<p>You are given a string <code>s</code> and an integer <code>repeatLimit</code>. Construct a new string <code>repeatLimitedString</code> using the characters of <code>s</code> such that no letter appears <strong>more than</strong> <code>repeatLimit</code> times <strong>in a row</strong>. You do <strong>not</strong> have to use all characters from <code>s</code>.</p>

<p>Return <em>the <strong>lexicographically largest</strong> </em><code>repeatLimitedString</code> <em>possible</em>.</p>

<p>A string <code>a</code> is <strong>lexicographically larger</strong> than a string <code>b</code> if in the first position where <code>a</code> and <code>b</code> differ, string <code>a</code> has a letter that appears later in the alphabet than the corresponding letter in <code>b</code>. If the first <code>min(a.length, b.length)</code> characters do not differ, then the longer string is the lexicographically larger one.</p>

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

<pre>
<strong>Input:</strong> s = &quot;cczazcc&quot;, repeatLimit = 3
<strong>Output:</strong> &quot;zzcccac&quot;
<strong>Explanation:</strong> We use all of the characters from s to construct the repeatLimitedString &quot;zzcccac&quot;.
The letter &#39;a&#39; appears at most 1 time in a row.
The letter &#39;c&#39; appears at most 3 times in a row.
The letter &#39;z&#39; appears at most 2 times in a row.
Hence, no letter appears more than repeatLimit times in a row and the string is a valid repeatLimitedString.
The string is the lexicographically largest repeatLimitedString possible so we return &quot;zzcccac&quot;.
Note that the string &quot;zzcccca&quot; is lexicographically larger but the letter &#39;c&#39; appears more than 3 times in a row, so it is not a valid repeatLimitedString.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;aababab&quot;, repeatLimit = 2
<strong>Output:</strong> &quot;bbabaa&quot;
<strong>Explanation:</strong> We use only some of the characters from s to construct the repeatLimitedString &quot;bbabaa&quot;. 
The letter &#39;a&#39; appears at most 2 times in a row.
The letter &#39;b&#39; appears at most 2 times in a row.
Hence, no letter appears more than repeatLimit times in a row and the string is a valid repeatLimitedString.
The string is the lexicographically largest repeatLimitedString possible so we return &quot;bbabaa&quot;.
Note that the string &quot;bbabaaa&quot; is lexicographically larger but the letter &#39;a&#39; appears more than 2 times in a row, so it is not a valid repeatLimitedString.
</pre>

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

<ul>
	<li><code>1 &lt;= repeatLimit &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

from collections import deque
class Solution:
    def repeatLimitedString(self, s: str, repeatLimit: int) -> str:
        
        slack_q = deque()
        s = sorted(s,reverse=True)
        
        ans = []
        i = 0
        ans.append(s[i])
        prev = s[i]
        total = 1
        i+=1
        #print(s)
        while(i < len(s)):
           # print(ans,prev,total,slack_q)
            if len(slack_q) == 0:
                if s[i] == prev:
                    if total < repeatLimit:
                        ans.append(s[i])
                        total+=1
                        prev = s[i]
                        i+=1
                    else:
                        slack_q.append(s[i])
                        prev = s[i]
                        i+=1
                else:
                    ans.append(s[i])
                    total = 1
                    prev = s[i]
                    i+=1
            else:
                if slack_q[0] != prev:
                    ans.append(slack_q[0])
                    prev = slack_q[0]
                    total = 1
                    slack_q.popleft()
                else:
                    if total < repeatLimit:
                        ans.append(slack_q[0])
                        prev = slack_q[0]
                        total+=1
                        slack_q.popleft()
                    else:
                        if s[i] == prev:
                            slack_q.append(s[i])
                            i+=1
                        else:
                            ans.append(s[i])
                            total = 1
                            prev = s[i]
                            i+=1
        while(len(slack_q)>0):
            top = slack_q[0]
            if top == prev:
                if total < repeatLimit:
                    ans.append(top)
                    slack_q.popleft()
                    total+=1
                    prev = top
                else:
                    slack_q.popleft()
            else:
                ans.append(top)
                slack_q.popleft()
                total = 1
                prev = top
        return "".join(ans) 

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