<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:
        result = 0
        for size in range(minSize, maxSize + 1):
            counts = {}
            letters = {}
            for i in range(len(s) - size + 1):
                current_substring = s[i:i+size]
                
                if not letters:
                    for l in current_substring:
                        letters[l] = letters.get(l, 0) + 1
                else:
                    # Decrement the counts for i-1th element
                    if letters[s[i - 1]] <= 1:
                        del letters[s[i - 1]]
                    else:
                        letters[s[i - 1]] -= 1

                    # Increment the counts for i+size-1th element
                    letters[s[i + size - 1]] = letters.get(s[i + size - 1], 0) + 1
                
                if len(letters) <= maxLetters:
                    counts[current_substring] = counts.get(current_substring, 0) + 1
            
            if counts:
                result = max(result, max(counts.values()))
        
        return result
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