<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 defaultdict
class Solution:
    def repeatLimitedString(self, s: str, repeatLimit: int) -> str:
        alphaMap = defaultdict(int)
        for c in s:
            alphaMap[c] = alphaMap[c] + 1
        
        lexLargestKeys = list(alphaMap.keys())
        lexLargestKeys.sort()
        lexLargestKeys.reverse()

        #print(alphaMap)
        #print(lexLargestKeys)

        def consume(aMap, keysList, c):
            # consume the character from the map
            # if there are no entries left in the map, remove it from the keysList
            aMap[c] = aMap[c] - 1
            if aMap[c] == 0:
                keysList.remove(c)

        constructedStr = ""
        repeated = 1
        prev_key = ''
        while len(lexLargestKeys) > 0:
            #print(lexLargestKeys, prev_key, repeated, constructedStr)
            k = lexLargestKeys[0]
            if repeated == repeatLimit and k == prev_key:
                if len(lexLargestKeys) == 1:
                    # no other characters to insert, so we just finish
                    break
                else:
                    # use the next one
                    k = lexLargestKeys[1]

            # add k to the string and update our local state (prev key and repeated)
            constructedStr += k
            if k == prev_key:
                repeated = repeated + 1
            else:
                prev_key = k
                repeated = 1

            consume(alphaMap, lexLargestKeys, k)
    
        return constructedStr
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