<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:
        unique = defaultdict(int)
        size = 0

        count = defaultdict(int)

        left = 0
        for i in range(len(s)):
            unique[s[i]] += 1
            size += 1
            while(len(unique) > maxLetters or size > maxSize):
                unique[s[left]]-=1
                if unique[s[left]] == 0: unique.pop(s[left])
                left += 1
                size -=1 
            if minSize <= size <= maxSize:
                # print(i, s[left : i + 1], minSize, size)
                # for i in range(minSize - size):
                for k in range(size - minSize+ 1):
                    # print(k, s[left + k  : i + 1])
                    count[s[left + k  : i + 1]] += 1
        

        # print(count)
        largest = 0
        for key in count.keys():
            if count[key] > largest:
                largest = count[key]
        return largest            


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