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 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:
 "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:

The length of string S will not exceed 12,000, and K is a positive integer.
String S consists only of alphanumerical characters (a-z and/or A-Z and/or 0-9) and dashes(-).
String S is non-empty.

Solution

(1) Java

class Solution {
    public String licenseKeyFormatting(String S, int K) {
        String[] strs = S.split("-");
        StringBuilder sb = new StringBuilder();
        for (String str : strs) {
            sb.append(str.toUpperCase());
        }
        String newstr = sb.toString();
        int groupNum = newstr.length()/K;
        int subNum = newstr.length()%K;
        StringBuilder rst =  new StringBuilder();
        int start = 0;
        int end = K;
        if (subNum != 0) {
            rst.append(newstr.substring(start, subNum)).append("-");
            start = subNum;
            end = subNum+K;
        }
        for (int i = 0; i < groupNum; i++) {
            rst.append(newstr.substring(start, end)).append("-");
            start = end;
            end += K;
        }
        if (rst.length() > 0)
            rst.deleteCharAt(rst.length()-1);
        return rst.toString();
    }
}

(2) Python

class Solution:
    def licenseKeyFormatting(self, S, K):
        """
        :type S: str
        :type K: int
        :rtype: str
        """
        if not S:
            return S
        rst = ""
        length = 0
        for i in range(len(S)-1, -1, -1):
            if S[i] != "-":
                rst += S[i].upper()
                length += 1
                if length%K == 0:
                    rst += "-"                   
        if rst and rst[-1] == "-":
            rst = rst[0:len(rst)-1]
        return rst[::-1]

482. License Key Formatting