<p>Given a string <code>s</code>, sort it in <strong>decreasing order</strong> based on the <strong>frequency</strong> of the characters. The <strong>frequency</strong> of a character is the number of times it appears in the string.</p>

<p>Return <em>the sorted string</em>. If there are multiple answers, return <em>any of them</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;tree&quot;
<strong>Output:</strong> &quot;eert&quot;
<strong>Explanation:</strong> &#39;e&#39; appears twice while &#39;r&#39; and &#39;t&#39; both appear once.
So &#39;e&#39; must appear before both &#39;r&#39; and &#39;t&#39;. Therefore &quot;eetr&quot; is also a valid answer.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;cccaaa&quot;
<strong>Output:</strong> &quot;aaaccc&quot;
<strong>Explanation:</strong> Both &#39;c&#39; and &#39;a&#39; appear three times, so both &quot;cccaaa&quot; and &quot;aaaccc&quot; are valid answers.
Note that &quot;cacaca&quot; is incorrect, as the same characters must be together.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;Aabb&quot;
<strong>Output:</strong> &quot;bbAa&quot;
<strong>Explanation:</strong> &quot;bbaA&quot; is also a valid answer, but &quot;Aabb&quot; is incorrect.
Note that &#39;A&#39; and &#39;a&#39; are treated as two different characters.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 5 * 10<sup>5</sup></code></li>
	<li><code>s</code> consists of uppercase and lowercase English letters and digits.</li>
</ul>

class Solution {
public:
    string frequencySort(string s) {
        vector<string> v(62);
        for (int i = 0; i < s.length(); i++) {
            char ch = s[i];
            if (ch >= 'a' && ch <= 'z') {
                v[ch - 'a'] += ch;
            } else if (ch >= 'A' && ch <= 'Z') {
                v[ch - 'A' + 26] += ch;
            } else {
                v[ch - '0' + 52] += ch;
            }
        }
        for (int i = 0; i < 61; i++) {
            int maxIndex = i;
            for (int j = i + 1; j < 62; j++) {
                if (v[maxIndex].length() < v[j].length()) {
                    maxIndex = j;
                }
            }
            swap(v[maxIndex], v[i]);
        }
        string str = "";
        for (int i = 0; i < 62; i++) {
            str += v[i];
        }
        return str;
    }
};
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