<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>

import heapq

class Solution:
    def repeatLimitedString(self, s: str, limit: int) -> str:
        # Create a dictionary to store frequency using -ord(char) as the key
        dic = {}
        ans = ""
        
        # Count the frequency of each character
        for i in range(len(s)):
            if -ord(s[i]) not in dic:
                dic[-ord(s[i])] = 1
            else:
                dic[-ord(s[i])] += 1

        # Convert the dictionary to a list of tuples and heapify it
        di = list(dic.items())
        heapq.heapify(di)
        
        while len(di) != 0:
            x = heapq.heappop(di)
            common = min(limit, x[1])  # Number of characters to take
            
            # Add the current character up to 'common' times
            ans += chr(-x[0]) * common
            
            # If there are more occurrences of this character than the limit
            if x[1] - common > 0:
                # Check if there are other characters left in the heap
                if len(di) > 0:
                    y = heapq.heappop(di)
                    # Add one of the next most frequent character
                    ans += chr(-y[0])
                    
                    # Decrease the count of this character and push it back if it has more occurrences
                    if y[1] - 1 > 0:
                        y = (y[0], y[1] - 1)
                        heapq.heappush(di, y)
                else:
                    return ans
                # Push the current character back with reduced count
                x = (x[0], x[1] - common)
                heapq.heappush(di, x)
        
        return ans


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