<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:
        
        n = len(s)
        beautySum = 0
    
        # i = starting index
        for i in range(n):
            
            # charFreqs[x] = num of occurences of x in curr substrings
            # freqs[x] = list of chars that have x has a freq
            charFreqs = defaultdict(int)
            freqs = defaultdict(set)
            
            # Add s[i] to freq dicts
            charFreqs[s[i]] = 1
            freqs[1].add(s[i])
                
            # j = ending index
            minFreq, maxFreq = 1, 1
            for j in range(i+1, n):
                
                # Add s[j] to freq dicts
                oldFreq = charFreqs[s[j]]
                charFreqs[s[j]] += 1
                freqs[oldFreq].discard(s[j])
                freqs[oldFreq+1].add(s[j])
            
                # Update min and max freq
                maxFreq = max(maxFreq, charFreqs[s[j]])
                if oldFreq == minFreq and len(freqs[oldFreq]) == 0:
                        minFreq += 1
                minFreq = min(minFreq, charFreqs[s[j]])
                        
                beautySum += (maxFreq - minFreq)
                        
        return beautySum
            
                
            
            
            
916