You are given a license key represented as a string s that consists of only alphanumeric characters and dashes. The string is separated into n + 1 groups by n dashes. You are also given an integer k.
We want to reformat the string s such that each group contains exactly k characters, except for the first group, which could be shorter than k but still must contain at least one character. Furthermore, there must be a dash inserted between two groups, and you should convert all lowercase letters to uppercase.
Return the reformatted license key.
Example 1:
Input: s = "5F3Z-2e-9-w", k = 4 Output: "5F3Z-2E9W" Explanation: The string s has been split into two parts, each part has 4 characters. Note that the two extra dashes are not needed and can be removed.
Example 2:
Input: s = "2-5g-3-J", k = 2 Output: "2-5G-3J" Explanation: The string s has been split into three parts, each part has 2 characters except the first part as it could be shorter as mentioned above.
Constraints:
1 <= s.length <= 105s consists of English letters, digits, and dashes '-'.1 <= k <= 104Problem Overview: You receive a license key string containing alphanumeric characters and dashes. The goal is to remove existing dashes, convert letters to uppercase, and regroup characters so the first group may be shorter while the rest contain exactly k characters separated by dashes.
Approach 1: Iterative Approach with StringBuilder (O(n) time, O(n) space)
This approach processes the string in a single forward pass after removing dashes. First, iterate through the original string and build a cleaned version containing only alphanumeric characters while converting letters to uppercase. Then compute the size of the first group using len % k. Append the first group (if non-zero) and continue appending the remaining characters in blocks of k, inserting dashes between groups. A StringBuilder or dynamic string buffer avoids repeated string concatenation, keeping the operation linear. This method is easy to reason about because the grouping logic follows the final formatted order.
The algorithm performs a simple scan and controlled appends, making it a good example of practical string manipulation. You touch each character once during cleaning and once during grouping, so the total runtime remains O(n) with O(n) auxiliary space for the result.
Approach 2: Reverse and Append Approach (O(n) time, O(n) space)
The reverse-building technique simplifies grouping logic by working from the end of the string. Iterate backward through the original string, skipping dashes and converting characters to uppercase. Maintain a counter that tracks how many characters have been added to the current group. Every time the counter reaches k, append a dash before continuing the next group.
Since characters are appended in reverse order, the result must be reversed at the end. The key insight is that grouping from the end guarantees that all groups except possibly the first naturally contain exactly k characters. This removes the need to precompute the first group length. The technique resembles a simple string traversal pattern often used in formatting and parsing tasks.
Recommended for interviews: Both solutions run in O(n) time and O(n) space, which is the expected optimal complexity. The reverse-and-append method tends to be cleaner because it avoids calculating the first group length explicitly. Still, demonstrating the forward iterative approach shows you understand grouping logic and controlled string construction. Interviewers typically expect a linear pass using a buffer structure rather than repeated string concatenation.
This approach involves first cleaning up the input string by removing all non-alphanumeric characters and converting it to uppercase. Then, starting from the end of the string, we build the new license key by appending groups of size k. This is achieved using a StringBuilder for efficient string manipulation.
We begin by removing all dashes and converting the string to uppercase. We calculate the remainder, which helps in determining the first group size that could be less than k. We use a loop to iterate through the cleaned-up string in steps of k and store these groups. Finally, we join the groups with a dash and return the result.
Python
JavaScript
Time Complexity: O(n), where n is the length of the string s. Space Complexity: O(n) due to the additional storage requirements for the new formatted string.
This method suggests reversing the cleaned-up string to easily extract groups from the end, and then reversing the result at the end. This eliminates the need for calculating the remainder at the start and allows straightforward appending of characters.
By reversing the process, we simplify grouping into steps of k directly from the end without needing any special handling for the first group. We maintain a counter to insert a dash after every k characters and finally reverse the accumulated result.
Time Complexity: O(n), where n is the length of s. Space Complexity: O(n) for storing the result string temporarily.
| Approach | Complexity |
|---|---|
| Iterative Approach with StringBuilder | Time Complexity: O(n), where n is the length of the string |
| Reverse and Append Approach | Time Complexity: O(n), where n is the length of |
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Iterative Approach with StringBuilder | O(n) | O(n) | When you want straightforward forward traversal and explicit control over the first group size |
| Reverse and Append Approach | O(n) | O(n) | When grouping from the end simplifies logic and avoids computing the first group length |
License Key Formatting • Kevin Naughton Jr. • 16,160 views views
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