<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:
        # Dictionary to store the count of each substring
        substring_counts = {}
        
        # Sliding window to generate substrings
        for size in range(minSize, maxSize + 1):
            window_start = 0
            window_end = size
            
            while window_end <= len(s):
                substring = s[window_start:window_end]
                unique_chars = len(set(substring))
                
                # Check if the substring has unique characters less than or equal to maxLetters
                if unique_chars <= maxLetters:
                    # Update the count of the substring in the dictionary
                    substring_counts[substring] = substring_counts.get(substring, 0) + 1
                
                window_start += 1
                window_end += 1
        
        # Find the maximum count of any valid substring
        max_count = 0
        for count in substring_counts.values():
            max_count = max(max_count, count)
        
        return max_count

# Example usage:
solution = Solution()
s = "aababcaab"
maxLetters = 2
minSize = 3
maxSize = 4
print(solution.maxFreq(s, maxLetters, minSize, maxSize))  # Output should be 2
108700