<p>Given a string <code>s</code>, return the maximum number of occurrences of <strong>any</strong> substring under the following rules:</p>

<ul>
	<li>The number of unique characters in the substring must be less than or equal to <code>maxLetters</code>.</li>
	<li>The substring size must be between <code>minSize</code> and <code>maxSize</code> inclusive.</li>
</ul>

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

<pre>
<strong>Input:</strong> s = &quot;aababcaab&quot;, maxLetters = 2, minSize = 3, maxSize = 4
<strong>Output:</strong> 2
<strong>Explanation:</strong> Substring &quot;aab&quot; has 2 occurrences in the original string.
It satisfies the conditions, 2 unique letters and size 3 (between minSize and maxSize).
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;aaaa&quot;, maxLetters = 1, minSize = 3, maxSize = 3
<strong>Output:</strong> 2
<strong>Explanation:</strong> Substring &quot;aaa&quot; occur 2 times in the string. It can overlap.
</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>1 &lt;= maxLetters &lt;= 26</code></li>
	<li><code>1 &lt;= minSize &lt;= maxSize &lt;= min(26, s.length)</code></li>
	<li><code>s</code> consists of only lowercase English letters.</li>
</ul>

class Solution:
    def maxFreq(self, s: str, maxLetters: int, minSize: int, maxSize: int) -> int:
        ans = ""
        freqs = {}

        curStr = []
        uniqueLetters = 0
        letters = [0] * 26
        for c in s:
            curStr.append(c)
            letters[ord(c)-ord('a')] += 1
            uniqueLetters += 1 if letters[ord(c)-ord('a')] == 1 else 0

            if len(curStr) > minSize:
                cc = ord(curStr[0]) - ord('a')
                letters[cc] -= 1
                uniqueLetters -= 1 if letters[cc] == 0 else 0
                curStr.pop(0)
            
            if len(curStr) == minSize and uniqueLetters <= maxLetters:
                ss = ''.join(curStr)
                if ss not in freqs:
                    freqs[ss] = 0
                freqs[ss] += 1
                print(ss, freqs[ss])
                if ans == '' or freqs[ans] < freqs[ss]:
                    ans = ss
            
        return 0 if ans == "" else freqs[ans]

18600