<p>You are given a string <code>s</code> that consists of lowercase English letters.</p>

<p>A string is called <strong>special</strong> if it is made up of only a single character. For example, the string <code>&quot;abc&quot;</code> is not special, whereas the strings <code>&quot;ddd&quot;</code>, <code>&quot;zz&quot;</code>, and <code>&quot;f&quot;</code> are special.</p>

<p>Return <em>the length of the <strong>longest special substring</strong> of </em><code>s</code> <em>which occurs <strong>at least thrice</strong></em>, <em>or </em><code>-1</code><em> if no special substring occurs at least thrice</em>.</p>

<p>A <strong>substring</strong> is a contiguous <strong>non-empty</strong> sequence of characters within a string.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;aaaa&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> The longest special substring which occurs thrice is &quot;aa&quot;: substrings &quot;<u><strong>aa</strong></u>aa&quot;, &quot;a<u><strong>aa</strong></u>a&quot;, and &quot;aa<u><strong>aa</strong></u>&quot;.
It can be shown that the maximum length achievable is 2.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abcdef&quot;
<strong>Output:</strong> -1
<strong>Explanation:</strong> There exists no special substring which occurs at least thrice. Hence return -1.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;abcaba&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> The longest special substring which occurs thrice is &quot;a&quot;: substrings &quot;<u><strong>a</strong></u>bcaba&quot;, &quot;abc<u><strong>a</strong></u>ba&quot;, and &quot;abcab<u><strong>a</strong></u>&quot;.
It can be shown that the maximum length achievable is 1.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= s.length &lt;= 50</code></li>
	<li><code>s</code> consists of only lowercase English letters.</li>
</ul>

class Solution:
    def count_occurance(self, subs, s):
        count = 0
        s_copy = s
        do = True
        while do:
            found = s_copy.find(subs)
            if found == -1:
                do= False
            else:
                s_copy = s_copy[found+1:]
                count += 1

        return count

    def maximumLength(self, s: str) -> int:
        # Generate all substrings from a string.
        all_substrings = [s[i:j] for i in range(len(s))
                          for j in range(i+1, len(s) + 1) if len(set(s[i:j])) == 1]
        print(all_substrings)
        processed_substrings = []
        qualified_substrings = []
        # print(all_substrings)
        for subs in all_substrings:
            if subs not in processed_substrings:
                processed_substrings.append(subs)
                if self.count_occurance(subs, s) >= 3:
                    qualified_substrings.append(subs)
        qualified_substrings.sort(key=lambda item: len(item), reverse=True)
        print(qualified_substrings)
        return len(qualified_substrings[0]) if qualified_substrings else -1
    

print('aaaa'.count('aa'))
224