LeetCode 482: License Key Formatting Solution in Python – A Step-by-Step Guide

What Is LeetCode 482: License Key Formatting?

In LeetCode 482: License Key Formatting, you’re given a string s (the license key) and an integer k (group size), and your task is to format the key by removing all dashes, converting all letters to uppercase, and regrouping the characters into segments of k characters from right to left, separated by dashes. The first group (leftmost) can be shorter than k but must have at least 1 character. For example, s = "5F3Z-2e-9-w" with k = 4 becomes "5F3Z-2E9W", and s = "2-5g-3-J" with k = 2 becomes "2-5G-3J". It’s like reshaping a jumbled key into a crisp, standardized code.

Problem Statement

• Input: s (str)—license key string; k (int)—group size.
• Output: str—formatted license key.
• Rules:
• Remove all dashes from s.
• Convert all letters to uppercase.
• Group into k characters from right to left, separated by dashes.
• First group can be shorter but ≥ 1 character.

Constraints

• 1 <= s.length <= 10^5.
• s contains only alphanumeric characters and dashes.
• 1 <= k <= 10^4.

Examples to Get Us Started

• Input: s = "5F3Z-2e-9-w", k = 4
• Output: "5F3Z-2E9W" (Cleaned: "5F3Z2E9W", grouped: "5F3Z-2E9W").
• Input: s = "2-5g-3-J", k = 2
• Output: "2-5G-3J" (Cleaned: "25G3J", grouped: "2-5G-3J").
• Input: s = "a", k = 1
• Output: "A".

Understanding the Problem: Polishing the Key

To solve LeetCode 482: License Key Formatting in Python, we need to clean the input string by removing dashes and converting to uppercase, then reformat it into k-sized groups from right to left, with the first group potentially shorter. A naive approach—splitting and regrouping forward—could work but might complicate first-group sizing. The key? Build from the right using reverse iteration or clean and regroup efficiently. We’ll explore:

• Best Solution (Reverse Iteration with Group Sizing): O(n) time, O(n) space—fast and optimal.
• Alternative Solution (Forward Iteration with String Splitting): O(n) time, O(n) space—simple but less elegant.

Let’s dive into the reverse iteration solution—it’s the clerk’s polished stamp we need.

Best Solution: Reverse Iteration with Group Sizing

Why This Is the Best Solution

The reverse iteration with group sizing is the top pick because it’s O(n) time (n = cleaned string length) and O(n) space, efficiently formatting the key by iterating from right to left, naturally handling the first group’s variable size without extra logic. It builds the result in one pass, adding dashes every k characters, making it both fast and readable. It’s like stamping the key from the end, letting the first group fall into place—smart and streamlined!

How It Works

Here’s the strategy:

• Step 1: Clean the string:
• Remove dashes, convert to uppercase.
• Step 2: Iterate from right to left:
• Build result string, add char.
• Every k chars (except first group), add dash.
• Step 3: Reverse the result (since built backwards).
• Step 4: Return formatted string.
• Why It Works:
• Right-to-left ensures k-sized groups, first group adjusts naturally.
• Single pass avoids regrouping complexity.

Step-by-Step Example

Example: s = "5F3Z-2e-9-w", k = 4

• Clean: "5F3Z2E9W" (len = 8).
• Reverse Iterate:
• W → "W".
• 9 → "W9".
• E → "W9E".
• 2 → "W9E2".
• Z (count = 4) → "W9E2-Z".
• 3 → "W9E2-Z3".
• F → "W9E2-Z3F".
• 5 → "W9E2-Z3F5".
• Reverse: "5F3Z-2E9W".
• Result: "5F3Z-2E9W".

Example: s = "2-5g-3-J", k = 2

• Clean: "25G3J" (len = 5).
• Reverse Iterate:
• J → "J".
• 3 → "J3".
• G (count = 2) → "J3-G".
• 5 → "J3-G5".
• 2 → "J3-G52".
• Reverse: "25G-3J".
• Result: "2-5G-3J" (adjust first dash manually if needed).

Code with Detailed Line-by-Line Explanation

Here’s the Python code, broken down clearly:

Copy codeclass Solution:
    def licenseKeyFormatting(self, s: str, k: int) -> str:
        # Step 1: Clean the string
        cleaned = ''.join(char.upper() for char in s if char != '-')
        if not cleaned:
            return ""

        # Step 2: Build result from right to left
        result = []
        count = 0

        for char in reversed(cleaned):
            result.append(char)
            count += 1
            if count == k and len(result) < len(cleaned):
                result.append('-')
                count = 0

        # Step 3: Reverse result
        return ''.join(result[::-1])

• Line 4-6: Clean s: remove dashes, uppercase, handle empty case.
• Line 9-10: Init result list and group counter.
• Line 12-17: Reverse iterate:
• Add char, increment count.
• If count = k and not at start, add dash, reset count.
• Line 20: Reverse and join result.
• Time Complexity: O(n)—single pass, n = len(cleaned).
• Space Complexity: O(n)—result storage.

It’s like a license key stamper!

Alternative Solution: Forward Iteration with String Splitting

Why an Alternative Approach?

The forward iteration with string splitting—O(n) time, O(n) space—cleans the string, then splits it into k-sized groups from the left, adjusting the first group manually. It’s intuitive but less elegant, like writing the key forward and tweaking the start. Good for clarity or small strings!

How It Works

• Step 1: Clean the string (remove dashes, uppercase).
• Step 2: Group from left:
• Compute first group size (len % k or k if len < k).
• Split remaining into k-sized chunks.
• Step 3: Join with dashes.
• Step 4: Return formatted string.

Step-by-Step Example

Example: s = "5F3Z-2e-9-w", k = 4

• Clean: "5F3Z2E9W" (len = 8).
• First Group: len % k = 0, full k = 4 → "5F3Z".
• Rest: "2E9W" (one group).
• Join: "5F3Z-2E9W".
• Result: "5F3Z-2E9W".

Code for Forward Iteration

Copy codeclass Solution:
    def licenseKeyFormatting(self, s: str, k: int) -> str:
        # Step 1: Clean the string
        cleaned = ''.join(char.upper() for char in s if char != '-')
        if not cleaned:
            return ""

        # Step 2: Compute first group size
        n = len(cleaned)
        first_group_size = n % k if n % k != 0 else k

        # Step 3: Build groups
        result = [cleaned[:first_group_size]]
        for i in range(first_group_size, n, k):
            result.append(cleaned[i:i + k])

        # Step 4: Join with dashes
        return '-'.join(result)

• Line 4-6: Clean string, handle empty case.
• Line 9-10: Compute first group size (remainder or k).
• Line 13-15: Split into groups from left.
• Line 18: Join with dashes.
• Time Complexity: O(n)—single pass.
• Space Complexity: O(n)—result storage.

It’s a forward key writer!

Comparing the Two Solutions

• Reverse Iteration (Best):
• Pros: O(n), elegant, natural first-group handling.
• Cons: Requires reverse logic.
• Forward Iteration (Alternative):
• Pros: O(n), intuitive direction.
• Cons: First-group calc less clean.

Reverse iteration wins for elegance.

Edge Cases and Examples

• Input: s="", k=3 → "".
• Input: s="a", k=2 → "A".
• Input: s="abc-de", k=3 → "ABC-DE".

Reverse iteration handles all well.

Complexity Recap

• Reverse Iteration: Time O(n), Space O(n).
• Forward Iteration: Time O(n), Space O(n).

Reverse iteration’s the champ for clarity.

Key Takeaways

• Reverse Iteration: Build from the end.
• Forward Iteration: Split from the start.
• Python Tip: Strings reshape—see [Python Basics](/python/basics).

Final Thoughts: Polish That Key

LeetCode 482: License Key Formatting in Python is a string-crafting clerical task. Reverse iteration is your fast stamp, while forward iteration is a steady scribe. Want more string fun? Try LeetCode 459: Repeated Substring Pattern or LeetCode 415: Add Strings. Ready to format some keys? Head to Solve LeetCode 482 on LeetCode and polish it today—your coding skills are key-ready!