<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){
    map<char,int>mapper;
    for(int i=0;i<s.size();i++){
        char ele=s[i];
    mapper[ele]++;
    }
    map<int,string>mapper2;
    for(auto i:mapper){
        int freq=i.second;
        char ch=i.first;
        mapper2[freq].push_back(ch);
    }
    string She="";
    for(auto i:mapper2){
    int count=i.first;
    string temp=i.second;
    int size=temp.size();
    for(int j=0;j<size;j++){
        char ch=temp[j];
        She=She+string(count,ch);
    }
    }
    reverse(She.begin(),She.end());
    return She;
    }
};

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