<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 upper(string s)
    {
      
        for(int i=0;i<s.size();i++)
        {
            if(s[i]>='a' and s[i]<='z')
            {
                s[i] = s[i] - 'a'+ 'A';  
            }
        }
        return s;
    }
    string remove(string s)
    {
        string temp;
        for(int i=0;i<s.size();i++)
        {
            if(s[i]!='-')
            {
                temp+=s[i];
            }
            else
            {
                continue;
            }
        }
        return temp;
    }
    string licenseKeyFormatting(string s, int k) {
        //
        int n = s.size();
    
         // this should be first part length
        // if remainder is zero then each part is of equal length
        string ans;
        // we also need to transform string to uppercase
        string newstring = upper(s);
        newstring = remove(newstring);
       
        int charandnumcnt = newstring.size();
        if(charandnumcnt==0)
        {
            return ans;
        }
        int remainder = charandnumcnt%k;
        // we will remove dashes
        
        int i=0;
      
        while(remainder!=0)
        {
            ans+=newstring[i];
            i++;
            remainder--;
            if(remainder==0)
            {
                ans+='-';
            }
           
        }
        
        // for first part we are done
        int cnt=0;
        
        for(i;i<charandnumcnt;i++)
        {
            while(cnt<k)
            {
                ans+=newstring[i];
                cnt++;
                i++;
            }
            if(cnt==k)
            {
                 ans+='-'; // adding the dash after each part
                cnt=0; // updating the cnt
                i--;  // at the end i was getting incremented two times so for that
            }
           
        }
            ans.pop_back();
             return ans;
        
    }
};
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