# 482. License Key Formatting

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
```

**Thoughts:**

1. Reverse the order to group the string (first group length is flexible)
2. When group length is equal to k, add a "-"

**Code**

```java
class Solution {
    public String licenseKeyFormatting(String S, int K) {
        StringBuilder sb = new StringBuilder();
        for(int i = S.length()- 1; i >= 0 ; i --){
            if(S.charAt(i) != '-'){
                sb.append(sb.length() % (K + 1) == K ?"-":"").append(S.charAt(i));
            }    

        }

        return sb.reverse().toString().toUpperCase();
    }
}
```

```cpp
class Solution {
public:
    string licenseKeyFormatting(string S, int K) {
        string res;
        for (auto i =S.rbegin(); i< S.rend(); i++){
          if(*i != '-') (res.size()%(K + 1) - K ? res: res+= '-') += toupper(*i);  
        }

        reverse(res.begin(), res.end());
        return res;
    }
};
```
