<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:
        self.substrings = []
        self.l = []
        self.s = list(s)
        for i in range(len(self.s)):
            self.helper(i)
        res = 0
        dp = {}
        for subs in self.substrings:
            if subs in dp:
                res += dp[subs]
            else:
                dp[subs] = self.calc_beauty(subs)
                res += dp[subs]
        return res
    
    def calc_beauty(self, s):
        d = collections.Counter(s)
        max_val = max(d.values())
        min_val = min(d.values())
        return max_val - min_val


    def helper(self, idx):
        if len(self.l) > 0:
            self.substrings.append("".join(self.l))
        if idx == len(self.s):
            return
        self.l.append(self.s[idx])
        self.helper(idx+1)
        self.l.pop()
9465