License Key Formatting


#1

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 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 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: ā€œ5F3Z-2E9Wā€

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: ā€œ2-5G-3Jā€

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.

#2

Process the string from the rear, add a - every K characters and change to upper case. Finally reverse it.

func licenseKeyFormatting(S string, K int) string {
	var sb strings.Builder
    t := 0
    for i:=len(S)-1;i>=0; i--{
        ch := S[i] 
        if ch == '-'{
            continue
        } else if ch >= 'a' && ch<='z' {
            ch = byte(int(ch) + int('Z')-int('z'))
        }
        
        sb.WriteByte(ch)
        t++
        if t==K {
            t = 0
            sb.WriteByte('-')
        }
    }
    
    res := sb.String()
    if len(res)>0 && res[len(res)-1] == '-' {
        res = res[:len(res)-1]
    }
    
    return reverse([]byte(res))
}

func reverse(b []byte) string {
	for l, r := 0, len(b)-1; l < r; l, r = l+1, r-1 {
		b[l], b[r] = b[r], b[l]
	}
	return string(b)
}