Sum of Beauty of All Substrings
<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>
Hash Table; String; Counting
/**
 * @param {string} s
 * @return {number}
 */
var beautySum = function (s) {
    let ans = 0;
    for (let i = 0; i < s.length; ++i) {
        const cnt = new Map();
        for (let j = i; j < s.length; ++j) {
            cnt.set(s[j], (cnt.get(s[j]) || 0) + 1);
            const t = Array.from(cnt.values());
            ans += Math.max(...t) - Math.min(...t);
        }
    }
    return ans;
};
