<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:
        hm = {}  # Frequency map of current window
        freqMap = {}  # Frequency map of valid substrings
        l = 0  # Left pointer for sliding window
        
        for r in range(len(s)):
            # Add the current character to the window frequency map
            if s[r] in hm:
                hm[s[r]] += 1
            else:
                hm[s[r]] = 1
            
            # Once the window reaches the size of minSize
            if r - l + 1 == minSize:
                # Check if the number of unique characters is valid
                if len(hm) <= maxLetters:
                    substring = s[l:r+1]  # The current window substring
                    if substring in freqMap:
                        freqMap[substring] += 1
                    else:
                        freqMap[substring] = 1
                
                # Now slide the window by removing the leftmost character
                hm[s[l]] -= 1
                if hm[s[l]] == 0:
                    del hm[s[l]]  # Remove the character entirely if its count is 0
                l += 1  # Move the left pointer to the right
        
        # Find the maximum frequency of any substring
        return max(freqMap.values(), default=0)

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