<p>You are given a license key represented as a string <code>s</code> that consists of only alphanumeric characters and dashes. The string is separated into <code>n + 1</code> groups by <code>n</code> dashes. You are also given an integer <code>k</code>.</p>

<p>We want to reformat the string <code>s</code> such that each group contains exactly <code>k</code> characters, except for the first group, which could be shorter than <code>k</code> but still must contain at least one character. Furthermore, there must be a dash inserted between two groups, and you should convert all lowercase letters to uppercase.</p>

<p>Return <em>the reformatted license key</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;5F3Z-2e-9-w&quot;, k = 4
<strong>Output:</strong> &quot;5F3Z-2E9W&quot;
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;2-5g-3-J&quot;, k = 2
<strong>Output:</strong> &quot;2-5G-3J&quot;
<strong>Explanation:</strong> 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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of English letters, digits, and dashes <code>&#39;-&#39;</code>.</li>
	<li><code>1 &lt;= k &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    string licenseKeyFormatting(string s, int k) {
        string str = "";
        for (auto c : s) {
            if (c == '-') continue;
            if (isdigit(c)) str += c;
            else str += toupper(c);
        }

        string res = "";
        reverse(str.begin(), str.end());
        for (int i = 0; i < str.size()/k; ++i) {
            res += str.substr(i*k, k) + '-';
        }
        if (str.size() % k != 0)
            res += str.substr(str.size()/k *k);
        else {
            if (!res.empty()) res.pop_back();
        }
        reverse(res.begin(), res.end());

        return res;
    }
};
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