# 0482-license-key-formatting


Try it on <a href='https://leetcode.com/problems/0482-license-key-formatting'>leetcode</a>

## Description
<div class="description">
<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>

</div>

## Solution(Python)
```Python
class Solution:
    def licenseKeyFormatting(self, s: str, k: int) -> str:
        total_character_count = sum([1 for c in s if c != '-'])
        first_group_size = total_character_count % k
        if first_group_size == 0:
            first_group_size = k
        ans = []
        i = 0
        count = 0
        n = len(s)
        while i  < n:
            if (count == first_group_size):
                count = 0

                break
            if s[i] != '-':
                count +=1
                ans.append(s[i].upper())
            i+=1
        if i>= n:
            return ''.join(ans)
        ans.append('-')

        while i < n:
            if s[i] != '-':
                if count == k:
                    ans.append('-')
                    count = 0 
                ans.append(s[i].upper())
                count+=1
            i+=1
        return ''.join(ans)



        
```