<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>

#me correct but time limit exceeded
# class Solution:
#     def beautySum(self, s: str) -> int:
#         l=[]
#         for x in range(len(s)):
#             for y in range(x+2, len(s)+1):
#                 l.append(s[x:y])
#         sum=0
#         for ind, x in enumerate(l):
#             set1=set()
#             for y in x:
#                 set1.add(x.count(y))
#             if len(set1)>1:
#                 sum+=max(set1)-min(set1)
#         return sum
class Solution:
    def beautySum(self, s: str) -> int:
        beauty_sum = 0
        for i in range(len(s)):
            for j in range(i + 2, len(s) + 1):
                substring = s[i:j]
                freq = {}  # Store character frequencies
                for char in substring:
                    freq[char] = freq.get(char, 0) + 1
                max_freq = max(freq.values())
                min_freq = min(freq.values())
                beauty_sum += max_freq - min_freq
        return beauty_sum
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