# License Key Formatting

**Easy**

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.

## Solution

Kind of verbose (using additional character list, and finding the length of first group before '-') - (13ms

```java
class Solution {
    public String licenseKeyFormatting(String S, int K) {
        int count = 0;
        List<Character> chs = new ArrayList<>();
        for (int i = 0; i < S.length(); i++) {
            if (S.charAt(i) != '-') {
                count++;
                chs.add(S.charAt(i));
            }
        }
        int groupOneLength = count % K;

        StringBuilder sb = new StringBuilder();
        for (int j = 0; j < count; j++) {
            if (j == groupOneLength && groupOneLength > 0) {
                sb.append('-');
            } else if (j > groupOneLength && (j - groupOneLength) % K == 0) {
                sb.append('-');
            }
            sb.append(upperCase(chs.get(j)));
        }
        return sb.toString();
    }

    private char upperCase(char ch) {
        if (ch >= 'a' && ch <= 'z') {
            ch = (char) ('A' + ch - 'a');
        }

        return ch;
    }
}
```

**Preferred (12ms, 92.13%)**

用两个index来操作，一个主循环i遍历输入字符串S，一个辅助计数cnt，记录生成的String的段落长度

```java
class Solution {
    public String licenseKeyFormatting(String S, int K) {
        StringBuilder sb = new StringBuilder();
        int cnt = 0;
        for(int i = S.length()-1; i>= 0; i--){
            char c = S.charAt(i);
            if(c == '-') continue;
            if(cnt == K){
                sb.append('-');
                cnt = 0;
            }
            sb.append(Character.toUpperCase(c));
            cnt += 1;
        }
        return sb.reverse().toString();
    }
}
```

A variation (Using cnt % (K + 1) == K as the timing for adding '-')

```java
class Solution {
    public String licenseKeyFormatting(String S, int K) {
        StringBuilder sb = new StringBuilder();
        int cnt = 0;
        for(int i = S.length()-1; i>= 0; i--){
            char c = S.charAt(i);
            if(c == '-') continue;
            if(cnt % (K + 1) == K){
                sb.append('-');
                cnt += 1;
            }
            sb.append(Character.toUpperCase(c));
            cnt += 1;
        }
        return sb.reverse().toString();
    }
}
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://aaronice.gitbook.io/lintcode/string/license-key-formatting.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.