<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:
        '''def count(s):
            d={}
            for i in s:
                if i in d:
                    d[i]+=1
                else:
                    d[i]=1
            d=dict(sorted(d.items(),key=lambda x:x[1],reverse=True))
            val=list(d.values())
            if(len(val)>1):
                return val[0]-val[len(val)-1]
            else:
                return 0'''
        a=''
        ans=0
        l={}
        for i in range(len(s)):
            d=defaultdict(int)
            for j in range(i,len(s)):
                    d[s[j]]+=1
                    mi=501
                    ma=0
                    for x in d.values():
                        mi=min(mi,x)
                        ma=max(ma,x)

                    ans+=ma-mi
        return ans
5377