<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:
    def licenseKeyFormatting(self, s: str, k: int) -> str:
        # If we are given single element string
        if len(s) <= 1 and s[0] != '-':
            return str(s.upper())
        elif len(s) <= 1 and s[0] == '-':
            return ""

        # Initialize empty list
        noDash = deque()

        # Iterate through the strings
        for c in s:
            # Check if the current character is lowercase
            if c.islower():
                # If it is, then convert it to uppercase
                c = c.upper()
            # Append the character to our list only if it is not a dash
            if c != '-':
                noDash.append(c)
        print(noDash)

        noDash = ''.join(noDash)
        n = len(noDash) # Get the length of 'noDash' string

        # Use modulo w n w/k to get the starting position

        # if 'n % k = 0' --> We have even 'k' groups so we can start greating our 'k' grouping from the 0th index
        ans = ""
        if n % k == 0:
            for i in range(0, n, k):
                group = noDash[i:k + i]
                if k + i >= n:
                    ans += group
                else:
                    ans += group + '-'
        # if n % k != 0 -> We do NOT have even 'k' groups so we append the first 
        else:
            # Get all characters up to but not including element at 'n%k'th index for our 'ans' string and add a '-' at the end
            ans += noDash[0:n % k] + '-'
            # Start our loop at the 'n%k' index
            for i in range(n % k, n, k):
                # Create our 'group' string w/ 'k' characters
                group = noDash[i:k + i]
                # If it is the last group, don't add the '-'
                if k + i >= n:
                    ans += group
                else:
                    # Add the group to our 'ans' string with a '-'
                    ans += group + '-'
        
        # Return our 'ans' string
        return ans
18700