<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:
        maxCounter = 0 
        storedCounter = defaultdict(int)

        s_i = 0 
        while s_i < len(s) or s_i + minSize < len(s):
            e_i = s_i + minSize
            curr_str = s[s_i: min(len(s), e_i)]
            curr_chars = set(curr_str)
            curr_size = len(curr_str)
            if curr_size < minSize or len(curr_chars) > maxLetters: 
                s_i += 1
                continue 
            storedCounter[curr_str] += 1
            while e_i < len(s) and curr_size < maxSize and len(curr_chars) < maxLetters: 
                curr_str += s[e_i]
                curr_size += 1
                curr_chars.add(s[e_i])
                storedCounter[curr_str] += 1
                e_i += 1
            s_i += 1
        if len(storedCounter.values()) > 0:
            return max(list(storedCounter.values()))
        return 0





            


        #aababcaab

        """
        aaba
        """

75900