<p>The <strong>beauty</strong> of a string is the difference in frequencies between the most frequent and least frequent characters.</p>

<ul>
	<li>For example, the beauty of <code>&quot;abaacc&quot;</code> is <code>3 - 1 = 2</code>.</li>
</ul>

<p>Given a string <code>s</code>, return <em>the sum of <strong>beauty</strong> of all of its substrings.</em></p>

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

<pre>
<strong>Input:</strong> s = &quot;aabcb&quot;
<strong>Output:</strong> 5
<strong>Explanation: </strong>The substrings with non-zero beauty are [&quot;aab&quot;,&quot;aabc&quot;,&quot;aabcb&quot;,&quot;abcb&quot;,&quot;bcb&quot;], each with beauty equal to 1.</pre>

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

<pre>
<strong>Input:</strong> s = &quot;aabcbaa&quot;
<strong>Output:</strong> 17
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;=<sup> </sup>500</code></li>
	<li><code>s</code> consists of only lowercase English letters.</li>
</ul>

class Solution:
    def beautySum(self, s: str) -> int:
        prefix = []
        prefixes = defaultdict(lambda : [0])
        freq = defaultdict(int)
        a_s = 0
        for c in s:
            freq[c] += 1
            for ch in range(97, 123):
                abc = chr(ch)
                prefixes[abc].append(freq[abc])
            
        res = 0
        for i in range(len(s)):
            for j in range(i + 1, len(s)):
                
                max_elem = -1
                min_elem = len(s) + 1
                for ch in range(97, 123):
                    abc = chr(ch)
                    max_elem = max(max_elem, prefixes[abc][j + 1] - prefixes[abc][i])
                    count = prefixes[abc][j+1] - prefixes[abc][i]
                    if count > 0:
                        min_elem = min(min_elem, count)
                res += max_elem - min_elem
        
        return res
            
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