# License Key Formatting

[https://leetcode.com/problems/license-key-formatting/description/](https://leetcode.com/problems/license-key-formatting/description/)

> You are given a license key represented as a string S which consists only alphanumeric character and dashes. The string is separated into N+1 groups by N dashes.
>
> Given a number K, we would want to reformat the strings such that each group containsexactlyK 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 all lowercase letters should be converted to uppercase.
>
> Given a non-empty string S and a number K, format the string according to the rules described above.
>
> **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.
> ```
>
> **Note:**
>
> 1. The length of string S will not exceed 12,000, and K is a positive integer.
> 2. String S consists only of alphanumerical characters (a-z and/or A-Z and/or 0-9) and dashes(-).
> 3. String S is non-empty.

## Thoughts

只允许第一组少于K个，因此第一组存N % K个即可。比较tricky的地方在于当第一组个数为0时第一个"-"不应该存在。

## Code

```
class Solution {
public:
    string licenseKeyFormatting(string S, int k) {
        string s;
        for (char c : S) {
            if (c != '-') {
                s.push_back(toupper(c));
            }
        }
        int first = s.length() % k;
        string res = s.substr(0, first);
        for (int i = 0; i < s.length() - first; ++i) {
            if ((i + first) && i % k == 0) {
                res.push_back('-');
            }
            res.push_back(s[first + i]);
        }

        return res;
    }
};
```

## Analysis

时间复杂度O(N)
