<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:
        dic = dict()
        ans = 0
        n = len(s)

        for i in range(n):
            dic = dict()
            for j in range(i,n):
                if s[j] in dic:
                    dic[s[j]] = dic[s[j]] + 1;
                else:
                    dic[s[j]] = 1
                high, low = self.getLowAndHigh(dic)
                if low > 0:
                    ans = ans + high - low
        return ans;

    def getLowAndHigh(self, dic):
        high = 0;
        low = 501;
        for key in dic:
            val = dic[key]
            high = max(high, val)
            low = min(low, val)
        # print(dic)
        # print("high",high)
        # print("low",low)
        return (high, low)
        
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