482. License Key Formatting

1. 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 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.

2. Example

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.

3. 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.

4. Solutions

String Formatting

n = i_key.size()
Time complexity: O(n)
Space complexity: O(n)

class Solution
{
public:
    // string licenseKeyFormatting(string S, int K)
    string licenseKeyFormatting(string &i_key, int length)
    {
        string keyWithoutDash;
        for(int i = 0; i < i_key.size(); i++)
        {
            if(i_key[i] != '-')
            {
                keyWithoutDash.push_back(toupper(i_key[i]));
            }
        }

        if(keyWithoutDash.empty()) return string("");

        int index = 0;
        string result;
        if(keyWithoutDash.size() % length != 0)
        {
            result = keyWithoutDash.substr(index, keyWithoutDash.size() % length) + string("-");

            index += keyWithoutDash.size() % length;
        }

        for(; index < keyWithoutDash.size(); index += length)
        {
            result += keyWithoutDash.substr(index, length) + string("-");
        }

        result.pop_back();

        return result;
    }
};