<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>

class Solution:
    def repeatLimitedString(self, s: str, repeatLimit: int) -> str:
        from collections import Counter
        import heapq
        
        # Count frequency of each character
        char_count = Counter(s)
        max_heap = []
        
        # Create a max heap based on character ascii values
        for char, count in char_count.items():
            heapq.heappush(max_heap, (-ord(char), char, count))
        
        result = []
        while max_heap:
            first_entry = heapq.heappop(max_heap)
            _, first_char, first_count = first_entry
            
            # Append as many of the first_char as possible up to repeatLimit
            limit = min(repeatLimit, first_count)
            result.append(first_char * limit)
            
            first_count -= limit
            
            # If there's still some of the first_char left, we need to place a different character
            if first_count > 0:
                if not max_heap:  # If no more characters are left, break
                    break
                
                second_entry = heapq.heappop(max_heap)
                _, second_char, second_count = second_entry
                
                # Append one of the second_char
                result.append(second_char)
                
                # Push back the remaining second_char
                if second_count > 1:
                    heapq.heappush(max_heap, (-ord(second_char), second_char, second_count - 1))
                
                # Push back the remaining first_char
                heapq.heappush(max_heap, (-ord(first_char), first_char, first_count))
            elif first_count == 0 and max_heap:  # If first_char is exhausted but there are more characters
                continue
        
        return ''.join(result)
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