most O(n) time. Therefore, the overall time complexity is O(n) + O(n) + O(n) = O(n).

The space complexity is O(n) for the output string stored in the ans list. However, if we ignore the space consumption of the answer string (as mentioned in the reference), the space complexity is O(1). The only additional variables used are n, cnt, i, and c, which all use constant space regardless of input size.

Common Pitfalls

1. Incorrectly Adding Trailing Dashes

The most common pitfall is adding an unnecessary dash at the end of the formatted string. This happens when the logic for determining whether to add a dash doesn't properly account for the end of the string.

Problematic approach:

# Wrong: Checking if we've completed a group without considering position if current_group_count == 0:  current_group_count = k  result.append("-") # This always adds a dash!

Why it fails: If the last character processed completes a group, this code will add a trailing dash that shouldn't be there.

Solution: Check if you're at the end of processing before adding a dash:

if current_group_count == 0:  current_group_count = k  # Only add dash if there are more characters to process  if index != total_length - 1:  result.append("-")

However, this check index != total_length - 1 can still be problematic because index refers to the position in the original string, which includes dashes. A more robust solution is to track whether there are more non-dash characters to process.

2. Mishandling Empty Input or All-Dash Input

Another pitfall is not handling edge cases where the input contains no alphanumeric characters.

Problematic scenario:

s = "---" # Only dashes k = 2 # The code might produce "" or fail

Solution: Add a check for empty result:

# After processing all characters if not result:  return ""

3. Off-by-One Error in First Group Calculation

A subtle pitfall is incorrectly calculating the first group size when the total character count is divisible by k.

Wrong approach:

first_group_size = char_count % k # This gives 0 when divisible!

Why it fails: When char_count is perfectly divisible by k, this gives 0, meaning the first group would have 0 characters, which violates the requirement.

Solution: Use the conditional operator or an if-statement:

first_group_size = char_count % k or k # OR first_group_size = char_count % k if first_group_size == 0:  first_group_size = k

4. Better Approach to Avoid the Trailing Dash Issue Entirely

Instead of checking positions, a cleaner approach is to track remaining characters:

class Solution:  def licenseKeyFormatting(self, s: str, k: int) -> str:  # Remove all dashes and convert to uppercase first  chars = [c.upper() for c in s if c != '-']   if not chars:  return ""   # Calculate first group size  first_group_size = len(chars) % k or k   # Build result with proper grouping  result = []  idx = 0   # Add first group  result.append(''.join(chars[:first_group_size]))  idx = first_group_size   # Add remaining groups  while idx < len(chars):  result.append(''.join(chars[idx:idx+k]))  idx += k   return '-'.join(result)

This approach completely avoids the trailing dash issue by:

1. Processing all characters first
2. Building complete groups
3. Joining groups with dashes only between them