<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>

from collections import Counter
class Solution:
    def beautySum(self, s: str) -> int:
        result=0
        
        for i in range(len(s)):
            for j in range(len(s)):
                temp=list(s[i:j+1])
                if not temp or len(temp)==1: continue
                counter=Counter(temp).most_common()
                diff=counter[0][1]-counter[-1][1]
                if diff>0:
                   result+=diff

        return result
                



        # def traverse(index,output):
        #     if index==len(s):
        #         # we have the substring
        #         # find beauty
        #         n=len(output)
        #         if not output or n==1: return
        #         counter=Counter(output).most_common()
        #         diff=counter[0][1]-counter[-1][1]
        #         if diff>0:
        #             self.result+=diff
        #             print(self.result,output)
        #         return
11324