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:

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.

string licenseKeyFormatting(string S, int K) { // time: O(n); space: O(n)
    string res;
    int cnt = 0;
    for (int i = S.length() - 1; i >= 0; --i) {
        if (S[i] == '-') continue;
        res += toupper(S[i]);
        if (++cnt % K == 0) {
            res += '-';
            cnt = 0;
        }
    }
    if (!res.empty() && res.back() == '-') res.pop_back();
    reverse(res.begin(), res.end());
    return res;
}

string licenseKeyFormatting(string S, int K) { // time: O(n); space: O(n)
    string res;
    for (int i = S.length() - 1; i >= 0; --i) {
        if (S[i] == '-') continue;
        res += (res.length() % (K + 1) ) == K ? "-" : "";
        res += toupper(S[i]);
    }
    reverse(res.begin(), res.end());
    return res;
}