Maximum Number of Occurrences of a Substring
<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>
Hash Table; String; Sliding Window
class Solution {
    public int maxFreq(String s, int maxLetters, int minSize, int maxSize) {
        int ans = 0;
        Map<String, Integer> cnt = new HashMap<>();
        for (int i = 0; i < s.length() - minSize + 1; ++i) {
            String t = s.substring(i, i + minSize);
            Set<Character> ss = new HashSet<>();
            for (int j = 0; j < minSize; ++j) {
                ss.add(t.charAt(j));
            }
            if (ss.size() <= maxLetters) {
                cnt.put(t, cnt.getOrDefault(t, 0) + 1);
                ans = Math.max(ans, cnt.get(t));
            }
        }
        return ans;
    }
}