<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:
        count=0
        hash=[0]*26
        for i in s:
            hash[ord(i)-97]+=1
        hash.sort()
        i=0
        while i!=len(hash):
            if hash[i]!=0:
                break
            else:
                i+=1
        beauty=hash[-1]-hash[i]
        #print(beauty)
        sub=[]
        for i in range(len(s)):
            str=""
            for j in range(i,len(s)):
                str+=s[j]
                sub.append(str)
        #print(sub)
        for i in range(len(sub)):
            new=[0]*26
            for j in sub[i]:
                #print(j)
                new[ord(j)-97]+=1
            new.sort()
            x=0
            while x!=len(new):
                if new[x]!=0:
                    break
                else:
                    x+=1
            val=new[-1]-new[x]
            count+=val
        return count
            


        

        
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